Chemical Bonding Class 11 Questions with Answers

Class 11 Chemistry Chemical Bonding And Molecular Structure Long Question And Answers

Question 1. How is crystalline NaCl formed from constituent elements?
Answer:

When a Na-atom combines with a Cl-atom, the electron lost by die electropositive Na-atom is gained by the electronegative Cl-atom, resulting in the formation of Na+ and Cl ions respectively, each having inert gas configuration. The oppositely charged ions are bound by a strong electrostatic force of attraction to form an ionic, crystalline solid, NaCl.

Chemical Bonding And Molecular Structure Question 5

In the crystal of NaCl, each Nation is surrounded by 6 Cl ions, and each Cl ion is surrounded by 6 Na+ ions. This results in the formation of a three-dimensional crystal, where the lattice sites are alternately occupied by Na+ and Cl ions.

Question 2. In which of the given molecules, the central atom does not obey the octet rule? ClF3, SF2, OsFg, BCl3, NH3, NO2
Answer:

 Chemical Bonding And Molecular Structure Question 26

Question 3. The melting point of MgBr2 is 700°C while that of AlBr3 is only 97°C. Give reason.
Answer:

According to Fajan’s rule, only the potential (phi) of the cations increases with an increase in cationic charge and a decrease in cationic radii. Consequently, the covalent character increases and the melting point of the corresponding salts decreases.

  1. In case of MgBr2 and AlBr3,
  2. The charge of Mg2+ is less than the charge of Al3+.
  3. Radius of Mg2+ is greater than the radius of Al3+.
  4. Thus, the melting point of AlBr3 is less than that of MgBr2

Question 4. What CuCl is more covalent than NaCl?
Answer:

If the charge and size of the cations remain constant, the cation with pseudo noble gas (18 electrons) configuration, as in the case of Cu+ (3s² 3p6 3d10) causes larger polarisation on the electron cloud of the anion than a cation with noble gas (8 electrons) configuration, as in case of Na+(2s22p6) because, (n-1) p electrons are more effective in shielding the outer electrons compared to the (n-1)d electrons.

As a result, an appreciable increase in electron charge cloud density between the two nuclei takes place, leading to an increase in the covalent character of the bond. Hence, CuCl is more covalent than NaCl.

Question 5. Arrange in increasing order according to the given properties and explain the order: MgCl2, AlCl3, NaCl, SiCl4 (melting point); LiBr, NaBr, KBr (melting point); (HI) MgCO3, CaCO3, BeCO3 (thermal stability); Hgl2, HgCl2 (intensity of color).
Answer:

SiCl4 < AlCl3 < MgCl2 < NaCl; [Ionic potential increases with either increase in charge on cation or cationic radius. As a result, the covalent character and melting point of the compounds formed increases.]

The order of melting point of the given bromides should be: LiBr < NaBr < KBr. Due to the increase in the size of the cation from Li+ to K+, the value of p increases.

So, the covalent character of the compounds increases. However, due to a decrease in lattice enthalpy from NaBr to KBr, the melting point decreases. Therefore, the correct order of melting point is LiBr < NaBr > KBr.

Question 6. Explain why AgCl is white whereas Agl is yellow If the degree of polarization of the anion is higher, then the electrovalent compound becomes colored (Example Pbl2 yellow) but if it is lower, then the compound is either white or colorless (Example PbCl2 is white)—why?
Answer:

The larger the anionic radius, the greater its tendency to get polarized. The higher polarisability of 1- ion, owing to its larger radius, facilitates the transition of electron (from anion to metal-ion) in the visible range, imparting a yellow color to Agl. On the other hand, the lower polarisability of the Cl ion, facilitates the transition of electrons in the UV range. Hence, AgCl appears white

Question 7. LiCl is soluble in organic solvents while the chlorides of other alkali metals are not. Explain.
Answer:

As we move down a group, the cationic radius increases, which decreases the polarising power of the cation, which ultimately decreases the covalent character of the compound. Since LiCl is the most covalent compound among all the alkali metal chlorides, it is soluble in organic (non-polar) solvents while the rest are not.

Question 8. Give reasons: SnCl2 is solid at room temperature while SnCl4 is liquid. Fel3 cannot be prepared. What is a coordinate covalent bond or coordinate bond?
Answer:

Sn4+ has a higher positive charge than Sn4+ liana greater polarising power than Sn2, Hence the covalency of the corresponding chlorides increases from SnCI2 to SnCI4, which results In a decrease In the melting point from SnCl2 to SnC2. Therefore SnCl2 Is a liquid while SnCl2 Is a solid at room temperature

Question 9. Aluminium chloride exists as a dimer—Explain.
Answer:

Chemical Bonding And Molecular Structure Question 40

In AlCl3, the Al-atom has only 6 electrons in its valence shell. It requires two more electrons to complete Its octet. So it accepts a lone pair of electrons from the Cl-atom of another AlCl3 molecule as shown above. Thus, AlCl3 exists as a dimer.

Question 10. AlCIg forms a dimer but BC13 cannot—Explain What do you understand by 1 bond length, 2 bond dissociation enthalpy, and 3 bond angle?
Answer:

The size of Al is much larger than that of B. Hence Al can easily accommodate 4 Cl-atoms around it. In AlCl3, as there are 6 electrons around the Al-atom, it completes its octet by accepting a lone pair of electrons from the Cl-atom of an adjacent molecule. As a result, AlCl3 exists as a chlorine-bridged dimer forming Al2Cl6.

On the other hand, B is comparatively smaller in size. Though it has an incomplete octet in BCl3, it cannot accommodate the fourth chlorine atom around it, owing to the large size of the Cl-atom. Thus, BCl3 does not exist as a dimer.

Question 11. Arrange the following compounds in increasing order of carbon-carbon bond strength and explain the order. CH2=CH2, CH3-CH3, HC=CH
Answer:

The increasing order of C—C bond strength is given by: C—C < C=C < C=C i.e., CH3—CH3 < CH2=CH2 < CH=CH Greater the bond multiplicity, the greater the bond dissociation enthal. In CH3—CH3, there is only an or -bond between the C-atoms whereas, CH2=CH2 and, CH=CH contain one and two n -bonds respectively, in addition to the cr -bond. So the energy required to break the carbon-carbon bonds increases in the order C—C < C—C < C=C.

Question 12. Bond angles in Pbr3(101.5°), PCl3 (100°), and PF3(97°) decrease with an increase in the electronegativities of the surrounding atoms. However, bond angles in BF2, BCl2,  and BBr3 do not change with a change in electronegativities of the surrounding atoms. Explain with reason.
Answer:

PX2 has a trigonal pyramidal geometry. With the increase in electronegativity of the surrounding halogen (X) atoms, bond pairs are oriented more towards halogen atoms, resulting in a decrease in bond pair-bond pair repulsions. Hence X—P—X bond angles decrease in the order: Pbr3(101.5°) > PCl(100°) > PF3(97°)

BX3 has a trigonal planar geometry, where 3 halogen atoms are located at 3 corners of an equilateral triangle. With the increase in the electronegativity of the halogen atom, bond pairs tend to concentrate more towards the halogen atoms resulting in a decrease in bond pair- bond pair repulsion. Since all 3 halogen atoms lie on the same plane, forming 3 equivalent B—X bonds, there is no change in the X—B —X bond angle (120°).

Question 13. The bond angle of H2O is greater than that of H2S —explain.
Answer:

Chemical Bonding And Molecular Structure Question 51

Both H2O and H2S have a tetrahedral geometry. Since Ip-bp repulsions are greater than bp-bp repulsions, these molecules attain a distorted tetrahedral geometry, where H —X—H (X = O or S) bond angles are less than the normal tetrahedral angle of 109°28.

Due to the higher electronegativity of the central O-atom than the S-atom, bp-bp repulsion is greater in the case of the O—H bond. Hence, the bond angle of H2O is greater than H2S.

Question 14. Arrange the following molecules/ions in order of decreasing —N —H bond angle and explain the order: NH3, NH+, NH2-
Answer: Chemical Bonding And Molecular Structure Question 54

NH3 has one lone pair and 3 bond pairs, NH2 has four bond pairs and NH2 has two lone pairs and two bond pairs.

According to VSEPR theory since the repulsions follow the order: Ip -Ip > Ip- bp > bp – bp, the bond angles (H—N—H) decrease in the order: NH+> NH3 > NH2-.

Question 15. Predict the state of hybridization of the central atom and the shape of each of the following species:
Answer:

Chemical Bonding And Molecular Structure Question 67

Question 16. Name the type of hybridization of the central atom that leads to each of the following geometries:
Answer:

  1. Square planar -dsp2
  2. Planar triangular -sp2
  3. Tetrahedral -sp3
  4. Linear-sp2
  5. Octahedral -sp3d2
  6. Trigonal bipyramidal-sp3d2

Question 17. Identify the state of hybridization of each carbon in:

  1. CH2=CH—CH=CH2
  2. CH2=C=CH2
  3. CH2=CH—CHO
  4. CH3—C= CH
  5. HCEEC—CHO

Answer: \(\stackrel{1}{\mathrm{C}} \mathrm{H}_2=\stackrel{2}{\mathrm{C}} \mathrm{H}-\stackrel{3}{\mathrm{C}} \mathrm{H}=\stackrel{4}{\mathrm{C}} \mathrm{H}_2 ; \mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3, \mathrm{C}_4-\text { all } s p^2hybridised.\)

Question 18. What are the possible geometrical shapes of covalent molecules of the general formula, AX2 and AX3 (X = a monovalent atom) when the central atom A has No lone pair of electrons, one lone pair of electrons, and two lone pairs of electrons?
Answer: AX2 (l) shape — linear, Example BeCl2

  1. shape — angular, Example CCl2
  2. shape — V-shaped, for Example H2O
  3. AX3 shape — trigonal planar, Example BF3
  4. shape —pyramidal, Example NH3
  5. shape — T-shaped, Example ClF3

Question 19. Why are the P —Cl bonds in PCl5 not the same length?
Answer:

In PCl5, two axial P—Cl bonds and three equatorial P—Cl bonds are present. An axial bond pair is repelled by three equatorial bond pairs at 90° and one axial bond pair at 180°.

Similarly, an equatorial bond pair is repelled by two axial bond pairs at 90° and two equatorial bond pairs at 120°. Thus, an axial bond pair is repelled by three electron pairs while an equatorial bond pair is repelled by two electron pairs. Thus, the axial bond pair suffers greater repulsion & hence slightly longer than equatorial bonds

Question 20. Which of the molecules Orions are iso-structural and why? BF3, NH+, CO2-, BF4, NO2, CH3+, CCl4 All the C-0 bond lengths – are not equal— explain.
Answer:

BF3, CH+3 — trigonal planar geometry. The central atom undergoes sp2-hybridisation forming 3cr bonds with the neighbouring atoms. H4, CCl4, BF4 — tetrahedral geometry. The central atom undergoes sp3-hybridisation forming 4σ bonds with tire neighbouring atoms. NO3-, CO3+2 — trigonal planar geometry. The central atom is sp2 -hybridized forming σ bonds and pi bonds.

Question 21. Which one among the following pairs is more electronegative and why?

  1. Csp or
  2. Carbon in CHl3 orcarbon in CHCl3,
  3. Na or Cl
  4. Carbon in C2H4 or C2H2

Answer: CS is more electronegative than Csp³ because, for hybrid orbitals, electronegativity increases with an increase in the s -the character of the hybrid orbital. 1

C in CHCl3 is more electronegative than C in CH3 because the electronegativity of an atom increases with an increase in the electronegativity of the atom bonded to it.

Chlorine (Cl) is more electronegative than sodium (Na) because, as we move from left to right in a given period, atomic size decreases, and effective nuclear charge increases. Hence electronegativity increases.

In C2H4, C is sp² hybridised while in C2H2 C is sp hybridised. Since electronegativity increases with an increase in the s -s-character of the hybrid orbitals, C in C2H2 is more electronegative than C in C2H4.

Question 22. How will you distinguish between the two geometrical isomers of l, 2-dichloroethane from their boiling points?
Answer:

The 2 geometrical isomers of 1,2 dichloroethene are cis- 1,2 dichloroethene and trans-1,2 dichloroethene. cis-1,2 dichloroethene has a definite dipole moment (μ≠ 0) whereas the dipole moment of trans-1,2 dichloroethene is found to be zero (μ= 0).

The ct’s-isomer is highly polar indicating strong dipole-dipole attractive forces among the molecules. Hence a large amount of energy is required to separate the molecules from each other. Therefore boiling point of cis-1,2 dichloroethene is higher than the trans-isomer.

Chemical Bonding And Molecular Structure Question 87

Question 23. Explain why the following molecules are non-polar:
Answer:

1,3,5-trinitrobenzene, the three NO2 groups are bonded to 3 alternate sp2 hybridized C-atom of the benzene ring. The three C-NO2 bond moments act at an angle of 120° to each other. Therefore, the net dipole moment of the molecule is zero (p = 0) and the molecule is non-polar trans-2,3-dichlorobut-2-ene.

Chemical Bonding And Molecular Structure Question 88

In trans-2,3-dichlorobut-2-ene, the two C —Cl and the two C —CH3 bond moments act in H3C opposite directions to balance each other, Because of this, the molecule possesses no net dipole moment. Hence, tram-2,3 dichloro but-2- ene is non-polar

Chemical Bonding And Molecular Structure Question 88.

Question 24. Predict the dipole moment of a molecule of the type, AB4 having square-planar geometry, a molecule of the type, AB2 having trigonal bipyramidal geometry, a molecule of the type, ABg having octahedral geometry, a molecule of the type, AB7 having pentagonal bipyramidal geometry.
Answer:

All the four A—B bond moments act at an angle of 90° with each other. Therefore the net dipole moment of AB4 is zero (p = 0).

Due to the symmetrical structure of the molecule, the equatorial bond moments cancel each other. Similarly, the axial bond moments cancel each other. Therefore the resultant dipole moment is zero (p = 0).

Question 25. Which one of each pair has a higher dipole moment and why?

  • CS2 and CO2;
  • NH3 and NF3;
  • CH3CH2Cl and CH2=CHCl;
  • 1,3,5- tribromobenzeneand 1,3-dibromobenzene.

Answer:

⇒ \(\text { (1) } \begin{array}{ll}\mathrm{S} \equiv \mathrm{C} \equiv \mathrm{S} & \mathrm{O}=\mathrm{C} \\\mathrm{CS}_2(\mu=0) & \mathrm{CoS}(\mu \neq 0)\end{array}\)

Question 26. NH3 molecules remain associated through intermolecular hydrogen bonding but there is no such association among HCl molecules even though electronegativities of N and Cl are the same. Explain.
Answer:

  1. Although the electronegativities of nitrogen and chlorine are the same, nitrogen can form hydrogen bonds but Cl cannot.
  2. This is because the N-atom is much smaller than the CIatom.
  3. Due to the large size of the Cl-atom, the electrostatic attraction between the Cl-atom of one molecule and the Hatom of another molecule becomes weak. Hence, Cl does not form hydrogen bonds while NH3 molecules undergo association by intermolecular hydrogen bonds.

Question 27. At normal temperature, o-hydroxybenzaldehyde is a liquid but p-hydroxybenzaldehyde is a solid. Give reason.
Answer:

In o-hydroxybenzaldehyde, the —OH and — CHO groups are situated at two adjacent C-atoms of the benzene ring and are involved in intramolecular hydrogen bonding. These molecules exist as discrete molecules and have a lower melting point. On the other hand, in p -hydroxybenzaldehyde, the —OH and — CHO groups are situated away from each other. Hence, intramolecular hydrogen bonding does not exist.

These molecules remain associated through intermolecular hydrogen bonding and hence have a high melting point. Thus the ortho-isomer exists as a liquid while the para-isomer exists as a solid.

Question 28. Arrange the following species in order of increasing stability and give reasons: Li2, Li+2, Li-2 are as follows [li (z=3)]:
Answer:

⇒ \(\mathrm{Li}_2-\mathrm{KK}\left(\sigma_{2 \mathrm{~s}}\right)^2 ;  \text { B.O. }=\frac{2-0}{2}=1 \)

⇒  \(\mathrm{Li}_2^{+}-\mathrm{KK}\left(\sigma_{2 \mathrm{~s}}\right)^1 ; \text { B.O. }=\frac{1-0}{2}=0.5 \)

⇒  \(\mathrm{Li}_2^{-}-\mathrm{KK}\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\sigma_{2 \mathrm{~s}}^{+}\right)^1 ; \text { B.O. }=\frac{2-1}{2}\)

= 0.5

The greater the bond order, the greater the bond dissociation enthalpy and hence greater the stability. Again stability decreases when excess electrons are present in a nonconjugate shell. Therefore the order of increasing stability of the given species is as follows:

⇒ \(\mathrm{Li}_2^{-}<\mathrm{Li}_2^{+}<\mathrm{Li}_2\)

Question 29. Inert gases are monoatomic. Explain in terms of MO theory.
Answer:

The molecular orbital energy level diagram for inert gases shows that the number of electrons in bonding molecular orbitals is equal to those in the antibonding molecular orbitals i.e., bond order (of inert gases) \(=\left(\frac{N_b-N_a}{2}\right)=0\) Therefore, all inert gases are monoatomic. For Example He (2); Electronic configuration of He2 is:

⇒ \(\mathrm{He}_2-\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\)

Bond Order \(=\frac{2-2}{2}=0\)

Question 30. The ionic frond between sodium and chloride ions is stronger than that between potassium and chloride ions. Explain.
Answer:

Since the atomic number of K (Z = 19)) Is higher Ilian that of (Z = 11), K+ ion Is larger than Na+ Ion. According to Fajan’s rule, KCl should be more Ionic than NaCl. However, due to the smaller size of, Na+ ion, the charge density in Na ion is higher than that of K+ ton.

As a consequence, the coulomblc forces of attraction between Na+ and Cl ions (the lattice energy) are more titan than between K+ and Clions. Therefore the ionic bond between Na+ and Cl ions is stronger than that between K+ and Cl ions.

Question 31. Silicon tetrachloride readily undergoes hydrolysis but carbon tetrachloride does not undergo hydrolysis under normal conditions. Explain.
Answer:

Since carbon (of the second period) has no vacant d -d-orbital, its maximum covalency is 4. On the other hand, silicon (of the third period) has vacant d -d-orbitals, and its maximum covalency is 6. As the Si -atom can extend its covalency to 6, SiCl4 undergoes ready hydrolysis to yield SiO2.

A lone pair of electrons from the O- atom of H2O is donated to the empty d -orbitals of Si, forming a coordinate intermediate that has a trigonal bipyramidal structure. The intermediate [SiCl4(H2O)] loses a molecule of HCl to form SiCl3(OH). In the same way, the other 3Cl -atoms are replaced by 3-OH groups to form orthosilicic acid [Si(OH)4] which finally loses 2 molecules of water to give SiO2.

Chemical Bonding And Molecular Structure Silicon Tetrachloride Readily Undergoes Hydrolysis

C-atom having no d -d-orbitals in its valence shell cannot extend its covalency beyond 4 so it does not undergo hydrolysis under normal conditions

Question 32. The second ionization enthalpy of Mg Is sufficiently high while the second electron affinity or electron gain enthalpy of oxygen is low (actually this value is positive), yet Mg2+  and O2-  ions form the Ionic compound, MgO. Explain with reasons.
Answer:

The sufficiently high second ionization enthalpy of Mg indicates that a large amount of energy is required to remove the second electron from the Mg -atom, Le., to convert Mg to Mg2+  ion. The second electron gain enthalpy of oxygen is positive indicating that the energy should be supplied to convert the O -atom into an O2-  ion.

Since both processes are endothermic, MgO is not expected to be produced through the formation of anionic bonds. But actually, it is produced and this is because of its high lattice energy (mainly due to comparable sizes of Mg2+  and O2- ions.

Question 33. Both sodium and hydrogen are electropositive elements. Sodium reacts with chlorine to form an electrovalent compound but hydrogen reacts with chlorine to form a covalent compound —explain.
Answer:

The ionization enthalpy of smaller H -atoms is sufficiently higher than that of larger Na -atoms.

Because of lower ionization enthalpy, sodium reacts with chlorine through the formation of Na+ ion to form the electrovalent compound, NaCl.

On the other hand, because of the much higher ionization enthalpy, hydrogen does not react with chlorine through the formation of H+. Instead both H+ and Cl atoms donate one electron each to form an electron pair and produce the covalent compound, HCl by sharing the electron pair equally.

Question 34. The Melting Point Of Cal2 Is Much Lower (575°C) That Of caf2 (1392°C) explained with reasons.
Answer:

According to Fajan’s rule, the tendency of a large-sized anion to be polarised is greater than that of a small-sized anion. So a compound containing a large-sized anion exhibits more covalent character than that with a small-sized anion.

Hence, Cal2 containing larger I- ion possesses a higher covalent character and melts at a relatively low temperature. On the other hand, CaF2 containing smaller F- ions possesses a much lower covalent character and melts at very high temperatures.

Question 35. The B — F bondin BF3 is shorter in length than the B — F bond in BFÿ — explain with reasons.
Answer:

The outermost shell of the central B-atom of the BF3 molecule contains the electrons. Since the B-atom has an incomplete octet, it participates in resonance with the F-atoms to complete its octet.

As a result, the B — F bonds acquire partial double bond character. On the other hand, the B -atom in the BF2 ion has a filled octet, and so it does not participate in resonance. Therefore, B — F bonds do not assume a double bond character. Hence, the B — F bonds in the BF3 molecule are shorter in length than those in the BF4 ion.

Chemical Bonding And Molecular Structure The B-f Bond

Question 64. The electronegativity of Br is less than that of F, yet BF3 is a weaker Lewis acid than BBr3
Answer:

B and F atoms are elements of the same period (second period), having comparable sizes. In BF3, the octet of B-atom is not filled up. To fulfill the octet, the B-atom participates in the resonance (n -n-backbonding) with the F- F-atoms. This resonance involving orbitals of comparable sizes (2p- 2p overlap) is very effective.

As a result, electron density on B-atom increases, and the tendency of BF3 to behave as a Lewis acid decreases. On the other hand, Bris is an element of the fourth period. In BBr3, effective n-back bonding involving orbitals of dissimilar sizes (2p- 4p overlap) does not take place. Hence, the electron density on B does not increase and therefore, BBr3 behaves as a stronger Lewis acid than BF3.

Question 36. Acetylene dissolves in acetone but not in water. explain the observation.
Answer:

Because of the considerable electronegativity of the sp -sp-hybridized C-atom of acetylene, the acetylenic hydrogen gets involved in intermolecular H-bonding with the O-atom of acetone. As a consequence, acetylene dissolves in acetone.

Since the energy of the H-bond formed is greater than the weak van der Waal’s attractive forces acting among the acetylene molecules and dipole-dipole attractive forces operating among the acetone molecules, the process of dissolution occurs easily.

Chemical Bonding And Molecular Structure The Proces Of Dissolution Occurs Easily

On the other hand, the intermolecular H-bonding between water molecules is stronger than the intermolecular H -H-H-bonding between water and acetylene molecules, if formed. So, acetylene shows no tendency to form H -bonds with water. Hence, acetylene does not dissolve in water.

Question 37. Arrange nitrogen dioxide molecule (NO2), nitronium ion (NO+2 ), and nitrite ion (NO-2) in increasing order of bond angle and explain the order.
Answer: NO2- ion:

Total number of electrons in the valence shell of the N -atom of the ion= [5 valence electrons of N -atom + 2 electrons of O -atom linked by a double bond + 1 electron of O -atom linked by a single bond] = [8 electrons or 4 electron pairs] = [2 cr -bond-pairs + 1 lone pair+ 1 n -bondpair].

Since the n-bond pair plays no role in determining the shape of the molecule, according to VSEPR theory, the three electron pairs will be oriented towards the comers of an equilateral triangle, and the shape of the ion having one lone pair is angular.

In this case, the lone pair-bond pair repulsion is greater than the die repulsion between two bond pairs of the two bonds having partial double bond character due to resonance. As a result, the O —N —O bond angle (115°) is less than the expected regular trigonal shape with a greater O—N—O bond angle (120°).

Chemical Bonding And Molecular Structure NO2 molecule

NO2 molecule:

Total number of electrons in the valence shell of the N -atom of the molecule = [5 valence electron of N-atom + 2 electrons of O-atom linked by a double bond + zero electrons of the O-atom linked by an o-ordinate covalent bond] = 7 electrons =[3 electron pairs + 1 odd electron] = [l(r-bond pair + 1 coordinate cr-bond pair + In’ -bond pair + 1 odd electron].

According to VSEPR theory, two bond pairs and the odd electron are arranged trigonally in a plane.

So, the shape of the NO2 molecule having an odd electron is angular. In this case, the repulsive force between the bond pairs of two bonds having partial double bond character is greater than the repulsive force acting between the bond pairs and the odd electron. As a result, the value of the O—N—0 bond angle (135°) is greater than that of the regular planar trigonal shape (120°).

Chemical Bonding And Molecular Structure NO+2 ions

NO2 ion:

Total number of electrons in the valence shell of the N -atom of the ion = 5 electrons of N -atom +4 electrons of two O -atoms linked by two double bonds -1 electron for the positive charge = 8 electrons = 4 electron pairs = 2 cr -bond pairs + 2a- -bond pairs.

The two n-bond pairs have no contribution toward the shape of the ion. According to VSEPER theory, the two bond pairs are oriented in opposite directions. Hence, the shape of the NO2 ion is linear and the value of the O —N —O bond angle is 180°. Therefore, the increasing order of the O —N —0 bond angle of the given species is: NO-2 < NO2 < NO+2.

Question 38. H2O is liquid while H2S is gas, though oxygen and sulfur, both belong to the same group of the periodic table
Answer:

  1. The oxygen atom is smaller and more electronegative than sulfur. Hence, in the H2O molecule, the O atom forms strong intermolecular H-bonds. However, in H2S molecules, S-atom, owing to its larger size and lesser electronegativity than Oatom, cannot form H-bonds.
  2. Strong intermolecular H-bonds bind H2O molecules in an associated state, while molecules of H2S are held by weak van der Waals forces. Therefore H2O is a liquid while H2S is a gas at room temperature.

Question 39. Hydrogen bonding between an F atom is stronger than that between H and O atoms. However, H2O is more viscous and its bp is greater than that of HF. Explain.
Answer:

Hydrogen bonding between H F is much stronger than that between H→O because F is more electronegative than O. However boiling point of H2O is much higher than that of HF because a single molecule of water can form four Hbonds with four other HaO molecules, while one H —F molecule can form only two H-bonds with HF molecule.

Because of this, H2O is more viscous than HF and its boiling point is higher.

Chemical Bonding And Molecular Structure Beacause Of This H2O is More Viscus Than HF And Its Boliing Point Is higher

Question 40. Why viscosity and boiling point of concentrated H2SO4 very high?
Answer:

The structure of sulphuric acid is as follows:

Chemical Bonding And Molecular Structure The structure of sulphuric acid

Each molecule of H2SO4 contains two OH groups and two doubly bonded oxygen atoms. Thus, each molecule of H2SO4 forms four H-bonds with other molecules. This causes extensive association among the H2SO4 molecules, increasing to boiling point.

Chemical Bonding And Molecular Structure The H2SO4 Molecules, Resulting in increase

The extensive intermolecular H-bonding enhances the intermolecular attraction among the different layers of the liquid, leading to an increase in viscosity.

Question 41. In the SF4 molecule, the lone pair of electrons occupies an equatorial position rather than an axial position, in the overall trigonal bipyramidal arrangement. Why?
Answer:

Depending on the position occupied by the lone pair, two structures of SF4 are possible

Chemical Bonding And Molecular Structure In SF4 Molecule The Lone Pair Of Electrons

In (1), there are three lone-pair-bond pair repulsions at 90° whereas in (2) there are only two lone pair-bond pair expulsions at 90°. lienee (2) Is more stable than (1), lienee the lone pair occupies the equatorial position In the SF4 molecule.

Question 42. Explains the shape of the Ion.
Answer:

The outer shell electronic configuration (In the ground state) of the central atom Is \(5 s^2 5 p_x^2 5 p_y^2 5 p_z^1 5 d^0\). It undergoes sp³d -hybridization. Out of the five sp³d hybrid orbitals, one is half-filled, one is empty and the remaining three arcs are filled.

The half-filled orbital forms a covalent bond with the iodine atom. The vacant orbital accepts an electron pair for the I- ion to form a coordinate bond. The remaining three fully-filled orbitals occupy equatorial positions. Thus, the geometry of three lone pairs and two bond pairs is trigonal bipyramidal and the shape of the I3 ion is linear.

Question 43. Indicate the type of bonds present in NH4NO3 and state the mode of hybridization of two N-atoms.
Answer:

NH4NO3 is an ionic compound in which the NH4+ ion is the cationic and NO3 is the anionic species. NH+ ion is formed by the combination of NH3 molecule and H+ ions through the dative bond.

N in NH3 ion is sp³ hybridized and has a tetrahedral geometry, while in NO3 ion, N is sp² hybridized and has a planar geometry. Thus three types of bonds, viz., ionic, covalent, and coordinate bonds are present in NH4NO3.

Chemical Bonding And Molecular Structure Indicate The Type Of Bonds Present In NH4NO3

Question 44. ClF3 exists, but FCI2 does not.
Answer:

The cl atom has empty d-orbitals. During horn) formation, the electrons from 3p -orbitals are promoted to 3d -orbitals

Chemical Bonding And Molecular Structure CIF3 Exists But FCL3 Does Not

In the first excited state, Cl-atom can exhibit a covalency of three. Hence CH3 is possible. F-atom cannot expand Its octet due to the absence of empty cl -orbitals in the 2nd energy level. Hence cannot exhibit covalency more than. Therefore FCl3 is not possible.

Question 45. The dipole moment of CH3Cl (p = 1.87D) is greater than that of CH3F (μ = 1.81D) even though the C — F bond is more polar than the C — Cl bond. Explain with reasons
Answer:

The dipole moment of a molecule, n = ex cl, where e = partial positive or negative charge developed on the bonded atoms and d = distance between the two charge centers. Because of the greater electronegativity of F compared to that of Cl, the charge developed in CH3F is higher than that in CH3Cl.

However, because of the larger size of the Cl -atom, the C—Cl bond length is greater than the C —F bond length, and consequently, the value of d in the case of CH3Cl is higher than that in the case of CH3F.In practice is found that the value of (ex d) for CH3F is lower than that for the CH3Cl molecule. Hence, CH3Cl has a greater dipole moment (p) than that of CH3F.

Question 46. Show by calculation that (lie dipole moment of methane of zero.
Answer:

Methane molecule is tetrahedral.

⇒ \(\begin{aligned}
& \mu_{\mathrm{CH}_3}=3 \mu_{\mathrm{C}-\mathrm{H}} \cos \left(180^{\circ}-109^{\circ} 28^{\prime}\right) \\
& =3 \mu_{C-H^{\prime}} \cos \left(70^{\circ} 32^{\prime}\right) \\
&
\end{aligned}\)

⇒ \(\begin{aligned}
& =\frac{3 \mu_{\mathrm{C}-\mathrm{H}}}{3}=\mu_{\mathrm{C}-\mathrm{H}} \\
\mu_{\mathrm{CH}_3} & =\mu_{\mathrm{C}-\mathrm{H}}=\mu_1 \text { (say) and } \theta=180^{\circ}
\end{aligned}\)

The resultant dipole moment

⇒ \(=\sqrt{\mu^2 \mathrm{C}-\mathrm{H}+\mu^2 \mathrm{CH}_3+2 \mu_{\mathrm{C}-\mathrm{H}} \cdot \mu_{\mathrm{CH}_3} \cos 180^{\circ}}\)

⇒ \(=\sqrt{\mu^2 \mathrm{C}-\mathrm{H}+\mu^2 \mathrm{CH}_3+2 \mu_{\mathrm{C}-\mathrm{H}} \cdot \mu_{\mathrm{CH}_3} \cos 180^{\circ}}\)

Alternative method:

For convenience, the four H-atoms of methane are designated as H1, H2, H3, And H4. The resultant moments of H1—C and H2—C bonds will be along the bisector of the H1—C—H2 angle, towards the carbon atom.

Similarly, the resultant of H3 —C and H4—C bond moments are along the bisector of the H3—C—H4 angle, towards the C-atom. These two resultants are equal in magnitude and opposite in direction. Hence, the molecule of methane has no resultant dipole moment.

Question 47. The boiling point of hydrogen fluoride is maximum among aU the halogen acids—explain.
Answer:

Fluorine is the smallest and the most electronegative element Hence F atom in the HF molecule forms strong intermolecular hydrogen bonds. On the other hand, since Cl, Br, and atoms are larger and less electronegative than F, their molecules are held by weak van der Waals forces.

As the hydrogen bond is stronger than the van der Waals forces, more energy is required to break the intermolecular hydrogen bonds of the associated HF molecules. Therefore the boiling point of HF is much higher than those of the other halogen acids. The boiling point of halogen acids follows the order: HF >HI > HBr > HCl

Question 48. Both CO2 and N2O are linear. However, N2O is polar while CO2 is non-polar—explain.
Answer:

The structures of CO2 and N2O are:

Chemical Bonding And Molecular Structure Both CO2 And N2O

The CO2 molecule is linear with a C atom at the center. The CO2 molecule contains 2 polar C—O bonds, as oxygen is more electronegative than carbon. in CO2 molecule, the two bond dipoles (Cδ+—Oδ-) =pl, actin opposite directions and cancel each other. As a result, the resultant dipole moment becomes zero. Thus CO2 molecule is non-polar.

On the other hand, the N2O molecule, though linear, contains a polar coordinate bond (N→O) at one end and a lone pair of electrons on the N-atom at the other end. Moment due to N— o bond and that due to lone pair acting in opposition but they do not cancel each other as they are not equal in magnitude. Hence N2O has a resultant dipole moment although it has a low value. Therefore N2O is a polar molecule.

Question 49. Although H2, Li2, and B2 molecules have the same bond order 1), they are not equally stable. Explain this observation and arrange them in order of decreasing stability.
Answer:

This observation can be explained as follows. The Li atom is much larger than the H atom. Hence, the Li—Li bond is much longer than the H —H bond (Li —Li bond length = 265 pm, H—H bond length = 74 pm).

Moreover, the Li2 molecule has two electrons in the antibonding σ*1s orbital while H2 has no electron in the antibonding orbitaL For these reasons, Li2 is less stable than H2 (bond energy of Li2 = H2O kJ. mol-1 while that of H2 = 438kJ. mol-1).

B atom is smaller in size than the Li-atom but larger than the H-atom. Hence the bond length of B2 is in-between (159 pm). Moreover, the B2 molecule has two electrons more in the bonding molecular orbitals [n(2px) and (2py)] Therefore, B2 is more stable than Li2 but less stable than H2 (bond energy of B2 = 290kJ – mol-1). Hence, the order of decreasing stability of these three molecules is H2 > B2 > Li2.

Class 11 Chemistry Chemical Bonding And Molecular Structure Short Question And Answers

Question 1. Draw Iewis dot structures of H3PO4 and CO2-3.
Answer:

Chemical Bonding And Molecular Structure Question 24

Question 2. Calculate the formal charge on N-atom in the HNO3 molecule
Answer: Formal charge on N-atom in HNO3

Chemical Bonding And Molecular Structure Question 25

No. of valence electrons of Natom] – [No. of unshared electrons \(-\frac{1}{2} x\) [No. of shared electrons] \(=5-0-\frac{1}{2} \times 8=5-4=+1\)

Question 3. In water, the first and second 0 —H bond dissociation enthalpies are SO2 and 427kJ mol-1 respectively. Determine the value of bond enthalpy of the O —H bond.
Answer:

The bond enthalpy of water is given by the average of bond dissociation enthalpies of the two Q —H bonds.

Bond enthalpy \(=\frac{502+427}{2}=464.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 4. Arrange the given compounds in order of their Increasing bond length: HCl, HI, HBr, HF. Explain the order.
Answer:

For halogen acids, the bond length increases, with an increase in the size of the halogen atom. Since the size of the halogen atoms increases in the order F < Cl < Br <I, the bond length increases in the order HF < HCl < HBr < HI.

Question 5. Why does the value of the bond angle increase with the increased electronegativity of the central atom in the ABV type of molecule?
Answer:

As the electronegativity of the central atom of a molecule of ABx type increases, the electron pair responsible for covalent bond formation will be attracted towards the central atom. Consequently, bp-bp repulsion increases leading to an increase in bond angle.

Question 6. Explain why the formation of a n-bond is not possible between a py and a pz-orbital.
Answer:

px  and py orbitals are mutually perpendicular, n -bonds are formed by the lateral overlap of two parallel p -orbitals. Since lateral overlap is not possible between two mutually perpendicular orbitals, a n -bond is not possible between a py and a pz orbital.

Chemical Bonding And Molecular Structure Question 62

Question 7. A homonuclear diatomic molecule contains 8 electrons. Predict whether the molecule will exist or not.

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\)

Nb = 4, Na = 4

Bond order \(=\frac{N_b-N_a}{2}=\frac{4-4}{2}=0\)

Hence, the molecule does not exist.

Question 8. If the electronic configuration of A atoms 1 s², comment on the stability of the A2 molecule and A2+ ion.
Answer:

⇒ \(\mathrm{A}_2:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2 \quad \mathrm{~A}_2^{+}:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^1\)

The higher the bond order, the higher the bond dissociation enthalpy and the greater the stability. Hence stability of A2+ > A2. Thus, A2 will have no existence.

Question 9. Arrange methanol, water, and dimethyl etherin in order of increasing boiling points and explain the order.
Answer:

The more molecules of the compound remain associated, the greater will be the boiling point of the compound. Water molecules having two —OH groups remain more associated by intermolecular H-bonding than methanol (CH3OH) molecules having only one —OH group.

Dimethyl ether (CH3OCH3) having no —OH group does not remain associated through H-bonding. Therefore, the boiling points of these liquids follow the order: of dimethyl ether < methanol < water.

Question 10. Explain why diamond melts at a very high temperature even though it is composed of covalently linked carbon atoms.
Answer:

In the crystal diamond, each sp3 -hybridized C -atom is bonded to four others by single covalent bonds (bond length 1.54A), and a large number of tetrahedral units are linked together to form a three-dimensional giant molecule. Strong covalent bonds extend in all directions. Due to this compact structure involving strong bonds, diamond is very hard and has a very high melting point.

Question 11. Which out of 1-butyne and 1-butene has a larger dipole moment and why?
Answer: The structures of1-butyne and1-butene are as follows:

⇒ \(\begin{array}{cc}
\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH} & \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2 \\
\text { 1-butyne } & \text { 1-butene }
\end{array}\)

As the sp hybridized terminal C-atom in 1-butyne is more electronegative than the sp² hybridized C-atom in 1-butene, the latter has a larger dipole moment than the forme.

Question 12. BaSG4 Is insoluble in water, even though it Is an ionic compound. Why?
Answer:

The lattice energy (i.e., the energy required to break its crystal lattice, by separating Ba2+  and SO2- ions) of BaSO4 is greater than its solvation energy (i.e., the energy released due to solvation of Ba2+ and SO2- ions by water molecules). Hence, BaSO4 is insoluble in water.

Question 13. Explain why all three nitrogen-oxygen bonds in the NO-3 ion are equal in length.
Answer:

The nitrate ion (NO-3) can be represented as a resonance hybrid of the following three equivalent resonance structures or canonical forms: Since the hybrid structure is an average structure, so all N —O bond lengths are equal as shown above.

Chemical Bonding And Molecular Structure The three Nitrogen Oxygen Bond In NO-3 ion are equal in length

Question 76. Give the structure of (CH3)3 N and [(CH3)3 Si]3N. Are they isostructural?
Answer: (CH3)3N is trimethyl amine. It has a pyramidal structure, while [(CH3)3Si]3N is planar.

Chemical Bonding And Molecular Structure The Stucture Of CH3

Thus, the two species are not isostructural. In (CH3)3N, N-atom is sp³ hybridised while in [(CH3)3Si]3N, the N-atom is sp² hybridised.

Question 14. Covalent bonds have definite orientations but electrovalent bonds have no definite orientations — explain
Answer:

Covalent bonds arc formed by the overlap of atomic orbitals having definite orientations. Consequently, covalent bonds have specific orientations. On the other hand, electrovalent bonds have no definite orientations because oppositely charged Ions attract each other from all possible directions by electrostatic forces.

Question 15. Explain why the dipole moment of CD3F (1.858D) is higher than thatofCH3F (1.847D)
Answer:

D is more electron-releasing than H. The Difference in electronegativity between C and F in CD3F is much higher than that between C and F in CH3F. Hence, CD3F is more polar than CH3F. Therefore, the dipole moment of CD3F is higher than that of CH3F.

Question 16. Arrange ethane, ethylene, and acetylene in order of their decreasing C—H bond length. Explain the order.
Answer: The 1 s -orbital of hydrogen overlaps with the sp3, sp2, and sp -hybrid orbitals of carbon in ethane, ethylene, and acetylene respectively to form C —H cr -bonds.

Since the size of these hybrid orbitals decreases in the order: sp3 > sp2 > sp2 the C —H bond length decreases in the order: C—H(C2H6) > C—H(C2H4) > C—H(C, H2)

Question 17. Which symmetry element presents a n -bond? What is meant by the hybridization of atomic orbitals?
Answer: A pi-bond possesses a plane of symmetry, which is also known as a nodal plane.

Chemical Bonding And Molecular Structure Question 63

Question 18. What will be the spatial distribution of sp³, sp², and sp hybrid orbitals?
Answer: 

  1. sp3 — In this case, each of the hybrid orbitals is directed toward the four corners of the tetrahedron.
  2. sp2— In this case, each of the hybrid orbitals is. directed towards the three corners of a triangle.
  3. sp—In this case, the two hybrid orbitals are linearly arranged.

Class 11 Chemistry Chemical Bonding And Molecular Structure Very Short Question And Answers

Question 1. What will be the nature of the compound formed between the metallic elements of groups 1 and 2 and non-metals of groups 6 or 7 of the periodic table?
Answer: Ionic compound;

Question 2. In terms of ionization and electron gain enthalpy, which type of atoms combine to form an ionic compound?
Answer: A metal atom with low ionization enthalpy and a non-metal atom with high electron-gain enthalpy;

Question 3. Write the Lewis symbols of magnesium and aluminium.
Answer: Mg, Al2;

Question 4. Write the structure of an anion which is isostructural with BF3 and a cation which is isostructural with CH4
Answer: NO3 (triangular planar), NH+ (tetrahedral)

Question 5. Give an example of a compound in which electrovalent, covalent, and coordinate covalent bonds are present.
Answer: NH4Cl

Question 6. Which one of CHCI3 and CCl4 is regular tetrahedral?
Answer: CCl4

Question 7. How many cr and n -bonds are present in CH2=CH—CH=CH2?
Answer: No. of bonds = 9 and no. of n -bonds = 2

Question 8. What is the hybrid state of the central atom in each of the following? BF-, NO2, PF5, CO-2
Answer: sp3, sp2 sp3d and sp respectively;

Question 9. Predict the shapes using VSEPR theory: IF?, C1F3, SF6, BeCl2.
Answer: IF2 Pentagonal bipyramidal; ClF3 –  T-shaped; SF6 – Octahedral; BeCl2 – linear

Question 10. How many resonance structures can be written for SO4-?
Answer: 6

Question 11. Arrange the halogen hydroids in order of their decreasing boiling points.
Answer: HF > HCl > HBr > HI;

Question 12. Find out the non-polar molecules among CH3Cl, SF6, SO2, C2H64 and HO—<g>—OH.
Answer: SF6

Question 13. Which one is less viscous, between HF and H2O?
Answer: HF; (Each HF molecule is involved in forming two H-bonds, whereas each H2O molecule is involved in forming four H-bonds.)

Question 14. Which one out of O2 and O2 exhibits the highest paramagnetism?
Answer: O2 (it has two unpaired electrons);

Question 15. How will you distinguish B2 from the following species having the same bond order: Li2, O2- and F2?
Answer: B2 is paramagnetic, but Li2 O2- and F2 are diamagnetic;

Question 16. Is there any change in bond order if the electron is added to the bonding molecular orbital?
Answer: Bond order will increase.

Question 17. The bond order of He+ ion is—\(-\frac{1}{2}, 1, \frac{3}{2}, \mathrm{O}\)
Answer: \(-\frac{1}{2}, 1, \frac{3}{2}, \mathrm{O}\)

Question 18. Give examples of two compounds in which there exists electrovalency, covalency and coordinate covalency.
Answer: Ammonium chloride (NH4Cl) Sodium fluoborate (NaBF4)

Question 19.  Is hybridization possible in an isolated atom?
Answer: Hybridization is not possible for isolated atoms. It occurs only when the atom takes part in bond formation.

Question 20. Which is the most electronegative element according to Pauling’s scale of electronegativity?
Answer:

According to the Pauling’s scale of electronegativity, fluorine is the most electronegative element with electronegativity = 4

Class 11 Chemistry Chemical Bonding And Molecular Structure Fill In The Blanks

Question 1. A _____________a covalent bond is formed between two atoms having different electronegativities.
Answer: Polar

Question 2. Pi bonds are_____________than sigma bonds.
Answer: Weaker

Question 3. In different resonating structures, the arrangement remains the same.
Answer: Atomic

Question 4. When______________atomic_ whereas orbitals when overlapping head-on, overlap laterally, bond formed bond is formed is.
Answer: sigma bond; pi bond

Question 5. AlCl3 is_____________compound, while PCl5 is compound in terms of the octet rule.
Answer: electron deficient, hypervalent

Question 6. The C.G.S unit of dipole moment is_____________whereas its SI unit is
Answer: Deb ye, Coulomb-metre (G-m)

Question 7. In general, the larger is the bond length, _____________ bond energy.
Answer: Smaller

Question 8. The energy of-bond is
Answer: 12.5 to 41.5 kJ mol-1

Question 9. The hybrid state of S in the SO3 molecule is
Answer: sp2

Question 10. The shape of the molecule contains 3 bond pairs and one lone pair around the central atom is
Answer: trigonal pyramids

Question 11. The bond order of CO molecule is ________whereas that of CO+ ion is
Answer: 3,3.5

Question 12. In ice, each O atom is surrounded by out of which, _____________H-atoms are bonded by covalent bonds, while bonds the rest.
Answer: four, two, H-bonds

Question 13. The shape of sulphur hexafluoride molecule is whereas that of sulphur tetrafluoride is _________.
Answer: Regular octahedral, distorted tetrahedral

Question 14. There are_____________π bonds in a nitrogen molecule.
Answer: Two

Question 15. The strongest hydrogen bond is formed between ____________and a hydrogen atom.
Answer: Fluorine

Question 16. A hydrogen bond is then a covalent bond.
Answer: Weaker

Question 17. The dipole moment of methyl alcohol Is._____________ that of CH3SH.
Answer: Higher

Question 18. d2sp3 hybridisation represents
Answer: Octahedral

Question 19. Among the three isomers of nitro phenol, the one that is the least soluble in water Is
Answer: Ortho-isomer

Question 20. Among N2O, SO2, 1+ and l2 , the linear species are and
Answer: N2O and I3-.

Class 11 Chemistry Chemical Bonding And Molecular Structure Warm-Up Exercise Question And Answers

Question 1. The elements belonging to which group(s) of the periodic table combine to form electrovalent compounds, and why?
Answer: Elements of groups 1 and 2 form electrovalent compounds because they are highly electropositive.

Question 2. Which elements exhibit variable electrovalency and why?
Answer: Transition elements. This is because the outermost electron of the ns -subshell and the penultimate (n-1)d subshell are involved in bonding.

Question 3. Why the ionic compounds do not exhibit isomerism?
Answer: Electrostatic force in an ionic compound is distributed uniformly in all directions. Thus, each compound holds a definite number of oppositely charged ions. Hence there are no discrete molecules in ionic compounds. Since ionic bonds are non-directional, ionic compounds do not exhibit isomerism.

Question 4. Why are the n -bonds weaker and more reactive than the cr -bonds?
Answer: The extent of axial overlapping is greater as compared to sideways overlapping. Hence n -bonds are weaker and more reactive than cr -bonds.

Question 5. Which type of ionic compounds exhibit isomorphism?
Answer: Isoelectronic ionic compounds

Question 6. What is solvation energy solvation enthalpy?
Answer: The process of orientation of polar solvent molecules around the ions of the polar solute molecules is called solvation and the energy released in this process is called solvation energy.

Question 7. What is the condition for dissolution of an ionic compound in a particular solvent?
Answer: An ionic compound is soluble (dissolves) in a particular solvent only if the solvation energy exceeds the lattice energy of the crystal (ionic compound).

Question 8. The ionic compounds are soluble in polar solvents but insoluble in non-polar solvents—why?
Answer: In the case of polar solvents, the solvation energy of ionic compounds is greater than lattice enthalpy. So, ionic compounds are soluble in polar solvents, but they are insoluble in non-polar solvents because solvation by nonpolar solvents is not possible for ionic compounds.

Question 9. Why do ionic compounds conduct electricity only in a solution or molten state and not in a solid state?
Answer: Ionic compounds are good conductors of electricity in solution or in the molten state as in these states, their ions are free to move. As the ions are charged, they are attracted towards electrodes and thus act as carriers of electric current.

Question 10. What are valence electrons? 
Answer: The electrons in the ultimate (or outermost) and in some cases, the penultimate shell of an atom that is responsible for chemical bonding are known as valence electrons

Question 11. Give the Lewis symbols of—(1) Br (2) N (3) O2- (4) S2- (5) N3-
Answer:

 Chemical Bonding And Molecular Structure Question 2

Question 12. Hydrogen bonds are usually longer than covalent bonds. Give an example where covalent and hydrogen bonds are equal in length. Explain.
Answer: In fluoride ion (HF2), the covalent bond and the hydrogen bond are equal in length and this is because the structure of this ion is a resonance hybrid of structures 1 and 2.

Question 13. Compare the stabilities of O2 and N2+ ions and comment on their magnetic nature.
Answer: Bond orders of N2 and O2 are 2.5 and 1.5 respectively. Hence, the N2+ ion is more stable. Both the ions contain unpaired electrons and hence are paramagnetic.

Question 14. Why does PClg form PCl3 and Cl2, on strong heating?
Answer:

PCl5 has 2 axial and 3 equatorial bonds. When PCl5 is heated, the weaker axial bonds break, forming PCl3

⇒ \(\mathrm{PCl}_5 \xrightarrow{\text { Heat }} \mathrm{PCl}_3+\mathrm{Cl}_2\)

Chemical Bonding Class 11 Chemistry Short Answer Questions

Chemical Bonding And Molecular Structure Short Answer Type Questions

Question 1. Name the energy that is released during the formation of an ionic crystal.
Answer: Lattice energy

Question 2. Out of NaCI and MgO, which one has higher lattice energy?
Answer: MgO (each ion carries two unit charge).

Question 3. Elements belonging to which groups of the periodic table combine to form electrovalent compounds?
Answer: Highly electropositive metals of groups- 1 and 2 combine with the electronegative elements of groups- 15, and 16 and 7 to form electrovalent compounds.

Question 4. A+ and B2+ ions are isoelectronic, then which of the following information regarding their size is correct: 

  1. A+ > B2+ 
  2. A+ < B2+
  3. A+ = B2+

Answer: A+>B2+

Question 5. Arrange the following isoelectronic ions in increasing order of their ionic radii: X+, Y2+, A, and B2-.
Answer: Y2+ < X- < A- < B2-

Question 6. Designate the following changes as exothermic or endothermic:

  • A(g) A+(g) + e
  • B(g) + e→ B—(g)
  • X(s) → X(g)
  • A+(g) + B(g)→ AB

Answer: Endothermic; and exothermic.

Question 7. What is the coordination number of A+ and B ions if the geometry of the ionic crystal (AB) is cubic?
Answer: The coordination number of A+ and B ions will be 8

Question 8. Out of NaF, KCl, and MgO, which of the compounds exhibit isomorphism?
Answer: NaF and MgO are the two isomorphous compounds [Na (2,8), F (2,8); Mg2+ (2,8), O3 (2,8)]

Question 9. What is the geometry of the ionic crystal if the value of r+/r(radius ratio of the monovalent cation and anion) is in the range 0.225-0.414?
Answer: Tetrahedral

Question 10. Out of Sn2+ and Sn4+, which one is more stable?
Answer: Sn2+ ion is more stable due to the inert pair effect.

Question 11. The values dielectric constants of the solvents, A and B are x and y respectively. If \(\frac{1}{x}<\frac{1}{y}\), t lies in which solvent an ionic compound is more soluble?
Answer: if \(\frac{1}{x}<\frac{1}{y}\). Consequently, an ionic compound is more soluble in solvent A possessing a higher value of the dielectric constant.

Question 12. For the salt CaF2, ΔH°lattice > ΔH°hyd. Predict whether this salt is soluble in water or not.
Answer: CaF2 is insoluble in water.

Question 13. Which out of NaCl and CHC13 reacts with AgNO3 solution to give a precipitate of AgCl?
Answer:
NaCl (being an electrovalent compound, it gives Cl).

Question 14. Which of the following are hypervalent compounds? CO2,CIF3, SO2, IF5.
Answer:
IF5 and ClF3 (the central atom has more than eight electrons in its valence shells)

Question 15. How many types of bonds are present in LiAlH4?
Answer:
3 types—electrovalent, covalent, and coordinate covalent bonds.

Question 16. Give examples of an anion and a cation which are isostructural with BF3 and CH4 respectively.
Answer:
NO-3 (trigonal planar) and NH+4 (tetrahedral).

Question 17. Arrange in order of decreasing size: sp, sp2, sp3
Answer:
sp3 > sp2 > sp.

Question 18. Give the hybridization of P in PCl5. Why are axial bonds longer as compared to equatorial bonds?
Answer:
sp3d. This is due to greater repulsion on the axial bond pairs by the equatorial bond pairs.

Question 19. Mention the change in hybridization (if any) of the Al -atom in the reaction: AlCl3 + Cl-> AlCl4.
Answer:
In AICI3, Al-atom is sp3 – hybridised, but in AlCl3, Al- atom is sp3,-hybridized.

Question 20. Is there any change in his hybridization of II and N atoms as a result of the following reaction?
\(\mathrm{BF}_3+\mathrm{NH}_3 \rightarrow \mathrm{F}_3 \mathrm{~B} \cdot \mathrm{NH}_3\)
Answer:
In the BF3 molecule, the B -atom is sp2 -hybridized, and In the Nil, molecule, the N -atom Is sp3 -hybridized. In the product molecule, both B and N atoms are sp3-hybridized.

Question 21. Although the O-atoms In water and diethyl ether are sp3 -hybridized, the H —O —H and Et —O —Et bond angles arc different —why?
Answer:
Because of steric hindrance between two relatively bulkier C2H5 groups, the Et—O—Et bond angle is greater than the H—O—H bond angle.

Question 22. Draw the resonance structures of N2O obeying the octet rule
Answer:
\(: \ddot{\mathrm{N}}=\stackrel{+}{\mathrm{N}}=\ddot{\mathrm{O}}: \longleftrightarrow: \mathrm{N} \equiv \stackrel{+}{\mathrm{N}}-\ddot{\mathrm{O}}:\)

Question 23. Which Of the following hybrid orbitals possess two types of angles?

  • sp3,
  • sp2,
  • sp,
  • sp3d,
  • sp3d2,
  • sp3d3

Answer: sp3d and sp3d3

Question 24. Which of the following arcs isostructural?
\(\text { (2) } \mathrm{SO}_4^{2-}, \text { (2) } \mathrm{NO}_3^{-}, \text {(3) } \mathrm{NH}_4^{+}, \text {(4) } \mathrm{CO}_3^{2-} \text {, (5) } \mathrm{PO}_4^{3-}\)
Answer: BF3, NO3¯, and CO3²¯ are isostructural (trigonal planar and all the central atoms are sp² -hybridized) and SO32-, NH+4, and PO4-3 are isostructural (tetrahedral and all the central atoms are sp³ -hybridized).

Question 25. Give an example of an anion that is Isostructural with BF3.
Answer: [BeF]3

Question 26. Out of SF6 and SCI2, S has higher electronegativity in which of the compounds and why?
Answer: The electronegativity of S in SF6 is higher because SF6 S has a higher oxidation state.

Question 27. Arrange in the order of increasing ionic character C—H, F—H, Br—II, Na—I, K—F, and Li—Cl
Answer: C—H < Br—H < F—H < Li —Cl < Na —I < K—F

Question 28. What is of a molecule, AB2 if it has a definite dipole moment?
Answer: \(\left(_B \backslash {A}\backslash_B\right)\).

Question 29. The dipole moment of HF is 2.0 D. Calculate its value in coulomb-metre (c.m)
Answer: 0.6674 x 10-29 c.m.

Question 30. N2O is polar even though it is linear- why?
Answer: The linear molecule N2O is polar asitisnot symmetrical.

Question 31. Arrange halobenzenes (C6Hg—X, X = F, Cl, Br, I) in the order of their increasing polarity.
Answer: C6H5F < C6H5Cl < C6H5Br < C6H5l.

Question 32. The dipole moment of a molecule, u = e x d. What is the value of d in the case of CCl4?
Answer: The resultant bond moment of the 2C—Cl bond is equal and opposite to that of the remaining 2C—Cl bonds. Hence there is no net dipole moment and so the value of u is zero.

Question 33. Arrange the following in the order of decreasing strengths: N —H–N, O —H—O, F—H—F.
Answer: F —H…F > O —H…O > N —H—N.

Question 34. Arrange the following interactions in the order of their increasing strengths: covalent bond, H- bonding, dipole-dipole interaction, and ran der Waals forces.
Answer: Wander Waals forces < dipole-dipole interaction < hydrogen bonding < covalent bond.

Question 35. Which of the following combinations of orbitals produce n -n-molecular orbitals?

  • 2pz– 2pz
  • 2ps + 2pz
  • 2z + 2pz
  • 2py + 2py

Answer: 2px– 2pX;2px + 2py

Question 36. What is the change in bond order, if an electron is added to a bonding MO?
Answer: Bond order increases with the addition of an electron to a bonding molecular orbital.

Question 37. According to MO theory which of the following combinations between the orbitals are not possible?

  • 2Pz > 2Pz
  • 2s > 2Py
  • ls, 2s
  • 2Px, 2Px

Answer: 2s > 2Py ;1s,2s

Question 38. Out of O and O2, which one has a greater ionization enthalpy and why?
Answer: O has greater ionization enthalpy. In O2, the first electron has to be removed from the π2px orbital, which has higher energy than the 2p -orbital of the O-atom.

Question 39. Sodium chloride is a solid with having high melting point but carbon tetrachloride is a liquid—why?
Answer: NaCl is an ionic compound. In the crystal of NaCl, the oppositely charged Na+ and Cl are held together by strong electrostatic forces of attraction. Hence a large amount of energy is required to bring the ions into the liquid state.

Hence, sodium chloride is a solid with having high melting point. Carbon tetrachloride (CCl4), on the other hand, consists of non-polar covalent molecules and the only force that operates among the molecules is the weak van der Waals force.

So, the molecules are weakly held together. Hence, carbon tetrachloride has a very low melting point and it is liquid at ordinary temperature.

Question 40. Sodium chloride is soluble in water but insoluble in benzene or hexane. Explain the observation.
Answer: Sodium chloride is an ionic solid and water is a polar solvent. Water molecules attract Na+ and Cl ions by their negative and positive poles respectively. As a result, the ions get detached from the crystal lattice and undergo solvation with the evolution of solvation energy. In this case, since the solvation energy is greater than lattice t energy, NaCl dissolves in water.

On the other hand, benzene and hexane are non-polar organic solvents and they cannot solvate Na+ and Cl ions. Hence, sodium chloride is’ insoluble in benzene or hexane.

Question 41. Nitrogen produces only NCI3 but phosphorus produces both PCl3 and PCl6. Give reasons
Answer: The valence shell electronic configurations of nitrogen and phosphorus are as follows:

Both nitrogen and phosphorus contain 3 unpaired electrons in their valence shells. Using these unpaired electrons, they can form three covalent bonds with chlorine. In this way, NCl3 and PCl3 are produced.

Due to the presence of a vacant 3d -orbital in phosphorus, one 3selectron can be promoted to 3d -orbital and the five unpaired electrons thus obtained can form five covalent bonds with five chlorine atoms to produce PCl5. On the other hand, nitrogen has no d -d-orbital in its second shell. Unlike phosphorus, it cannot extend its covalency to five and hence, it cannot produce NCl5.

Question 42. Aqueous solution of hydrogen chloride is strongly acidic but the solution of hydrogen chloride in benzene is not at all acidic —why?
Answer: Polar H —Cl molecules undergo complete ionization in water and produce H+ ions. Hence, aqueous solution of HCl becomes strongly acidic. On the other hand, HCl molecules do not undergo ionization in non-polar benzene. Because of this, HCl in benzene is not at all acidic.

⇒ \(\mathrm{H}_2 \ddot{\mathrm{O}}: \stackrel{\delta+}{+}+\overbrace{\mathrm{H}}^{\delta-} \mathrm{Cl} \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

Question 43. MgO has a higher lattice energy than NaF. Why?
Answer: The lattice energy of an anionic compound increases with an increase in charge of the ions and decreases with the sum of the ionic radii. The charges on the two ions in MgO (Mg2+ and O2 ions) are twice those on the two ions in NaF ( Na+ and F ions). The sum of ionic radii of Mg2+ and O2 — ions (0.64 A+1.40 A = 2.04 A) is less than that of Na+ and F ions (0.95A+ 1.36A = 2.31A). Therefore, MgO has a higher lattice energy than that of NaF.

Question 44. SnCl4 is a covalent compound whereas SnCl2 is an ionic compound —why
Answer: According to Fajan’s rule, a cation having a small size and high charge exerts a large polarising effect on the neighboring anion resulting in the development of a considerable amount of covalent character in the compound.

Sn4+ ions having a high charge and small size compared to those of Sn2+ ions polarise Cl ion to a greater extent. Therefore, SnCl4 behaves as a covalent compound whereas SnCl2 is an ionic compound.

Question 45. The ionic radius of Na+ is less than the atomic radius of Na+ but the ionic radius of Cl— is greater than the atomic radius of Cl —why?
Answer: The Na+ ion has 11 protons in its nucleus but it has 10 extranuclear electrons. Since the number of electrons is less than that of protons, these electrons are attracted by the nucleus to a greater extent. Hence, the ionic radius of Na+ is smaller than that of having the same number of protons (11) and electrons (11).

On the other hand, the Cl ion has 17 protons in its nucleus but has 18 extranuclear electrons. Since the number of protons is less than that of electrons, the electrons are not strongly attracted by the nucleus. To be relieved of the strain of the electron-electron repulsion, the electron cloud gets more diffused. Hence, the ionic radius of Clis greater than the atomic radius of Cl.

Question 46. Arrange Al3+, Na+, and Mg2+ ions in the decreasing order of their ionic radii and explain the order
Answer: The decreasing order of ionic radii of these three ions is: \(r_{\mathrm{Na}^{+}}>r_{\mathrm{Mg}^{2+}}>r_{\mathrm{Al}^{3+}}\) These three ions are isoelectronic and they contain 10 electrons each in their extra-nuclear shells.  Since the magnitude of nuclear charge gradually increases from Na to Al, the nuclear attractive force for the same number of electrons increases. As a result, their ionic radii gradually decrease.

Question 47. Arrange O2+, N3, and F ions in the decreasing order of their ionic radii and explain the order.
Answer: The ionic radii of these three ions decrease in the order:

⇒ \(r_{\mathrm{N}^{3-}}>r_{\mathrm{O}^{2-}}>r_{\mathrm{F}^{-}}\) ions 316 isoelectronic and they contain 10 electrons each in their extra-nuclear shells. As the number of protons goes on increasing from N3 to F ions, a nuclear attractive force for the same number of electrons gradually increases and as a consequence, their ionic radii gradually decrease i.e., their ionic radii follow the above sequence.

Question 48. Explain the following order of thermal stability of the carbonates of the alkaline earth metals: \(\mathrm{BaCO}_3>\mathrm{SrCO}_3>\mathrm{CaCO}_3>\mathrm{MgCO}_3\)

Answer: As the ionic potential of the metal ion increases, its attraction for the electron of the O-atom of CO ion increases. As a consequence, the tendency of thermal decomposition of the metal carbonate (MCO) to produce metal oxide (MO3) and carbon dioxide (CO2) increases. As the ionic potential increases progressively from Ba2+ to Mg2+, the thermal stability of the carbonates follows the given order.

Question 49. HCl is volatile but NaCl is not. Explain.
Answer: The volatility compound depends on its boiling point which in turn depends on the intermolecular attractive forces. The only attractive force that operates among the polar covalent HCl molecules is dipole-dipole attraction.

Since this attractive force is relatively weak, the boiling point of HO is low and it is volatile (at ordinary temperature, it is a gas). On the other hand, in the electrovalent compound NaCl, the oppositely charged Naÿ and Clions are held together by strong electrostatic forces of attraction in the crystal lattice, and because of this, at ordinary temperature, NaO is a non-volatile solid.

Question 50. What type of bond is formed between two elements, A and B if both of them are highly electronegative or they have huge differences in their electronegativities? Give examples.
Answer: If both the elements A and B are highly electronegative, the bond formed between them would be covalent For example, the bond between N and O in the NCl3 molecule is covalent However, if the elements differ widely in their electronegativities, the bond formed between them would be ionic. For example, in NaCl, the bond between Na and Cl isionic.

Question 51. Out of AlF3 and AICl3, which one is more covalent and why?
Answer: The electron cloud of a large anion is easily distorted by small cations and this results in greater covalency in the compound. The larger Cl ion is polarised to a greater extent than the smaller F ion and because of this AICl3 is more covalently natural than AlF3.

Question 52. Indicate the nature of chemical bonding present in each of the following compounds:

  • CH3OH,
  • NH3
  • CaC12
  • CaH2
  • K2O
  • CO2
  • Al2O2
  • NH4Cl
  • HCl
  • Mg3N2
  • Ci2O
  • NaBH4

Answer:

  • Covalent,
  • Covalent,
  • Electrovalent
  • Electrovnt lent,
  • Electrovalent
  • Covalent,
  • Electrovalont, electrovalent.
  • Covalent and co-ordinate covalent, covalent.
  • Electrovalent,
  • Covalent,
  • Electrovalent
  • Colvent and coordinate covalent.

Question 53. Which one between p and sp –orbitals have more directional characteristics and why?
Answer: The directional nature of the sp-orbital is greater than that of the orbital. Because the two lobes of p -p-orbitals are similar in size and possess the same electron density while one of the two lobes of sp -orbitals is larger and has higher electron density

Question 54. In a certainpolar solvent, PCI- undergoes ionization as follows \(2 \mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_4^{+}+\mathrm{PCl}_6^{-}\) Predict geometrical shapes of all the species involved.
Answer: The geometrical shapes of the molecules or ions can be predicted from the hybridization state of the central atom.

⇒ \(\text { For } \mathbf{P C l}_5: H=\frac{1}{2}[5+5-0+0]=\frac{10}{2}=5\)

Therefore, the central P-atom is sp3d-hybridized. Consequently, the molecule is a trigonal bipyramidal shape.

⇒ \(\text { For } \mathbf{P C l}_4^{+}: H=\frac{1}{2}[5+4-1+0]=\frac{8}{2}=4 \text {. }\)

Therefore, the central P-atom is sp3 -hybridized. Consequently, the geometrical shape of this ion is tetrahedral.

⇒ \(\text { For } \mathbf{P C l}_6^{-}: H=\frac{1}{2}[5+6-0+1]=\frac{12}{2}=6 \text {. }\)

Therefore, the central P-atom is sp3d2 hybridized. Consequently, the geometrical shape of the ion is octahedral.

Question 55. MgCl2 is linear but SnCl2 is angular—explain
Answer: For \(H=\frac{1}{2}[2+2-0+0]=\frac{4}{2}=2,\) i.e.,, the central Mg -atom is sp -hybridised. Therefore, the molecule is linear. On the other hand, for SnCl2,\(H=\frac{1}{2}[4+2-0+0]=\frac{6}{2}=3,\) i.e. the central Sn-atom is sp2 -hybridised. Therefore, the two bond pairs and one lone pair present in the molecule are directed toward the comers of an equilateral triangle. Hence, the SnCl2 molecule is angular.

Question 56. Arrange the following in the increasing order of their polarities and explain with reason: B — Cl, Ba — Cl, Br — Cl, Cl — Cl
Answer: The increasing order of polarity of these bonds is: Cl — Cl<Cl — Br < Cl — B < Cl — Ba. The greater the difference in electronegativity between the covalently bonded atoms, the greater the polarity of the bond. The electronegativities of B, Ba, Br, and Cl follow the order: of Cl > Br > B > Ba. Consequently, the polarity of the bonds formed by these atoms with Cl-atom progressively increases.

Question 57. Identify the three isomeric chlorotoluenes having dipole moments: 1.35D, 1.9D, and 1.78D.
Answer: Methyl group ( —CH3) can exert a +1 effect while Cl-atom exerts a -I effect. In p-chlorotoluene, -CH3 and -Cl group moments act linearly. So its dipole moment is equal to the sum of the two group moments. Hence, the dipole moment of this compound is maximum. -CH3 group moment is considered to act in the direction of the ring. Soin m-chlorotoluene, these two group moments act at an angle of 60°.

Hence, the dipole moment of m-chlorotoluene is somewhat less than that of its p-isomer. In o-chlorotoluene, these two group moments act at an angle of 120° and hence the dipole moment of this isomer is less than that of the m-isomer. Thus, dipole moments of 1.35D, 1.78D and 1.90D correspond to o-, m- and p-chlorotoluenes.

Chemical Bonding And Molecular Structure Methyl Group

Question 58. Which has the least dipole moment— 1-butene, cis- 2-butene, draws-2-butene and 2-methylpropene?
Answer: The structure of the given compounds is as follows

Chemical Bonding And Molecular Structure Trans2Butene

Question 59. What do you mean by hydrogen bond donor and hydrogen bond acceptor? Define protic and aprotic solvents. Give examples.
Answer: H-bonding represented as X—H—Y denotes the interaction between a donor species and an acceptor species. X—H is considered as a donor and Y is an acceptor of H.

Chemical Bonding And Molecular Structure H Bonding Represtend As X,H,Y

Protic solvent: The solvent whose molecules can act as hydrogen-bond donors are called protic solvents. Water (H2O), ethanol (G2H5OH), acetic acid (CH3COOH), etc., are some common examples ofprotic solvents.

Aprotic solvent: The solvent whose molecules can’t act as a bond donor is known as aprotic solvent. Diethyl ether (C2H5OC2H5), methylene chloride (CH2C12), hexane [CH3(CH2)4CH3], etc., are some examples of aprotic solvents.

Question 60. The boiling point of water (100°C) is much higher than that of HF(19.5°C), even though they have similar molecular masses. Explain.
Answer: Each water molecule is involved in intermolecular H -bonding with four other water molecules but each HF molecule is involved in intermolecular H -bonding with two other HF molecules.

Therefore, the degree of molecular association in water is much higher than that in HF. Therefore, the boiling point of water is much higher than that of HF.

Chemical Bonding And Molecular Structure Intermolecular H-Bonding In Water

Question 61. Explain the following observations: Glycerol [HOCH2CH(OH)CH2OH] is a highly viscous liquid. When 30 mL of water is added to 30 mL of ethanol, the volume of the mixture becomes less than 60 ml.
Answer: A glycerol molecule contains three —OH groups. Hence glycerol molecules remain extensively associated through intermolecular H-bonding. This accounts for the high viscosity of glycerol.

The intermolecular H-bonding formed between ethanol and water is stronger than that formed in ethanol itself. When water is added to ethanol, the stronger intermolecular H-bonding between ethanol and water leads to a contraction of volume. So, when 30 mL of water is added to 30 mL of ethanol, the volume of the mixture becomes less than 60 mL.

Question 62. Arrange water, methanol, and dimethyl ether in increasing order of their viscosity and give reasons in favor of that order.
Answer: The greater the extent of intermolecular H-bonding, the greater the intermolecular attraction among different layers, and hence, the greater the viscosity of the compound.

Water molecules having two -OH groups have greater intermolecularH-bonding than methanol (CH3OH) molecules having only one -OH group. Dimethyl ether (CH3OCH3) having no -OH group does not remain associated through H-bonding. Therefore, the viscosity of these liquids follows the order: of dimethyl ether < methanol < water.

Question 63. At equilibrium, acetylacetone (CH3COCH2COCH3) exists mainly (80%) in enol form— explain.
Answer: The enol-form of acetylacetone is stabilized by resonance. Besides this, its stability is further increased by strong intramolecular bonding. Similar stability in the case of keto form is not possible. Hence, at equilibrium, acetylacetone exists mainly in the enol form.

Chemical Bonding And Molecular Structure Intermolecular Enol form of acetyl acetone

Question 64. State the useful rule related to the solubility of a compound. Unlike ethane, ethanol dissolves in water. Explain and discuss in terms of energy change.
Answer: The useful rule relevant to the solubility of a compound is “like dissolves like” which means a polar solute dissolves in a polar solvent & a non-polar solute dissolves in a non-polar solvent Ethanol molecules get involved in intermolecular Hbonding with water molecules. So, ethanol readily dissolves in water.

On the other hand, non-polar ethane (CH3CH3) molecules are not capable of forming H -bonds with water molecules so ethane does not dissolve in water. Alternatively, it can be said that H-bonds between different ethanol and different water molecules are replaced by very similar H-bonds (in terms of energy) between water and ethanol molecules.

So, the dissolution of ethanol in water takes place readily. Since non-polar ethane molecules are not solvated by polar water molecules, no solvation energy is liberated. Thus, the energy required to separate the water molecules (associated with strong H -bonding) is not available, and so stronger H -bonds are not replaced by any other intermolecular forces. Hence, ethane does not dissolve in water.

Question 65. Inert gases do not generally participate in chemical reactions—explain with reason.
Answer: Because of the very stable electronic configuration of the outermost shell, inert gases tend to exhibit the least chemical reactivity. The two possible reasons that can explain the stable electronic configuration, as well as the poor reactivity of inert gases, are as follows:

Since there are no odd electrons in the valence shell of inert gases, they have the least tendency to form a covalent bond by forming an electron pair i.e., they produce no covalent compounds.

Since the s-and p -p-orbitals of the valence shell are filled with electrons, the ionization enthalpy of inert gases is very high while their electron gain enthalpy is negligibly small. Hence, they exhibit the least tendency to form ionic compounds.

Question 66. The bond dissociation enthalpy of N2 is higher than that of N2 but the bond dissociation enthalpy of is higher than that of O2. Explain.
Answer:

⇒ \(\begin{aligned}
& \mathrm{N}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\sigma_{2 p_z}\right)^2 \\
& \mathrm{~N}_2^{+}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\sigma_{2 p_z}\right)^1 \\
& \mathrm{O}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^1 \\
& \mathrm{O}_2^{+}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1
\end{aligned}\)

⇒ \(\begin{array}{|c|c|c|c|c|}
\hline \text { B.O. } & \mathbf{N}_2 & \mathbf{N}_2^{+} & \mathbf{O}_2 & \mathbf{O}_2^{+} \\
\hline \frac{\boldsymbol{N}_b-\boldsymbol{N}_a}{2} & \frac{8-2}{2}=3 & \frac{7-2}{2}=2.5 & \frac{8-4}{2}=2 & \frac{8-3}{2}=2.5 \\
\hline
\end{array}\)

The greater the value of bond order, the higher will be the value of bond dissociation enthalpy. Hence, the bond dissociation enthalpy of N is higher than that of N2 but the die bond dissociation enthalpy of O2 is lower than that of O2.

Question 67. Arrange the following species in order of their increasing bond lengths and explain: C2, C2, and C2-2.
Answer:

⇒ \(\begin{aligned}
& \mathrm{C}_2: K K^*\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2, \\
& \mathrm{C}_2^{-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\sigma_{2 p_z}\right)^1 \\
& \mathrm{C}_2^{2-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\sigma_{2 p_z}\right)^2
\end{aligned}\)

⇒ \(\begin{array}{|c|c|c|c|}
\hline \text { B.0. } & \mathrm{C}_2 & \mathrm{C}_2^{-} & \mathrm{C}_2^{2-} \\
\hline \frac{N_b-N_a}{2} & \frac{6-2}{2}=2 & \frac{7-2}{2}=2.5 & \frac{8-2}{2}=3 \\
\hline
\end{array}\)

Since bond length is inversely proportional to bond order, the increasing sequence of bond length of the given species: C2 < C2 < C2

Question 68. Explain the following order of bond dissociation enthalpies: F—F < Cl—Cl < 0=0 < N = N
Answer: Since the size of the F -atom is smaller than that of the Cl -atom, the F—F bond length is less than the Cl—Cl bond length. Therefore, the die F —F bond dissociation enthalpy is expected to be higher than the Cl—Cl bond dissociation enthalpy. But actually, the reverse is true. In the F2 molecule, each F -atom contains three lone pairs of electrons. Because of their proximity, diese unshared electron pairs exert strong repulsive forces towards each other.

Similar forces of repulsion caused by the same number of lone pairs are much less in the case of Cl2 because of the larger size of Cl. So despite the smaller bond length, the die bond dissociation enthalpy of the F —F bond is less than the Cl—Cl bond dissociation enthalpy.

In the O2 molecule, the two 0 -atoms are attached by a double bond. Due to the presence of two unshared pairs of electrons in each 0- atom, the force of repulsion is relatively lower. So, 0=0 bond dissociation enthalpy is higher than the Cl —Cl bond dissociation enthalpy.

In an N2 molecule, the two N -atoms are linked together by a triple bond. Each N -atom contains only one unshared electron pair which causes minimum force of repulsion. Thus, the bond dissociation enthalpy of the N = N bond is the highest. Hence, the bond dissociation enthalpies of the bonds present in die given molecules follow’ the given sequence.

Question 69. Determine the shapes of the given molecules or ions:
\(\text {(1) } \mathrm{POCl}_3\left(2) \mathrm{CH}_3^{-} \text {(3) } \mathrm{CH}_3 \text { (4) } \mathrm{PO}_4^{3-} 6 \text (5) \mathrm{~F}_2 \mathrm{O}\right.\)
Answer: The p-atom in POCl, molecule is attached to the 0-atom by a double bond and to three Cl -atoms by three single bonds. The number of electrons in the valence shell of P-atom = 5 + 2 + 1 + 1 + 1 = 10. Out of these 10 electrons or 5 electron pairs,1 electron pair is involved in the formation of the bond which has no role in determining the shape of the molecule. According to VSEPR theory, four electron pairs are oriented tetrahedrally, i.e., the shape of the molecule is tetrahedral.

C-atom in CH3 ion is bonded to three H-atoms by three single bonds. The number of electrons in the valence shell of C-atom = 4 + 3 + 1 (for the negative charge) = 8. Out of these 8 electrons or 4 electron pairs, there are three bond pairs and one lone pair. According to VSEPR theory, the ion is pyramidal.

C-atom in the CH3 ion is bonded to three H -atoms by three single bonds. Thus, the number of electrons present in the valence shell of C =4 + 3-1 (for the positive charge) = 6. Because of the presence of 6 electrons or 3 electron pairs, the ion is trigonal planar.

The central P-atom in PO- ion is bonded to three O-atoms by three single bonds and one O-atom by a coordinate covalent bond. Thus, the number of electrons present in the valence shell of P-atom = 5 + 3 + 0=8 or 4 electron pairs (three covalent cr bond pairs and one coordinate cr bond pair). According to VSEPR theory, the four electron pairs are oriented tetrahedrally, i.e., the ion is tetrahedral.

The central O -atom in the F2O molecule is bonded to two F -atoms by two single bonds. The number of electrons in the valence shell of O -atom =6+1 + 1 =8. Out of these 8 electrons or 4 electron pairs, two are bond pairs and two are lone pairs. According to VSEPR theory, the shape of the molecule is angular or V-shaped.

Question 70. The molecule of any compound composed of two dissimilar elements Is always polar—justify the statement.
Answer: The statement is not always true. A diatomic molecule (A-B) consisting of two dissimilar elements of different electronegativities is always polar because there is no possibility of cancellation of the bond moment HCl, HF, etc., are examples of such molecules.

A poly-atomic molecule with two dissimilar elements may or may not be polar. The polarity of such a molecule depend of such a molecule depends on its geometrical shapes If the molecule is linear having a structure like A—B—A (i.e., the individual bond moments cancel out each other), it is polar and if not, it is polar.

The linear CO2 molecule, for example, is nonpolar because the two equal and opposite C2O bond moments cancel out each other. However, the angular H2O molecule is polar because the two O—H bond moments do not cancel out each other and there exists a resultant moment.

Question 71. Explain the following observations:

  1. H+2ion is more stable than H2 ion even though the bond orders of both ions are the same.
  2. When a magnet is dipped in a jar containing liquid oxygen, some oxygen molecules cling to it.
  3. O2 is more paramagnetic than O2.
  4. The molecular ion, HeH- does not exist

Answer: The antibonding,\(\sigma_{1 s}^*\) molecular orbital of the H2 ion contains one electron but the antibonding crÿs molecular orbital of the H2 ion contains no electron. Therefore, H2 ion is more stable than H2 ion, even though their bond orders are the same.

The electronic configuration oxygen molecule (O2) shows the presence of two unpaired electrons. This suggests that the molecule is paramagnetic. Thus, when a bar magnet is dipped in a jar of liquid oxygen, some molecules cling to

The electronic configurations of O2 and O2 show that the former contains two unpaired electrons (one in each \(\pi_{2 p_x}^*\) and \(\pi_{2 p_y}^*\) while the latter contains one unpaired electron in 7r2p. Therefore, O2 is more paramagnetic than O+2.

Total number of electrons in HeH- = 2 (for He) + 1 (for H ) + 1 (for the negative charge) = 4. Its electronic configuration is \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\) while the latter contains one unpaired electron in \(\pi_{2 p_x}^*\) Therefore, 02 is more paramagnetic than O+2.

Total number of electrons in HeH- = 2 (for He) + 1 (for H ) + 1 (for the negative charge) = 4. Its electronic configuration is \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\) So, bond
order \(=\frac{2-2}{2}=0\). hence, he- does not exist.

Question 72. Is it correct to say that bond order always increases with the loss of electrons? Explain your answer.
Answer: The statement is not always correct When an electron is expelled from a bonding MO, the bond order decreases, but when an electron is expelled from an antibonding MO, the bond order increases.

For example, loss of an electron from \(\mathrm{H}_2\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^0 \text { to from } \mathrm{H}_2^{+}\left(\sigma_{1 s}\right)^1\left(\sigma_{1 s}^*\right)^0\) results in decrease of bond order. However, loss of an electron from \(\mathrm{H}_2^{-}\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^1 \text { to form } \mathrm{H}_2\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^0\) results in increase of bond order. Bond order of \(H_2^{+}=\frac{1}{2}(1-0)=\frac{1}{2}\mathrm{H}_2=\frac{1}{2}(2-0)=1 ; \mathrm{H}_2^{-}=\frac{1}{2}(2-1)=\frac{1}{2} \text {. }\)

Question 73. Identify polar and non-polar molecules from the following: Cl2, CHCl3, NH3, and BCl3, The dipole moment of the NF-, molecule is less than that of the NH3 molecule. Explain.
Answer: Polar Molecules: CHCH3 NH3,Non-polar Molecules: Cl2, BCI3

Question 74. X is the central atom of the XO2 molecule. If the dipole moment of the molecule is zero, indicate the hybridization state of X.
Answer: \(s p[O=X=O]\)

Question 75. Arrange the following compounds in increasing order of boiling point: HF, H2O, NH3
Answer: \(\mathrm{NH}_3<\mathrm{HF}<\mathrm{H}_2 \mathrm{O}\)

Question 76. Arrange the following molecules in increasing order of the number of lone pairs of electrons. H2O, PCl3, H2O, BF3.
Answer: Considering the lone pair of the central atom: BF3(0/p) < PCl3(l/p) < H2O(2/p)

Considering the lone pairs of all the atoms present in the molecule:

⇒ \(\begin{aligned}
& \mathrm{H}_2 \mathrm{O}<\mathrm{BF}_3<\mathrm{PCl}_3 \\
& (2 l p) \quad(9 l p) \quad(10 l p)
\end{aligned}\)

Question 77. Mention the state of hybridization of the central atom of the following molecules/ions: CO2, PH4, ClO3, CS2 Or, Write the resonating structures of the ClO4 ion
Answer:

⇒ \(\begin{array}{|c|c|c|c|c|}
\hline \text { Molecule/Ion } & \mathrm{CO}_3^{2-} & \mathrm{PH}_4^{+} & \mathrm{ClO}_3^{-} & \mathrm{CS}_2 \\
\hline \begin{array}{c}
\text { Hybridisation of } \\
\text { central atom }
\end{array} & s p^2 & s p^3 & s p^3 & s p \\
\hline
\end{array}\)

Question 78. Mention the nature of bonding of the following molecules/ions: CaH2, BH4, Na2O2, SiH4
Answer:

⇒ \(\begin{array}{|c|c|c|c|c|}
\hline \begin{array}{c}
\text { Molecule/ } \\
\text { Ions }
\end{array} & \mathrm{CaH}_2 & \mathrm{BH}_4^{-} & \mathrm{Na}_2 \mathrm{O}_2 & \mathrm{SiH}_4 \\
\hline \begin{array}{c}
\text { Nature of } \\
\text { bonding }
\end{array} & \text { ionic } & \begin{array}{c}
\text { covalent and } \\
\text { coordinate }
\end{array} & \begin{array}{c}
\text { ionic } \\
\text { and } \\
\text { covalent }
\end{array} & \text { covalent } \\
\hline
\end{array}\)

Question 79. Between N2O2, and NO2 molecules, which one is more polar? Explain.
Answer:

Chemical Bonding And Molecular Structure N2O Linear Molecule

N2O is a linear molecule. In this molecule the N bond moment and the moment due to the unshared electron pair act in opposite directions which decreases the dipole moment making the molecule less polar. However, the resulting dipole moment is large in NO2 due to its angular nature.

Question 80. Arrange the following ions in the increasing order of their ionic radii. F, Mg2+,Al3+, O2-
Answer: Al3+ < Mg2+ < F < O2-

Question 81. Arrange the following molecules in increasing order of their dipole moments: NH3, NF2, CBr4
Answer: CBr4<NF3<NH3

Question 82. How many (cr) and (a) bonds are present in buta-1,3- diyne?
Anwer: H—C=C—C=C—H <x -bonds: 5; n -bonds: 4 Buta-l,3-diyne

Question 83. What are the different types of bonds present in ammonium bromide molecules?
Answer: Ionic, covalent, and coordinate bonds.

Chemical Bonding And Molecular Structure ionic covalent and coordinate bonds

Question 84. Draw the resonating structures of sulfate ion
Answer:

Chemical Bonding And Molecular Structure The resonating structures of sulphate ions

Question 85. Which of the H2O or H2S molecules has a greater bond angle? Explain. Between o-nitrophenol and p-nitrophenol, which one has a greater boiling point? Explain.
Answer: As the electronegativity O-atom is higher than S, the bond angle of H2O is larger than that of H2S.

Chemical Bonding And Molecular Structure The resonating structures of sulphate ions

Question 86. Which of the following is not paramagnetic—

  1. N2+
  2. Li2
  3. O2
  4. H2+

Answer: \(\mathrm{Li}_2:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\)

Question 87. Which bond among the following is the least ionic—
Answer: In F—F, there is no electronegativity difference

Question 88. When hydrogen combines with oxygen, a polar covalent product is formed—explain. What types of bonds are present in KHF2
Answer: Hydrogen reacts with oxygen to form H2O as a covalent compound. Due to angular structure the resultant of the two O —H bond moments and the moment contributed by the lone pair act in the same direction. Hence, the H2O molecule is polar.

K+[F…..H —F]- Ionic, covalent and hydrogen bond.

Question 89. Write canonicals of ClO4 ion.
Answer:

⇒ \(\begin{aligned}
& \mathrm{H}_2:\left(\sigma_{1 s}\right)^2, \mathrm{~B} \cdot \mathrm{O}=\frac{1}{2}(2-0)=1 \\
& \mathrm{He}_2:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2, \mathrm{~B} \cdot \mathrm{O}=\frac{1}{2}(2-2)=0
\end{aligned}\)

As the bond order is zero, the He2 molecule has no existence.

Question 90. In which of the following conversions there are changes of hybridization and shape
Answer: BF3 molecule exhibits trigonal planar geometry with each F—B—F bond angle 120°. B-atom is sp2 hybridised.

Chemical Bonding And Molecular Structure BF3 molecule

BF4 on the other hand exhibits tetrahedral geometry with each F —B —F bond angle 109.5°. B-atom is sp3 – hybridized.

Question 91. What are the types of hybridization of NH4, CO2-, H2S, and SO2? The C—0 bond is polar but CO2 does not have a dipole moment. Why?
Answer: In NH+4 and H2S the central atoms (N and S) undergoes sp3 hybridization. In CO2-3 the central atom C is sp2 hybridised. In SF6, the central atom S undergoes sp3d2 hybridization.

Question 92. The bond order of the He2+ ion is—
Answer: There is no electron present in the He2+ ion. Thus bond order = 0

Question 93. Which is not paramagnetic of the following— N+  CO  O- NO

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 p_z}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\)

Question 94. Arrange die following compounds according to their increase of melting point: NaCl, MgCl2, AlCl3 Between NH2 and NF3 which one is more polar and why?
Answer: AlCl3 < MgCl2 < NaCl

Question 95. Why does the PCl5 exist but NCl5 does not? Why is BaSO4 not soluble in water?
Answer: For a compound to be soluble in H2O, its lattice enthalpy should be low compared to its hydration enthalpy. Since both Ba2+ and SO -are large, they stabilize each other to form a strong lattice. This leads to the insolubility of BaSO4 in water.

Question 96. Which has the smallest bond length—
Answer: O+2. Its bond order is 2.5 (maximum among the given species) and we know bond-length \(\propto \frac{1}{\text { bond order }}\)

Question 97. What is the hybridization state of the central atom in 1-3

  1. sp3
  2. dsp2
  3. sp3d2
  4. sp3d

Answer: 4. sp3d

Question 98. Both Br(g) and NO2(g) are reddish-brown gaseous substances. How will you chemically distinguish between them? What will be the order of covalent character of the following compounds?

  1. LIP
  2. LiCl
  3. LiBr
  4. Lil

Answer: 1. LIP

Question 99. The state of hybridization of the central atom of which of the following is sp3d2

  1. SP4
  2. PCI5
  3. SP6

Answer: 3. SP6

Question 100. Which of the following is the correct order of repulsive interaction of lone pair (Ip) and bond pair (bp) of electrons —

  1. Ip- Ip > Ip- bp > bp- bp
  2. Ip- bp > Ip- Ip > bp -bp
  3. bp- bp > Ip-Ip >Ip- bp
  4. Ip- Ip > bp- bp > Ip -bp

Answer: 1. Ip- Ip > Ip- bp > bp- bp

Question 101. Draw the canonicals of CO3. Why is the boiling point of H2O greater than that of II2S?
Answer: Due to the presence of extensive intermolecular H bonding in H2O, it has a higher boiling point than H2S.

Question 102. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Answer:

⇒ \(\begin{array}{|c|c|c|c|c|c|c|}
\hline \text { Element } & \mathbf{M g} & \mathbf{N a} & \mathbf{B} & \mathbf{O} & \mathbf{N} & \mathbf{B r} \\
\hline \begin{array}{c}
\text { Atomic } \\
\text { No. }
\end{array} & 12 & 11 & 5 & 8 & 7 & 35 \\
\hline \text { E.C. } & 2,8,2 & 2,8,1 & 2,3 & 2,6 & 2,5 & 2,8,18,7 \\
\hline \begin{array}{c}
\text { Lewis } \\
\text { dot } \\
\text { structure }
\end{array} & \ddot{\mathrm{Mg}} & \dot{\mathrm{Na}} & \cdot \dot{\mathrm{B}} \cdot & \mathbf{\text { Ö: }} & \mathbf{: \mathrm { N } :} & \mathbf{: \mathrm { Br } :} \\
\hline
\end{array}\)

Question 103. Write Lewis symbols for the following atoms and ions: S and S2-; Al and Al3+, H and H
Answer:

⇒ \(\begin{array}{|c|c|c|c|}
\hline \text { Element } & 16^{\mathbf{S}} & { }^{13} \mathbf{A l} & \mathbf{1}^{\mathbf{H}} \\
\hline \text { E.C. } & 2,8,6 & 2,8,3 & 1 \\
\hline \text { Atom } & : \dot{\mathrm{s}}: & \cdot \dot{\mathrm{Al}} \cdot & \mathrm{H} \cdot \\
\hline \text { Ion } & {\left[\:_0\right]^{2-}} & {[\mathrm{Al}]^{3+}} & {[\mathrm{H}:]^{-}} \\
\hline
\end{array}\)

Question 104. Although the geometries of NH3 and H2O molecules are distorted tetrahedral, the bond angle in water is less than that of ammonia. Discuss.
Answer: The difference in the bond angles of water (H2O) and ammonia (NH3) is due to the difference in the number of lone pairs and bond pairs in these two species. In NH3, N has 1 lone pair and 3 bond pairs while in H2O, O has 2 lone pairs and 2 bond pairs. The lone pair-bond pair repulsion in H2O is much more than in NH3. Hence, the bond angle around the central atom in H2O is relatively smaller than that in NH3.

Question 105. Explain the important aspects of resonance concerning the CO2-3 ion
Answer: When a single structure cannot explain all the properties of a molecule, some probable hypothetical structures (canonical structures) are used. The actual structure is called the resonance hybrid which is an intermediate of the canonical structures. This is termed resonance.

Carbonate ion (CO2-3) can be represented by a combination of the following 3 resonating structures in which CO2-3 is the resonance hybrid of the three canonical structures 1,2 and 3.

Question 106. H3PO3 can be represented by structures 1 and 2. Can these two structures be taken as the canonical forms of the resonance hybrid representing H2PO3? If not give reasons for the same.
Answer: These cannot be taken as canonical forms, since the position of the atoms has been changed.

Question 107. Although both CO2 and H2O are triatomic molecules, the shape of the H2O molecule is bent while that of CO2 is linear. Explain this based on the dipole moment.
Answer: The bond moments of two C=0 bonds in CO2 cancel each other, indicating the linear structure of CO2. However, the H2O molecule has a net dipole moment (0). Thus, the bond moments of two O —H bonds do not cancel each other. As a result, H2O has a bent structure.

Question 108. Define electronegativity. How does it differ from electron gain enthalpy?
Answer: The electronegativity of an element is the tendency or ability of its atom to attract the shared pair of electrons towards itself in a covalent bond. On the other hand, electron gain enthalpy is the energy released, when one mole of gaseous atoms of the element accepts electrons from gaseous un negative ion.

Question 109. Arrange in order of increasing ionic character in the molecules: LiF, KaO, N2, SO2, and CIF3.
Answer: The ionic character of a molecule depends on the difference in electronegativities of the atoms involved in bond formation and also on the geometrical arrangements of the bonds. Therefore, the increasing order of ionic character is given by— N2 < SO2 < ClF3 < K2O < LiF.

Question 110. The skeletal structure of CH6COOH as shown below is 4.8.2, correct, but the sonic of the bonds is shown incorrectly. Write the correct Lewis structure for acetic acid.
Answer: The skeletal structure of CH3COOH is correct, but according to Lewis’s theory the arrangement of electrons is not perfect. The correct Lewis structure of CH3COOH

Question 111. Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its center. Explain why CH4 is not square planar.
Answer: According to VSEPR theory, shared electron pairs around the central atom in a covalent molecule lie apart at the farthest possible distance to minimize the electrostatic forces of repulsion between them. For tetrahedral geometry, the bond angle is 109°28 while for square planar geometry, the bond angle is 90°.

In square planar geometry, repulsions between the bond pairs are greater than that in the tetrahedral geometry. Consequently, methane assumes tetrahedral geometry instead of square planar geometry. However, for square planar geometry, the required hybridization is dsp² which is not possible for carbon as it has no orbitals in the valence shell.

Question 112. Explain why the BeH2 molecule has a zero dipole moment although the Be —H bonds are polar.
Answer: The BeH2 molecule is linear (H —Be —H). Its bond angle is 180°. The two Be —H bond moments are equal and opposite and hence cancel out each other.

Question 113. Describe the change in hybridization (if any) of the AI in the following reaction. AlCl3 + Cl AlCl-3
Answer: Using the equation \(H=\frac{1}{2}[V+X-C+A]\)

⇒ \(\text { In } \mathrm{AlCl}_3, \mathrm{H}=\frac{1}{2}(3+3-0+0)=3\)

∴ Hybrid state of Al = sp2

⇒ \(\text { In }\left[\mathrm{AlCl}_4\right]^{-}, \mathrm{H}=\frac{1}{2}(3+4-0+1)=4\)

∴ Hybrid State of Al= sp3

Question 114. Is there any change in the hybridization of B and N atoms as a result of the following reaction?

⇒ \(\mathrm{BF}_3+\mathrm{NH}_3 \rightarrow \mathrm{F}_3 \mathrm{~B} \cdot \mathrm{NH}_3\)

Answer: N in NH3 is the donor and B and BF3 are the acceptor. The hybrid state of B in BF3 is sp2 and that of Nin NH3 is sp3. In the compound F3B.NH3, both and B atoms are surrounded by 4 bond pairs. Thus both have a hybrid state of sp3. Hence, during combination, the hybrid state of B changes from sp2 to sp3 but that remains the same.

Question 115. Considering the x -x-axis as the intermolecular axis which out of the following will not form a sigma bond and why?

  1. Is, Is
  2. Is, 2px-,
  3. 2py, 2py
  4. Is, 2s.

Answer: σ -bond is formed between two atoms by the end-to-end (head-on) overlap of atomic orbitals. Hence form σ -bond. In the case of (3), if x -is considered as an intermolecular axis, then the n -bond will form due to lateral overlapping between py orbitals.

Question 116. Write the significance of a plus and a minus sign shown in representing the orbitals.
Answer: An orbital is a pictorial representation of wave function, where the ‘+’ and- ‘ sign represents the opposite phases of the wave.

The bonding molecular orbital is formed by a combination of ‘+’ with ‘+’ or ‘ with part of the electron waves whereas the antibonding molecular orbital is formed by the combination of ‘+’ with part.

MCQs for Class 11 Chemical Bonding and Molecular Structure

Chemical Bonding And Molecular Structure Multiple Choice Questions

Question 1. The Sp3d2-hybridization of the central atom of a molecule would be

  1. Wouldsquareleadplanarto— geometry
  2. Tetrahedral geometry
  3. Trigonal bipyramidal geometry
  4. Octahedral geometry

Answer: 4. Octahedral geometry

One s, three p, and two d -orbitals mix up together to form six equivalent Sp3d2-hybrid orbitals. The molecules, in which these orbitals are involved, have octahedral geometry.

Question 2. Which of the following is paramagnetic

  1. N2
  2. NO
  3. CO
  4. O3

Answer: 2. No

From a consideration of electron distribution in molecular orbitals of NO, it is known that it has one unpaired electron. So, NO molecule is paramagnetic.

Question 3. In the electron-dot structure, calculate the formal charge from left to right nitrogen atom, \(\ddot{\mathrm{N}}=\mathrm{N}=\ddot{\mathrm{N}}-\)

  1. -1,-1+1
  2. -1,+1,-1
  3. +1,-1,-1
  4. +1,-1,+1

Answer: 2. -1,+1,-1

Formal charge = No. of valence electrons in the atom No. of unshared electrons \(-\frac{1}{2}\) No. of shared electrons.

Chemical Bonding And Molecular Number of Shared Electrons

  • Formal charge on N-atom (1) =5-4- (4 ÷ 2) = -1
  • Formal charge on N-atom (2) = 5- 0- (8 ÷ 2) = 1
  • Formal charge onN-atom (3) = 5- 4- (4÷ 2) = -1

Question 4. Which of the following Compounds has the maximum volatility

Chemical Bonding And Molecular The Following Compounds Has Maximum Volatility

Answer: 3. Chemical Bonding And Molecular The Following Compounds Has Maximum Volatility. In O-hydroxy carboxylic acid, the —OH and —COOH groups are situated at two adjacent carbon atoms of the ring and are involved in intramolecular H-bond formation.

So, these molecules exist as discrete molecules and consequently, the compound has maximum volatility.

Chemical Bonding And Molecular The Following Compounds Has Maximum Volatility..

Question 5. The number of acid protons in H3PO3 is

  1. 0
  2. 1
  3. 2
  4. 3

Chemical Bonding And Molecular Number Of -OH Group

Answer: 3. 2 Number of the —OH group in H3PO3 is 2 and henceitis dibasic in nature.

Question 6. In 2-butene, which of the following statements is true—

  1. C1 —C2 bondis a sp3-sp3 tr – bond
  2. C2— C3 bond is a sp3-sp2 or- bond
  3. C1—C2 bond is a sp3-sp3 r-bond
  4. C1—C2 bond is a sp2-sp2 cr-bond

Answer: 3. C1—C2 bond is a sp3-sp2 r-bond

Chemical Bonding And Molecular In 2- Butene

Question 7. The paramagnetic behavior of B2 Is due to the presence of—

  1. 2 unpaired electrons πb MO
  2. 2 unpaired electrons in π* MO
  3. 2 unpaired electrons in σ* MO
  4. 2 unpaired electrons in σb MO

Answer: 1. The electronic configuration of B2 (10 electrons) is \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p x}\right)^1\left(\pi_{2 p y}\right)^1\)

Since the molecule contains two unpaired electrons in πbMO, it is paramagnetic.

Question 8. The state of hybridization of the central atom and the number of lone pairs over the central atom in POCl3 are-

  1. sp,O
  2. sp2O
  3. sp3,O
  4. dsp2,1

Answer: 3. The Geometrical shape of the POC13 molecule is tetrahedral. In POC13 central P-atom is sp3 hybridised and has no lone pair.

Question 9. CO is practically non-polar since—

  1. The cr -electron drift from c to o is almost nullified by the n -electron drift from 0 to c
  2. The tr -electron drift from o to c is almost nullified by the 7r -electron drift from c to o
  3. The bond moment is low
  4. There is a triple bond between c and O

Answer: 1. The cr -electron drift from c to o is almost nullified by the n -electron drift from 0 to c

Chemical Bonding And Molecular Co is practically Non- polar since

Oxygen donates an unshared pair of electrons to carbon and helps it to complete its octet by forming a dative n bond with it. As a result, a much stronger n -n-moment acts from oxygen to a carbon atom, and this moment is almost canceled by the cr -moment and the weak n -moment acting in the opposite direction. Hence, the polarity of the molecule is verylow.

Question 10. The increasing order of the O —N —0 bond angle in the species NO2, NO+2, and NO2is—

  1. \(\mathrm{NO}_2^{+}<\mathrm{NO}_2<\mathrm{NO}_2^{-}\)
  2. \(\mathrm{NO}_2<\mathrm{NO}_2^{-}<\mathrm{NO}_2^{+}\)
  3. \(\mathrm{NO}_2^{+}<\mathrm{NO}_2^{-}<\mathrm{NO}_2\)
  4. \(\mathrm{NO}_2<\mathrm{NO}_2^{+}<\mathrm{NO}_2^{-}\)

Answer: None; the correct order is \(\mathrm{NO}_2<\mathrm{NO}_2^{-}<\mathrm{NO}_2^{+}\)

Chemical Bonding And Molecular Oxygen Donates An Unshared Pair Of Electron To Carbon

Question 11. The ground state electronic configuration of the CO molecule is-

  1. \(1 \sigma^2 2 \sigma^2 1 \pi^2 3 \sigma^2\)
  2. \(1 \sigma^2 2 \sigma^2 3 \sigma^2 1 \pi^2 2 \pi^2\)
  3. \(1 \sigma^2 2 \sigma^2 1 \pi^2 3 \sigma^2 2 \pi^2\)
  4. \(1 \sigma^2 1 \pi^4 2 \sigma^2 3 \sigma^2\)

Answer: 4. \(1 \sigma^2 1 \pi^4 2 \sigma^2 3 \sigma^2\)

The ground state outer electronic configuration of the CO molecule is \(1 \sigma^2 1 \pi^4 2 \sigma^2 3 \sigma^2\)

Question 12. In diborane, the number of electrons that account for bonding die bridges is

  1. Six
  2. Two
  3. Eight
  4. Four

Answer: 4. Four

Chemical Bonding And Molecular In Diborance, The Number Of Electrons

In diborane, each bridging B——H——B bond is formed by two electrons. Hence, four electrons account for bonding the bridges.

Question 13. In O2 and H2O2, the O — O bond lengths are 1.2lA and 1.48A respectively. In ozone, the average O —O bond length is

  1. 1.28A
  2. 1.18A
  3. 1.44A
  4. 1.526A

Answer: 1. Bond length is nearly average of O—O length in

Chemical Bonding And Molecular Bond Length Is Nearly Avrage

Question 14. In SOC12, the Cl—S—Cl and Cl—S—O bond angles are—

  1. 130°, 115°
  2. 106°, 96°
  3. 107°, 108°
  4. 96°, 106°

Answer: 4. 96°, 106°

IN SOC12, the Cl —S —Cl bond angle is 96° and the Cl — S —O bond angle is 106°, since multiple bonds create more repulsions than single bonds.

Question 15. The structure of XeF6 is experimentally determined to be a distorted octahedron. Its structure according to VSEPR theory is—

  1. Octahedron
  2. Trigonal bipyramid
  3. Pentagonal bipyramid
  4. Tetragonal bipyramid

Answer: 3. Pentagonal bipyramid

In XeF6, Xe is surrounded by 6 bond pairs and one lone pair. So, according to VSPER theory, the geometry (geometry of electron pairs) is pentagonal bipyramid.

Question 16. In the case of heteronuclear diatomics of the type AB, where A is more electronegative than B, bonding MO resembles the character of A more than that of B. The statement—

  1. Is False
  2. Is True
  3. Cannot Be Evaluated Since the Data Is Not Sufficient
  4. Is True Only FOr Certaqin Systems

Answer: 2. Cannot Be Evaluated Since Data Is Not Sufficient

Chemical Bonding And Molecular More Electronegative

As A is more electronegative, there is less energy difference between the atomic orbital of A and bonding M.O. Hence, bonding M.O. resembles A more closely.

Question 17. The bond angle in NF3 (102.3°) is smaller than NH3 (107.2°). This is because of—

  1. Large size off compared to
  2. The large size compared to
  3. Opposite polarity of n in the two molecules
  4. Small size compared to a ton

Answer: 3. In NF3 molecules, the N—F bond pair is drawn

more towards the more electronegative F-atom. But in the NH3 molecule, the N—H bond pair is drawn more towards the more electronegative N-atom. Therefore, the extent of bp-bp repulsion in NH3 is more than that in NF3. As a consequence, the bond angle in NH3(107.2°) is greater than that of NF3(102.3°).

Question 18. The compound that will have a permanent dipole moment among the following is

Chemical Bonding And Molecular Permant Dipole Moment Means A zero value of dipole moment..

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 1. Permanent dipole moment means a non-zero value of dipole moment. So, only compound (1) has a permanent dipole moment.

Chemical Bonding And Molecular Permant Dipole Moment Means A zero value of dipole moment

Question 19. Among the following structures, the one which is not a resonating structure of others is—

Chemical Bonding And Molecular The Following Structures The One Which Is Not A Resonating Structture Of Other Is

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 

Chemical Bonding And Molecular The One Which Is Not A Resonating Structure Of Other Is

Structure (4) is not a resonance structure because it involves shifting a pair of electrons as well as an H-atom

Question 20. The correct order of decreasing length ofthe bond as indicated by the arrow in the following structures is—

Chemical Bonding And Molecular The Order Of Decreasing Length Of The Bond

  1. 1>2>3
  2. 2>13
  3. 3>2>1
  4. 1>3>2

Answer: 3. In general, C=C and C—C bond lengths are respectively 1.33A and 1.54A

Chemical Bonding And Molecular In General C=C ANd C-C Bond Lenghts

In structure I, n -electrons are delocalized over Ci — C2 and C2— C3 bonds. In structure II a pair of n electrons are delocalised Over C5-C6, C6-C7C6-C8

Question 21. The correct order of decreasing H—C—H angle in the following molecules is

Chemical Bonding And Molecular The Correct Order Of Decreasing g H-C-H Angle In The Foloowing Molecules Is

  1. 1>2>3
  2. 2>1>3
  3. 3>2>1
  4. 1>3>2

Answer: Overlapping is maximum when orbitals overlap “endon” i.e., via a -bonding, n -bonds overlap laterally. The overlap in cyclopropane is neither end-on nor lateral but in between. So, it is intermediate between cr -and n bonding.

Chemical Bonding And Molecular The Correct Order Of Decreasing g H-C-H Angle In The Foloowing Molecules Is,.

So, in order of decreasing H —C —H angle: 2 >1 > 3

Question 22. The correct order of decreasing length ofthe bond as indicated by the arrow in the following structures is—

Chemical Bonding And Molecular The Number Of Sp Hybridised Carbon.

  1. 1>2>3
  2. 2>1>3
  3. 3>2>1
  4. 1>3>2

Answer: 3. 3>2>1

Chemical Bonding And Molecular The Number Of Sp Hybridised Carbon

Question 23. Thenumberoflone pairs of electrons on the central atoms of H2O, SnCl2 PCl3, and XeF2 respectively, are

  1. 2,1,1,3
  2. 2,2,1,3
  3. 3,1,1,2
  4. 2,1,2,3

Answer: 4. Number of lone pairs of electrons present in the hybrid orbital, L = H- X- D [Where H: no. of orbitals involved in the hybridization, X: no. of monovalent atoms surrounding the central atom, D: no. of bivalent atoms attached to the central atom.

Question 24. The number of car and n -bonds present between the two carbon atoms of calcium carbide are respectively

  1. 1σ, 1π- bond
  2. 1σ, 2π- bond
  3. 2σ, 1π- bond
  4. \(1 \sigma, 1 \frac{1}{2} \pi \text {-bond }\)

Answer: 2. 1σ, 2π- bond

In calcium carbide, two carbon atoms are bonded by a triple bond. Thus between two carbon atoms, 1 cr, and 2/r bonds are present.

Question 25. Which of the following molecules have a shape like CO

  1. HgCl2
  2. SnCl2
  3. C2H2
  4. NO2

Answer: 1. HgCl2

Chemical Bonding And Molecular Structure The Following Molecules Have Shape lIke CO2

Question 26. The ground state magnetic property of B2 and C2 molecules will be—

  1. B, paramagnetic and C2 diamagnetic
  2. B, diamagnetic C2 paramagnetic
  3. Botii are diamagnetic
  4. Both are paramagnetic

Answer: 1. B, paramagnetic and C2 diamagnetic

MO electronic configuration of B2

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^1\left(\pi_{2 p_y}\right)^1\)

Mo electronic configuration of C2

⇒ \(\operatorname{KK}\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\)

In BO, there are unpaired electrons present which is paramagnetic. But in C9 there is no impaired electron and henceitis diamagnetic.

Question 27. The shape of XeFg- is—

  1. Square pyramidal
  2. Triangularbipyramidal
  3. Planar
  4. Pentagonal bipramidal

Answer: 3. Planar

⇒ \(\text { Here, } \mathrm{H}=\frac{1}{2}(8+5-0+1)=7\)

Central atom Xe: sp3d2 -hybridized

No. of lone pair of electronic Xe, L = (7-5-0) = 2

therefore XeF5 ionic planar.

Question 28. Which statements are correct for the peroxide ion—

  1. It has five filled anti-bonding molecular orbitals
  2. It is diamagnetic
  3. It has bond order one
  4. It is isoelectronic with neon

Answer: 1. MO electronic configuration of 0|~ (peroxide ion):

\(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\) \(\left(\pi_{2 p x}\right)^2\left(\pi_{2 p y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2\)

Bond Order \(=\frac{N_b-N_a}{2}=\frac{10-8}{2}=1\)

Question 29. Which ofthe following has the strongest H-bond

  1. H-O…shape
  2. S-H….O
  3. F-H….F
  4. F-H….O

Answer: 3. F-H….F

Since fluorine is the most electronegative element F —H bond is more polar which forms the strongest H-bonding [F—H-—F] among the given compounds.

Question 30. B cannot form which ofthe following anions

  1. \(\mathrm{BF}_6^{3-}\)
  2. BH-4
  3. B(OH)-4
  4. BO-2

Answer: 1. \(\mathrm{BF}_6^{3-}\)

The stability of hydrides of group-15 decreases from NH3 to BiH3 due to an increase in the size of the central atom.

Question 31. Which of the following statements is wrong—

  1. Nitrogen cannot form an-inbound
  2. Single n-nbond is weaker than the single p-p bond
  3. N2O4 has two resonance structures
  4. The stability of hydrides increases from nh3 to bih3 due to the increase in size ofthe central atom.

Answer: 4. The stability of hydrides increases from NH3 to BiH3 due to the increase in size ofthe central atom.

Question 32. The Square Of IF7 is

  1. Square pyramid
  2. Trigonal bipyramid
  3. Octahedral
  4. Pentagonal bipyramid

Answer: 4. Pentagonal bipyramid

In IF7, the hybridization of the central atom is sp3d3 and its structure is a pentagonal bipyramid.

Question 33. The hybridization orbitals of N-atoinin NO-3, NO+2, and NH+4 are respectively—

  1. sp, sp2, sp3
  2. Sp2,sp,sp3
  3. Sp,sp3,s2
  4. sp2, sp3 sp

Answer: 2. Sp2,sp,sp3

Chemical Bonding And Molecular Structure The Hybridisation of orbitals of N-atom

Question 34. Among die following, die maximum covalent character is shown by

  1. FeCl2
  2. SnCl2
  3. AlCl3
  4. MgCl2

Answer: 3. AlCl3

The ionic potential (0) of the cations increases with the increase in cationic charge and decrease in cationic radii. Thus, the resulting compound possesses a more covalent
character. So, A1C13 exhibits maximum covalency.

Question 35. Iron exhibits +2 and +3 oxidation states. Which of the following statements about incorrect—

  1. Ferrous compounds are relatively more ionic than ferric compounds
  2. Ferrous compounds are less volatile than the corresponding ferric compounds
  3. Ferrous compounds are more easily hydrolyzed than the corresponding ferric compounds
  4. Ferrous oxide is more basic than ferric oxide

Answer: 3. Ferrous compounds are more easily hydrolyzed than the corresponding ferric compounds

The tendency of hydration increases with a decrease in the size of a cation. Ferrous ions are larger than ferric ions. Consequently, the ferric ion will be more easily hydrolyzed than the ferrous ion.

Question 36. Ortho-nitrophenol is less soluble in water than p – and m nitrophenols because—

  1. O-nitrophenol shows intramolecular-bonding
  2. O-nitrophenol shows intermolecular-bonding
  3. The melting point of o-nitrophenol is less than that of mand p-isomers
  4. O-nitrophenol is more volatile than those of m-and prisoners

Answer: 1. o-nitrophenol shows intramolecular H-bonding

In o-nitrophenol, —OH & —N02 groups are situated at two adjacent carbon atoms of the ring and involved in intramolecular bonding. So, o-nitrophenol is less soluble in water than p- and m- m-nitrophenol.

Question 37. The molecule having the smallest bond angle is

  1. AsCl3
  2. SbCl3
  3. PCl3
  4. NCL3

Answer: 2. SbCl3

The bond angle of a molecule increases with the increases in electronegativity or with the decrease in size of the central atom. Hence, the correct order of bond angle is: SbCl3 < ASC13 < PC13 < NCl3

Question 38. In which of the following pairs the two species are not isostructural

  1. PCI+6 and SiCl4
  2. PF5 and BrF5
  3. AlFg- and SF6
  4. CO6– and NO-3

Answer: 2. PF5 and BrF5

In molecule PF5, \(H=\frac{1}{2}\) [5 + 5- 0 + 0] = 5. Hybridisation of central atom(P): sp3d. Number of p lone pairs in the central atom (P): 0. Shape of the molecule: Trigonal bipyramidal.

For BrF5, Hybridisation of central atom (Br): sp3d2. Number of loner pairs in the central atom (Br): 1. Shape ofthe molecule: square pyramidal.

Question 39. The stability of the species Li2, Li2, and Li2 increases in the order of

  1. Li2<Li+2<Li-2
  2. Li22< Li+2 < Li2
  3. Li2<Li2<Li+2
  4. Li2<Li2<Li2

Answer: 2. Li2< Li-2 < Li2

⇒ \(\mathrm{Li}_2:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\)

⇒ \(\begin{aligned}
& \mathrm{Li}_2^{+}:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^1 \\
& \mathrm{Li}_2^{-}:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^1
\end{aligned}\)

\(\begin{array}{|c|c|c|c|}
\hline \mathbf{B . 0} & \mathbf{L i}_{\mathbf{2}} & \mathbf{L i}_{\mathbf{2}}^{+} & \mathbf{L i}_{\mathbf{2}}^{-} \\
\hline \frac{N_b-N_a}{2} & \frac{2-0}{2}=1 & \frac{1-0}{2}=0.5 & \frac{2-1}{2}=0.5 \\
\hline
\end{array}\)

Question 40. In which ofthe following pairs of molecules/ions, both the species are not likely to exist—

  1. \(\mathrm{H}_2^{+}, \mathrm{He}_2^{2-}\)
  2. \(\mathrm{H}_2^{-}, \mathrm{He}_2^{2-}\)
  3. \(\mathrm{H}_2^{2+}, \mathrm{He}_2\)
  4. \(\mathrm{H}_2^{-}, \mathrm{He}_2^{2+}\)

Answer: 3. \(\mathrm{H}_2^{2+}, \mathrm{He}_2\)

⇒ \(\mathrm{H}_2^{2+}:\left(\sigma_{1 s}\right)^0\)

⇒ \(\mathrm{He}_2:\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\)

\(\begin{array}{|c|c|c|}
\hline \text { B.o. } & \mathbf{H}_2^{2+} & \mathbf{H e}_{\mathbf{2}} \\
\hline \frac{N_b-N_a}{2} & 0 & \frac{2-2}{2}=0 \\
\hline
\end{array}\)

Question 41. Which one of the following molecules is expected to exhibit diamagnetic behavior—

  1. C2
  2. N2
  3. O2
  4. S2

Answer: 1. C2

Question 42. The correct statement for the molecule, Csl3, is—

  1. It contains Cs+, I and lattice I2 molecule
  2. It is a covalent molecule
  3. It contains Cs+ and I2 ions
  4. It contains Cs3+ and I ions

Answer: 3. It contains Cs+ and I-3 ions

Cs cannot show a +3 oxidation state. So, Csl3 is formulated as Cs+ and I3 ions. It is a typical ionic compound.

Question 43. For which of the following molecules is significant μ≠ 0-

Chemical Bonding And Molecular Structure The Following Moleucles Signifiancant

  1. 3 and 4
  2. only 1
  3. 1 and 2
  4. only 3

Answer: 1. 3 and 4

Question 44. Which of the following alkaline earth metal sulfates has its hydration enthalpy greater than its lattice enthalpy—

  1. BaSO4
  2. SRSO4
  3. CaSO4
  4. BeSO4

Answer: 1. BaSO4

In the NO4 ion, the N atom is sp -hybridized. O=N=0

Question 45. In which of the following molecule Orion the hybridization state of the-atom is sp —

  1. NO+2
  2. NO-2
  3. NO-3
  4. NO2

Answer: 4. NO2

There is no unpaired electron in the MO electronic configuration of the CO molecule. Thus, it is not paramagnetic.

Question 46. Which one of the following is not paramagnetic-

  1. O2
  2. B2
  3. NO
  4. CO

Answer: 4. CO

KC1 is an electrovalent compound, it exists as K+ and Cl- ions

Question 47. The total number of one pair of electrons I-3 ion is

  1. 9
  2. 3
  3. 13
  4. 6

Answer: 1. 9

Question 48. Which ofthe following compounds contain(s) no covalent bond(s)—KC1, PH3, O2, B2H6, H2SO4

  1. KCL
  2. KCl, B2H6
  3. KClB2H6,PH3
  4. KCl, H2SO4

Answer: 1. KCL

KC1 is an electrovalent compound, it exists as K+ and Cl- ions

Question 49. According to molecular orbital theory, which of the following will not be a viable molecule—

  1. H-2
  2. H2-2
  3. He2+2
  4. He+2

Answer: 2. MO electronic configuration of H2-2 \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}\right)^2\)

Therefore Bond Order \(=\frac{2-2}{2}=0\)

Question 50. In which of the following pairs of molecules/ions, do the central atoms have sp2 hybridization—

  1. NO-2 And NH3
  2. BF3 And NO-2
  3. NH-2 And H2O
  4. BF3 And NH-2

Answer: 2. BF3 And NO-2

\(\begin{array}{|c|c|c|}
\hline \begin{array}{c}
\text { Molecule / } \\
\text { ion }
\end{array} & \text { Value of } \mathbf{H} & \begin{array}{c}
\text { Type of } \\
\text { hybridisation }
\end{array} \\
\hline \mathrm{BF}_3 & \mathrm{H}=\frac{1}{2}(3+3-0+0) & s p^2 \\
& =3 & \\
\hline \mathrm{NO}_2^{-} & \mathrm{H}=\frac{1}{2}(5+0-0+1) & s p^2 \\
\hline
\end{array}\)

Question 51. Considering the state of hybridization ofC-atoms, find out the molecule among the following which is linear-

  1. CH3—CH2—CH2—CH3
  2. CH3—CH=CH—CH3
  3. CH3—C=C—CH3
  4. CH2=CH—CH2—C=CH

Answer: 3. CH3—C=C—CH3

In the case of sp3, sp2, and sp hybridized carbons, the bond angle is 109°28′; 120° and 180° respectively. So, only image- is linear (excluding H-atoms)

Question 52. Which ofthe following structures is the most preferred and
hence of the lowest energy for SO3-

Chemical Bonding And Molecular Structure The Following Structures Is The Most Prefered and hence of lowest energy for SO3

Answer: 4. Has a maximum number of covalent bonds and hence is of the lowest energy.

Question 53. The correct order of increasing bond length of C—H C—O, C—C, and C=C is

  1. C—H < C=C < C—O < C—C
  2. C—C < C=C < C—0 < C—H
  3. C—0<C—H<C—C<C=C
  4. C—H<C—0<C—C<C=C

Answer: 4. C—H<C—0<C—C<C=C

⇒ \(\begin{aligned}
& \mathrm{C}-\mathrm{H}<\mathrm{C}=\mathrm{C}<\mathrm{C}-\mathrm{O}<\mathrm{C}-\mathrm{C} \\
& 107 \mathrm{pm} \quad 134 \mathrm{pm} \quad 141 \mathrm{pm} \quad 154 \mathrm{pm}
\end{aligned}\)

Question 54. Which of the two ions from the list given below have the geometry that is explained by the same hybridization of orbitals, NO2, NO3, NH2, NH4, SCN

  1. NO2 and NO3
  2. NH+4 and NO3
  3. SCN and NH2
  4. NO2 and NH2

Answer: 1. Increasing the order of bond length is

⇒ \(\begin{array}{|c|c|c|c|c|c|}
\hline \text { Ions } & \mathrm{NO}_3^{-} & \mathbf{N O}_2^{-} & \mathbf{N H}_2^{-} & \mathbf{N H}_4^{+} & \mathbf{S C N}^{-} \\
\hline \text { Hybridisation } & s p^2 & s p^2 & s p^3 & s p^3 & s p \\
\hline
\end{array}\)

Question 55. Which has the minimum bond length—

  1. O+2
  2. O2
  3. O22-
  4. O2

Answer: 1. O+2

⇒ \(\mathrm{O}_2^*: \mathrm{KK}\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_1}\right)^2\left(\pi_{2 p}\right)^2\left(\pi_{2 p}^*\right)^1\)

⇒ \(\begin{aligned}
& \mathrm{O}_2^{-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^1 \\
& \mathrm{O}_2^{2-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x^*}^*\right)^2\left(\pi_{2 p_y}^*\right)^2
\end{aligned}\)

⇒ \(\mathrm{O}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^1\)

Question 56. Four diatomic species are listed below. Identify the correct order in which the bond order is increasing

  1. \(\mathrm{NO}<\mathrm{O}_2^{-}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
  2. \(\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
  3. \(\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}\)
  4. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

Answer: 4. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

NO(7=8=15)

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\)

Question 57. During the change of O2 to O2 ion, the electron adds on which one ofthe following orbitals—

  1. π* -orbitals
  2. π – orbitals
  3. σ* orbital
  4. σ- orbital

Answer: 1. π* -orbitals

⇒ \(\mathrm{O}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^1\)

⇒ \(\mathrm{O}_2^{-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p y}^*\right)^1\)

Thus, the electron goes into the π*-orbital.

Question 58. The pair of species with the same bond order is

  1. \(\mathrm{O}_2^{2-}\)
  2. \(\mathrm{O}_2^{+}, \mathrm{NO}^{+}\)
  3. NO, CO
  4. N2, O2

Answer: 1. \(\mathrm{O}_2^{2-}\)

⇒ \(\mathrm{O}_2^{2-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2\)

⇒ \(\mathrm{B}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^1\left(\pi_{2 p_y}\right)^1\)

\(\begin{array}{|c|c|c|}
\hline \text { Bond order } & \mathbf{O}_2^{2-} & \mathbf{B}_2 \\
\hline \frac{N_b-N_a}{2} & \frac{8-6}{2}=1 & \frac{4-2}{2}=1 \\
\hline
\end{array}\)

Note: B.O. of O2 = 2.5, N0+ = 3, NO = 2.5 CO = 3, N2 = 3 and O2 = 2]

Question 59. Which of the following species contains three bond pairs and one lone pair around the central atom—

  1. H2O
  2. BF2
  3. NH-2
  4. PCl2

Answer: 4. PCl3

Chemical Bonding And Molecular Structure The Following The Species Contains Threes Bond Pairs And One Lone Pair Around

Question 60. Bond order of 1.5 is shown by—

  1. O+2
  2. O-2
  3. O2-2
  4. O2

Answer: 2. O-2

⇒ \(\mathrm{O}_2^{+}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\)

⇒ \(\begin{aligned}
& \mathrm{O}_2^{+}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^1 \\
& \mathrm{O}_2^{2-}: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x^*}^*\right)^2\left(\pi_{2 p_y}^*\right)^2 \\
& \mathrm{O}_2: K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y^*}\right)^1
\end{aligned}\)

\(\begin{array}{|c|c|c|c|c|}
\hline \begin{array}{c}
\text { Bond } \\
\text { order }
\end{array} & \mathbf{O}_2^{+} & \mathbf{O}_2^{-} & \mathbf{0}_2^{2-} & \mathbf{O}_2 \\
\hline \frac{N_b-N_a}{2} & \frac{8-3}{2}=2.5 & \frac{8-5}{2}=1.5 & \frac{8-6}{2}=1 & \frac{8-4}{2}=2 \\
\hline
\end{array}\)

Question 61. The following pairs are isostructural-

  1. BC13 and BrCl3
  2. NH3 and NO3-
  3. NF3 and BF3
  4. BF4 and NH+4

Answer: 4. BF2 and NH+4

If some bond pairs and lone pairs are the same for the given pairs, they are isostructural.

Chemical Bonding And Molecular Structure If Number Of Bond Pairs And LOne PAirs Are Same For The Given Pairs They Are Isostructural

Chemical Bonding And Molecular Structure If Number Of Bond Pairs And LOne PAirs Are Same For The Given Pairs They Are Isostructural.

Question 62. Which contains no n -bond—

  1. SO2
  2. NO2
  3. CO2
  4. H2O

Answer: 4. H2O

Chemical Bonding And Molecular Structure The contains no pi bond

Question 63. Which ofthe following lanthanoid ions is diamagnetic (Atnos. Ce=58, Sm=62,Eu=63, Yb70)-Chemical Bonding And Molecular Structure If Number Of Bond Pairs And LOne PAirs Are Same For The Given Pairs They Are Isostructural.

  1. EU2+
  2. Yb2+
  3. Ce2+
  4. Sm2+

Answer: 4. Sm2+

⇒ \(\begin{aligned}
& ; \mathrm{Sm}^{2+}(\mathrm{Z}=62):[\mathrm{Xe}] 4 f^6 \\
& \mathrm{Yb}^{2+}(\mathrm{Z}=70):[\mathrm{Xe}] 4 f^{14} \\
& \mathrm{Ce}^{2+}(\mathrm{Z}=58):[\mathrm{Xe}] 4 f^1 5 d^1 \\
& \mathrm{Eu}^{2+}(\mathrm{Z}=63):[\mathrm{Xe}] 4 f^7
\end{aligned}\)

Question 64. Which of the following is paramagnetic—

CN

NO+

CO

O2

Answer: 4. O2

MO electronic configuration of O-2(17):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi 2_{p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^1\)

Owing to the presence of one unpaired electron, it is paramagnetic in nature.

Question 65. Which of the following is a polar molecule—

  1. SiF4
  2. XeF4
  3. BF3
  4. SF4

Answer: 4. SF4

SF4 has sp3d -hybridization and see-saw shape with 4 bond pairs and 1 lone pair and resultant u=0.

Question 66. XeF2 is isostructural with-

  1. SbCl3
  2. BaCL2
  3. TeF2
  4. ICI-2

Answer: 4. ICI-2

Question 67. Which species has a plane triangular shape-

  1. N3
  2. NO-3
  3. NO2
  4. CO2

Answer: 2. NO-3

Question 68. Which has the maximum dipole moment—

  1. CO2
  2. CH4
  3. NH3
  4. NF3

Answer: 3. NH3

Chemical Bonding And Molecular Structure The Maximum Dipole moment

In NH3, H is less electronegative than N and hence dipole moment of each N —H bond is towards N and creates a high net dipole moment.

Question 69. The formation ofthe oxide ion O2-(g), form oxygen atom requires first an exothermic and then an endothermic step as shown below

⇒ \(\begin{aligned}
& \mathrm{O}(\mathrm{g})+e^{-} \rightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta H_f^0=-141 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{O}^{-}(\mathrm{g})+e^{-} \rightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta H_f^0=+780 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Thus, the process of formation of O2- in the gas phase is unfavorable even though O2- is isoelectronic with neon. It is because—

  1. Electron repulsion outweighs the stability gained by achieving noble gas configuration
  2. O-ion has a comparatively smaller size than o-atom
  3. Oxygen is more electronegative
  4. The addition of electronic o results in large-size ion

Answer: 1. Electron repulsion outweighs the stability gained by achieving noble gas configuration

Electron repulsion predominates over the stability gained by achieving noble gas configuration. Hence, the formation of O2- in the gas phase is unfavorable.

Question 70. In which of the following pairs, both the species are not isostructural—

  1. SiCl4, Pcl+4
  2. Diamond, siC
  3. NH3,PH3
  4. XeF4, XeO4

Answer: 4. XeF4, XeO4

Chemical Bonding And Molecular Structure The Follwing Pairss

Diamond and silicon carbide (SiC), are both isostructural because their central atom is sp3 hybridized and both have tetrahedral arrangements.

Chemical Bonding And Molecular Structure The Follwing Pairss.

Both NH3 and PH3 have pyramidal geometry.

XeF4 has sp3d2 hybridisation while XeO4 has sp3 hybridisation.

Chemical Bonding And Molecular Structure The Follwing Pairss..

Hence, XeF4 and XeO4 are notisostructural.

Question 71. Decreasing order of stability is-

  1. \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)
  2. \(\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}>\mathrm{O}_2>\mathrm{O}_2^{+}\)
  3. \(\mathrm{O}_2>\mathrm{O}_2^{+}>\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}\)
  4. \(\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}>\mathrm{O}_2^{+}>\mathrm{O}_2\)

Answer: 1. \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

Order of stability ∞ bond order.

therefore The order of stability ofthe given species,

⇒ \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

Bond order: 2.5 2 1.5 1

Question 72. Which one of the following compounds §hows the presence of intramolecular hydrogen bond-

  1. HCN
  2. Cellulose
  3. Conc.Acetic Acid
  4. H2O2

Answer: 2. Cellulose

Only cellulose can perform intramolecular bonding whereas the other compounds can perform intermolecular-bonding only.

Question 73. Which of the following pairs of ions are isoelectronic and isostructural-

  1. \(\mathrm{ClO}_3^{-}, \mathrm{SO}_3^{2-}\)
  2. \(\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)
  3. \(\mathrm{ClO}_3^{-}, \mathrm{CO}_3^{2-}\)
  4. \(\mathrm{SO}_3^{2-}, \mathrm{CO}_3^{2-}\)

Answer: 2. \(\mathrm{CO}_3^{2-}, \mathrm{NO}_3^{-}\)

Both CO23 and NO3 have a total of 32 electrons and both are triangular planar shape

Question 74. Among the following, which one is a wrong statement—

  1. Pn-dn bonds are present in SO2
  2. SeF4 and CH4 have the same shape
  3. 1+3 has bent geometry
  4. PH3 and BiCl5 do not exist

Answer: 2. SeF4 and CH4 have same shape

The shape of CH4(sp3) is regular tetrahedron. However, in SeF4, the Se atom is sp3d-hybridized -hybridised and the presence of two lone pairs makes the molecule see-saw shaped.

Question 75. The hybridizations of atomic orbitals of nitrogen in NO+2 NO-3 and NH+4 respectively are-

  1. Sp2, sp3 and sp
  2. sp, sp2 and sp3
  3. sp2, sp and sp3
  4. sp, sp3 and sp2

Answer: 2. sp, sp2 and sp3

⇒ \(\mathrm{NO}_2^{\oplus}(s p) \text {-linear; } \mathrm{NO}_3^{-}\left(s p^2\right) \rightarrow \text { trigonal planar }\mathrm{NH}_4^{\oplus}\left(s p^3\right) \text {-tetrahedral }\)

Question 76. Consider the molecules CH, NH3, and H20. Which of the given statements is false—

  1. The h—c—h bond angle in ch4, the h—n—h bond angle in nh3 & the h—o —h bond angle in h,0 are all greater than 90°
  2. The h—o —h bond angle in h20 is larger than the h—c—h bond angle in ch4
  3. The h—o —h bond anglein h20 is smaller than the h —n—h bond anglein nh3
  4. The h—c—h bond angle in ch4 is larger than the h—n—h bond anglein nh

Answer: 2. The h—o —h bond angle in h20 is larger than the h—c—h bond angle in ch4

Question 77. Predict the correct order among the following—

  1. Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair
  2. Lone pair-lone pair > bond pair-bond pair > lone pairlonepair
  3. Bond pair-bond pair > lone pair-bond pair > lone pairlonepair
  4. Lone pair-bond pair > bond pair-bond pair > lone pairlonepair

Answer: 1. Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair

According to VSEPR theory, Ip — Ip repulsion > Ip- bp repulsion > bp-bp repulsion.

Question 78. The charge-forming electron pair in the carbonation CH3C=C exists in

  1. sp -orbital
  2. 2p-orbital
  3. sp3 -orbital
  4. sp2 -orbital

Answer: 1. sp -orbital

CH3CHC-, triply bonded carbon atoms are sp hybridized. Thus forming an electron pair is in sp orbital

Question 79. Match the compound given in column 1 with the hybridization and shape given in column 2 and mark the correct option:

Chemical Bonding And Molecular Structure Match The Column 1 and 2 The Compound

Code: 1 2 3 4

  1. 4-1- 2- 3
  2. 1- 3- 4- 2
  3. 1- 2- 4-3
  4. 4- 3- 1 – 2

Answer: 2. 1- 3- 4- 2

Question 80. Which of the following pairs of species have the bond order

  1. O2, NO+
  2. CN, CO
  3. N2,O2
  4. CO, NO

Answer: 2. Total no. of electrons present in CN = 14

  • Total no. electrons present in CO = 14
  • MO electronic configuration of CO

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 p_z}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\)

MO electronic Configuration of CN-

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\sigma_{2 p_z}\right)^2\)

⇒ \(\text { Bond order of } \mathrm{CO}=\frac{1}{2}(8-2)=3\)

⇒ \(\text { Bond order of } \mathrm{CN}^{-}=\frac{1}{2}(8-2)=3\)

Question 81. The species, having angles of 120° is—

  1. ClF3
  2. NCL3
  3. BCl3
  4. PH3

Answer: 3. BCl3

Question 82. Match the interhalogen compounds of column I with the geometry in column 2 and assign the correct code:

Chemical Bonding And Molecular Structure Match The Column 1 and 2

Code: (1) (2) (3) (4)

  1. 3-1- 4- 2
  2. 5- 4- 3- 2
  3. 4- 3 – 2- 1
  4. 3- 4 -1 -2

Answer: 1. 3-1- 4- 2

  • XX ⇒ linear (sp3d)
  • XX3 ⇒ T-shape (sp3d)
  • XXg ⇒ square pyramidal (sp3d2)
  • XXy ⇒ pentagonal bipyramidal (sp3d3)

Question 83. Consider the following species: CN+, CN-, NO, and CN Which one of these will have the highest bond order—

  1. CN
  2. NO
  3. CN+
  4. CN

Answer: 4. CN

  • \(\text { B.O. of } \mathrm{NO}=\frac{10-5}{2}=2.5\)
  • \(\text { B.O. of } \mathrm{CN}^{-}=\frac{10-4}{2}=3 \text {; }\)
  • \(\text { B.O. of } \mathrm{CN}=\frac{9-4}{2}=2.5 \text {; }\)
  • \(\text { B.O. of } \mathrm{CN}^{+}=\frac{8-4}{2}=2\)

Question 84. In the structure of C1F3, the number of lone pairs of electrons on central atom ‘Cl’ is-

  1. Three
  2. One
  3. Four
  4. Two

Answer: 4. Two

Chemical Bonding And Molecular Structure The Structure Of CLF3

Question 85. The decreasing order of bond angle is—

  1. BeCl2 > NO2 > SO2
  2. BeCl2 > SO2 > NO2
  3. SO2 > BeCl2 > NO2
  4. SO2>NO2>BeCl2

Answer: 1. BeCl2 > NO2 > SO2

Compound \(\begin{aligned}
& \mathrm{BeCl}_2>\mathrm{NO}_2>\mathrm{SO}_2 \\
& 180^{\circ} \quad 132^{\circ} \quad 119.5^{\circ} \\
&
\end{aligned}\)

Question 86. The dipole moment is minium in—

  1. NH3
  2. NF3
  3. SO2
  4. BF3

Answer: 4. BF3

BF3 has zero dipole moment.

Question 87. In BF3, the B —F bond length is 1.30 A when BF3 is allowed to be treated with mMe3 N, it forms an adduct, Me3 N→BF3, and the bond length of —F in the adduct is-

  1. Greater than 1.30A
  2. Smaller than 1.30A
  3. Equal to 1.30A
  4. None of these

Answer: 1. Greater than 1.30A

In BF3, there is backbonding in between fluorine and boron due to the presence of -orbital in boron.

Chemical Bonding And Molecular Structure Back Bondingback bonding imparts double-bond characteristics

As BF3 forms adduct, the back bonding Is no longer present and thus double bond characteristic disappears. Hence, the bond becomes a bit longer than earlier (1.30A).

Question 88. The total number of antibonding electrons present in O2 will be

  1. 6
  2. 8
  3. 4
  4. 2

Answer: 1. MO electronic configuration of O2

\(\begin{aligned}
& \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2 \\
& \left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^1
\end{aligned}\)

Hence, the correct B.O. is O+2 → O-2 → 2-2

Question 89. Which ofthe following represents the correct bond order

  1. \(\mathrm{O}_2^{+}<\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)
  2. \(\mathrm{O}_2^{+}<\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)
  3. \(\mathrm{O}_2^{2-}>\mathrm{O}_2^{+}>\mathrm{O}_2^{-}\)
  4. \(\mathrm{O}_2^{+}>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

Answer: 4. \(\mathrm{O}_2^{+}>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

Question 90. In the O3 molecule, the formal charge on the central O-atom is—

  1. 0
  2. -1
  3. -2
  4. +1

Answer: 4. +1

Lewis gave the structure of the O3 molecule as

Chemical Bonding And Molecular Structure Lewis gave the structure

Using the relation, Formal charge = [Total no, of valence electrons in the free atom] – [Total no. of non-bonding (lone pair) electrons]-\(\frac{1}{2}\)[Total no. of bonding (shared) electrons]

The formal charge on central O -atom i.e., no. 1 = +1

Question 91. Four diatomic species are listed below in different sequences. Which of these represents the correct order of their increasing order

  1. \(\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}<\mathrm{NO}<\mathrm{O}_2^{-}\)
  2. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)
  3. \(\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}<\mathrm{He}_2^{+}\)
  4. \(\mathrm{NO}<\mathrm{C}_2^{2-}<\mathrm{O}_2^{-}<\mathrm{He}_2^{+}\)

Answer: 2. \(\mathrm{He}_2^{+}<\mathrm{O}_2^{-}<\mathrm{NO}<\mathrm{C}_2^{2-}\)

According to molecular orbital theory, the energy level ofthe given molecules are

⇒ \(\mathrm{C}_2^{2-}-K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\sigma_{2 p_z}\right)^2\)

⇒ \(\begin{aligned}
& \text { B.O. }=\frac{1}{2}[10-4]=3 \\
& \mathrm{He}_2^{+}-\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^1
\end{aligned}\)

⇒ \(\begin{aligned}
& \text { B.O. }=\frac{1}{2}[2-1]=\frac{1}{2}=0.5 \\
& \mathrm{O}_2^{-}-K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2 \\
& \quad\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^1
\end{aligned}\)

⇒ \(\begin{aligned}
& \text { B.O. }=\frac{1}{2}[10-7]=1.5 \\
& \text { NO : }-K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2 \\
& \qquad\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\left(\pi_{2 p_y}^*\right)^0
\end{aligned}\)

⇒ \(\text { B.O. }=\frac{1}{2}[8-3]=2.5\)

So, the correct order of their increasing bond order is He+2 < O-2 < NO < C²-2

Question 92. Which of the following molecules has more than one lone pair

  1. SO2
  2. XeF2
  3. SiF4
  4. CH4

Answer: 2. XeF2

Chemical Bonding And Molecular Structure The following molecules has more than one lone pair

Question 93. The ASF5 molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are—

  1. \(d_{x^2-y^2}, d_{z^2}, s, p_x, p_y\)
  2. \(d_{x y}, s, p_x, p_y, p_z\)
  3. \(d_{x^2-y^2}, s, p_x, p_y\)
  4. \(s, p_x, p_y, p_z, d_z^2\)

Answer: 4. \(s, p_x, p_y, p_z, d_z^2\)

AsF5 has sp3d hybridisation. In sp2d hybridization, the dz2 orbital is used along with the ‘s’ and three ‘p’ orbitals to form three equatorial bonds and two equally strong axial bonds for a trigonal bipyramid

Question 94. H2O is polar, whereas BeF2 is not because—

  1. Electronegativity ofF is greater than that of O
  2. H2O involves H-bonding, whereas, BeF2 is a discrete molecule
  3. H2O is angular and BeF2 is linear
  4. H2O is linear and BeF2 is angular

Answer: 3. H2O is angular and BeF2 is linear

Because of the linear shape, dipole moments cancel each other in BeF2 (F—Be—F) and thus, it is non-polar, whereas H2O is V-shaped and hence, it is polar

Chemical Bonding And Molecular Structure Beacuse Of The Linear

Question 95. Which of the following have the same hybridization but are not isostructural

  1. C1F3 and l-3
  2. BrF3 and NH3
  3. CH4 and NH+4
  4. XeO3 and NH3

Answer: 1. C1F3 and l-3

  1. C1F3 (sp3d, T-shape); I3 (sp3d, linear)
  2. BrF3 (sp3d, T-shape); NH3 (sp3, pyramidal)
  3. CH4 (sp3, Tetrahedral); NH3 (sp3, Tetrahedral)
  4. XeO3 (sp3, Pyramidal); NH3 (sp3, Pyramidal)

Question 96. Which of the following pairs have different hybridization and the same shape—

  1. \(\mathrm{NO}_3^{-} \text {and } \mathrm{CO}_3^{2-}\)
  2. SO2 and NH2-
  3. XeF2 and CO2
  4. H2O and NH3

Choose The Correct Option

  1. 1 and 4
  2. 2 and 4
  3. 2 and 3
  4. None of these

Answer: 3. 2 and 3

  • NO-3 (sp2, trigonal planar);
  • CO2-3 (sp2, trigonal planar);
  • Same hybridization and the same shape.
  • SO2 (sp2, bent); NH2 (sp2, bent)

Different hybridization but the same shapes.

  • XeF2 (sp2d, linear); CO2 (sp, linear)
  • Different hybridization but the same shapes.
  • H2O (sp3angular); NH3 (sp3, pyramidal)

Question 97. Which of the following is correct regarding bond angles—

  1. SO2
  2. H2S<SO2
  3. SO2<H2S
  4. SBH3<NO+2

Choose the correct one

  1. 1 and 4
  2. 2, 1 and 4
  3. 1 and 3
  4. None of these

Answer: 1. 1 and 4

⇒ \(\begin{array}{ccccr}
\mathrm{SbH}_3 & \mathrm{H}_2 \mathrm{O} & \mathrm{H}_2 \mathrm{~S} & \mathrm{SO}_2 & \mathrm{NO}_2^{+} \\
91.3^{\circ} & 104.5^{\circ} & 109.5^{\circ} & 120^{\circ} & 180^{\circ}
\end{array}\)

Question 98. Which pair of molecules does not have an identical structure-

  1. I-3,BeF2
  2. O3,SO2
  3. BF2,ICI3
  4. BrF-4,XeF4

Answer: 3. BF3, ICI3

  1. BF3 —Trigonal planar
  2. IC13 —T-shape

Question 99. Which ofthe following order is correct—

  1. A1C13 < MgCl2 < NaCl: polarising power
  2. CO > CO2 > HCO-2 > CO2-3: bond length
  3. BeCl2 < NF3 < NH3 :dipole moment
  4. H2S > NH3 > SiH4 > BF3: bond angle

Answer: 3. A1C13 < MgCl2 < NaCl: polarising power

The polarising power of cations increases with the increasing charge.

⇒ \(\stackrel{(+1)}{\mathrm{NaCl}}<\stackrel{(+2)}{\mathrm{MgCl}_2}<\stackrel{(+3)}{\mathrm{AlCl}_2}\)

⇒ \(\text { Bond order } \propto \frac{1}{\text { Bond length }}\)

\(\text { Bond order }=\frac{\text { Bond order of each } \mathrm{C}-\mathrm{O} \text { bond }}{\text { Total no. of resonating structures }}\)

⇒ \(\begin{aligned}
& \mathrm{CO} \rightarrow \mathrm{C} \equiv \mathrm{O} \Rightarrow \frac{2+1}{1}=3.0 \\
& \mathrm{CO}_2 \rightarrow \mathrm{O}=\mathrm{C}=\mathrm{O} \Rightarrow \frac{2+2}{2}=2.0
\end{aligned}\)

Chemical Bonding And Molecular Structure Bond Order

Hence, the decreasing order of bond length is, CO2-3 > HCO2> CO2 > CO

Chemical Bonding And Molecular Structure Correct order of bond angle

The correct order of bond angle:

BF3 > SiH4 > NH3 > H2S
120° 109°28/ 107° 94°

Question 100. Which of the following has the maximum % of s- s-character

  1. N2H2
  2. N2H4
  3. NH3
  4. NH-2

Answer: 1. N2H2

Chemical Bonding And Molecular Structure the following has the maximum % of s- character

Question 101. Which of the following pairs does not have the same bond order-

  1. N2 and Cn-
  2. o+2 and no
  3. F-2 and O+2
  4. B22 and CN+

Answer: 1. N2 and Cn-

M.O. electronic configuration of N2(14):

⇒ \(\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_z}\right)^2\)

⇒ \(\text { B.O. }=\frac{1}{2}\left(\mathrm{~N}_b-\mathrm{N}_a\right)=\frac{1}{2}(10-4)=3.0\)

M.O. electronic configuration of CN-(14):

⇒ \(\begin{aligned}
& \quad\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_z}\right)^2 \\
& \text { B.O. }=\frac{1}{2}(10-4)=3.0
\end{aligned}\)

M.O. electronic configuration of O2(15):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\)

⇒ \(\text { B.O. }=\frac{1}{2}(10-5)=2.5\) \text { B.O. }=\frac{1}{2}(10-5)=2.5

M.O. electronic configuration of NO(15):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^1\)

⇒ \(\text { B.O. }=\frac{1}{2}(10-5)=2.5\)

M.O. electronic configuration of F-2( 19):

⇒ \(\begin{aligned}
& K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2 \\
& \left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2\left(\sigma_{2 p_z}\right)^1 \\
&
\end{aligned}\)

⇒ \(\text { B.O. }=\frac{1}{2}(10-8)=1.0\)

M.O. electronic configuration of O²2-(18):

⇒ \(\begin{aligned}
& K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p_z}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\left(\pi_{2 p_x}^*\right)^2\left(\pi_{2 p_y}^*\right)^2 \\
& \text { B.O. }=\frac{1}{2}(10-8)=1.0
\end{aligned}\)

M.O. electronic configuration of B²-2(12):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2\)

⇒ \(\text { B.O. }=\frac{1}{2}(8-4)=2.0\)

M.O. electronic configuration of CN+(12):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2 ; \text { B.O. }=\frac{1}{2}(8-4)=2.0\).

WBCHSE Class 11 Chemistry Hydrogen Notes

Hydrogen Introduction

  • Symbol: H
  • Molecular formula: H2
  • Atomic mass: 1.008
  • Electronic configuration: S1
  • Atomic number:1
  • Position in the periodic table: group -1 (IA) or group-17 (VII A), first period
  • Oxidation number: +1, -1

Position of hydrogen in the periodic table: Hydrogen is the first element in the periodic table to have atomic number 1, i.e., its electronic configuration is Is1. Because of hydrogen’s resemblance with alkali metals as well as with halogens, it can either be placed with the alkali metals in group-1 (IA) or with the halogens in group-17 (VII A) in the periodic table. The dual behavior of hydrogen is due to its electronic configuration.

Hydrogen Resemblance with alkali metals

From the above discussion, it is clear that hydrogen is unique in its behavior and it is not justified to place it either with the alkali metals of group-1 or with halogens of group-17. Thus, the position of hydrogen in the periodic table is anomalous (sometimes it is referred to as a ‘rogue element’) and it may be best placed separately in the periodic table. In the modern periodic table (IUPAC), hydrogen has been placed separately [a position in between group -l(IA) and 17(VHA)].

Occurrence Of Hydrogen

Hydrogen is the most abundant element in the universe (70% of the total mass of the universe). Hydrogen is not found in the atmosphere due to its lightweight. It is the third most abundant element on the earth’s surface. In a free state, it occurs in traces in volcanic gases and in the outer atmosphere of the sun and stars of the universe. In combined state, it exists mainly as water, natural gas, and petroleum. It is also an important constituent of organic matter in plants and animal tissues, carbohydrates, proteins, etc.

The extremely high temperature 4He + 0 energy of the sun is due to the nuclear helium positron fusion of hydrogen atoms liberating a large amount of energy. this energy. this energy is the main source of solar energy.

Isotopes Of Hydrogen

Hydrogen has three isotopes. These are protium or hydrogen, deuterium or heavy hydrogen, and tritium.

Hydrogen Isotopes of hydrogen

  • The three isotopes of hydrogen have the same chemical properties since they have the same electronic configuration (1s1), but differ from one another only in the number of neutrons in the nucleus.
  • However, because of different bond dissociation enthalpies, they have different rates of chemical reactions. Due to many differences in their atomic masses, they differ considerably in their physical properties.
  • The difference in properties arising due to differences in atomic masses is called the isotopic effect. Diatomic molecules containing only a protium atom (H2), only a deuterium atom (D2), and only tritium atoms (T2) are called diprotium, deuterium, and tritium respectively.
  • The term dihydrogen is used for the mixture of H2, D2, and HD concerning their natural abun¬ dance. The term hydron is used for the mixture of proton (H+) and deuteron (D+) concerning their natural abundance.
  1. Tritium can be artificially synthesized by bombarding the isotope of lithium \(\left({ }_3^6 \mathrm{Li}\right)\)or nitrogen by neutron.
  2. Deuterium is widely used as a tracer in determining the mechanism of organic reactions.
  3. Tritium gas is stored by converting it into a UT3 complex. When UT3 is heated at 673K it releases T2. It is used in the research on nuclear fusion reactions.

Water

Water is an important hydride of oxygen. In 1781, Cavendish first prepared water by exploding a mixture of 2 volumes of hydrogen and 1 volume of oxygen and proved that water is a compound that consists of the elements hydrogen and oxygen. Although water is the most abundant in the world, it is not always available in pure form. Hence, water in its natural state is not always fit for consumption and needs to be purified for drinking and laboratory use.

Structure Of Water Molecule

  • In water molecules, the two H-atoms are bonded to the O-atom by two covalent bonds. The oxygen atoms in water are sp3 -sp3-hybridized. Each of the covalent O —H bonds is formed by the axial overlap of the Is orbital of the H-atom and the sp3 -hybrid orbital of the O-atom.
  • The two bond pairs and the two lone pairs of electrons around the oxygen atom assume a tetrahedral arrangement. As a consequence, the H20 molecule has a bent structure.
  • Since the lone pair-lone pair and the lone pair-bond pair repulsions are greater than the bond pair-bond pair repulsive interaction, the H —O —H bond angle decreases from 109o28′ to 104.5°.
  • A molecule of water has a bent shape and hence the resultant of the two O —H bond moments adds to the moments produced by the lone pairs.
  • As a result, water molecules possess dipole moment (μ= 1.84D), i.e., it is a polar molecule. The bent structure of water, the orbital overlap picture of water, and the water molecule dipole are shown respectively.

Hydrogen Structure of water molecule

Water has higher melting and boiling points:

  • Since the electronegativity of the smaller oxygen atom is much higher, the O—H bond is considerably polar. This causes water molecules to remain associated through intermolecular hydrogen bonding.
  • Because of intermolecular H-bonding, water is liquid at room temperature and its melting and boiling points are relatively much higher than those of the hydrides of the other elements of group 16 having higher molecular mass.

The density of water is the highest at 4°C:

As the temperature is increased beyond 0°C, the open cage-like structure starts breaking due to cleavage of some H -bonds and ice starts melting. This causes water molecules to move into the holes or vacant spaces and to come closer to each other resulting in a decrease in volume and thereby increase in density. This continues till 4°C, when the density becomes maximum (1.00g . cm-3). Beyond this temperature, more H-bonds cleave due to the increased kinetic energy of the molecules. As a result, the expansion of water starts, and its density starts decreasing. At 100°C, most of the H -bonds break and water starts boiling.

Structure of ice: In a hexagonal crystal of ice each O-atom is tetrahedrally surrounded by four neighboring O-atoms. This gives rise to a three-dimensional structure having a large vacant space similar to an open cage. Each O-atom in the crystal is connected to four H-atoms — out of which two H-atoms are covalently bonded while the other two H atoms are bonded by weak hydrogen bonds. The density of ice is less than water because of the vacant space in its crystal structure. So, ice floats on water. Actually, 11 cm3 of water solidifies to form 12 cm3 ice.

Hydrogen Structure of ice

Properties Of Water

Physical properties

Pure water is a tasteless, odorless, and colorless liquid. Its physical properties are given below along with the physical properties of heavy water.

Hydrogen Physical properties

Chemical Properties

1. Nature: Water is a neutral oxide and neutral to litmus.

2. Solvent property: Water is an excellent solvent.

  • Being a highly polar compound (μ = 1.84D), water can stabilize ions by ion-dipole interactions. Again, its dielectric constant is much higher (∈ = 78.39) so, its ability to decrease the forces of attraction between oppositely charged ions is much higher.
  • For these reasons, water can dissolve many ionic or electrovalent compounds. All sodium, potassium, and ammonium salts [exception potassium perchlorate (KClO4), ammonium perchlorate (NH4ClO4), and all metal nitrates [exception: bismuth subnitrate, Bi(OH)2NO3] are soluble in water.
  • Water can dissolve many covalent compounds such as alcohols, amines, sugars, etc., by forming H-bonds with them. A molecule of water is capable of both accepting and donating protons.
  • So, water can dissolve some polar covalent compounds (For example HCl, NH3, etc.) by acid-base reactions. Because of such versatile solvent properties, water is called a ‘universal solvent.

3. Stability: Water is a very stable compound because it has a higher negative value of enthalpy of formation (A/7y° = -285.9kJ. mol-1). It does not dissociate even at much higher temperatures. Only 0.02% of it dissociates at 1200°C.
\(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g)\)

4. Acid-base character: Water is an amphoteric compound because it acts both as an acid and a base. According to the Bronsted-Lowry concept, it acts as an acid with NH3 by donating a proton and as a base with HCl by accepting a proton. Since water acts as a proton donor as well as a proton acceptor, it is called an amphoteric amphiprotic solvent.

⇒ \(\begin{aligned} & \mathrm{H}_2 \mathrm{O}(l)+\mathrm{NH}_3(a q) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q) \\ & \text { acid-1 base-2 } \quad \text { acid-2 } \quad \text { base- } 1 \\ & \end{aligned}\)

⇒ \(\underset{\text { base-1 }}{\mathrm{H}_2 \mathrm{O}(l)}+\underset{\text { acid-2 }}{\mathrm{HCl}}(a q) \rightleftharpoons \underset{\text { acid-1 }}{\mathrm{H}_3 \mathrm{O}^{+}(a q)}+\underset{\mathrm{Cl}^{-}(a q)}{\text { base-2 }}\)

Usually, water acts as a base in the presence of an acid stronger than it and acts as an acid in the presence of a base stronger than it. Because of such amphoteric character, water undergoes self-ionisation or autoprotolysis as follows:

⇒ \(\begin{aligned} & \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \\ & \text { acid-1 base-2 acid-2 base-1 } \\ & \text { (Acid) (Base) (Conjugate acid) (Conjugate base) } \\ & \end{aligned}\)

As the degree of self-ionization of water is much lower, the electrical conductivity of pure water is very low.

5. Oxidation-reduction reaction: Besides the acid-base reaction, water can undergo redox reactions.

Hydrogen Water acting as an oxidising agent

Hydrogen Water acting as a reducing agent

6. Hydrolytic reactions: Water can hydrolyze many metallic and non-metallic oxides, hydrides, nitrides, carbides, phosphides, and some other salts.

⇒ \(\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{CO}_3(a q)\)

⇒ \(\mathrm{SO}_2(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{SO}_3(a q)\)

⇒ \(\mathrm{P}_4 \mathrm{O}_{10}(s)+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 4 \mathrm{H}_3 \mathrm{PO}_4(a q)\)

⇒ \(\mathrm{Na}_2 \mathrm{O}(s)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{NaOH}(a q)\)

⇒ \(\mathrm{CaH}_2(s)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(a q)+2 \mathrm{H}_2(g)\)

⇒ \( \mathrm{Ca}_3 \mathrm{P}_2+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 3 \mathrm{Ca}(\mathrm{OH})_2(a q)+2 \mathrm{PH}_3(g)\)
[Phosphine]

⇒ \( \mathrm{CaC}_2(s)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_2(a q)+\mathrm{C}_2 \mathrm{H}_2(g)\)
[Acetylene]

⇒ \( \mathrm{Al}_4 \mathrm{C}_3(s)+12 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 4 \mathrm{Al}(\mathrm{OH})_3(a q)+3 \mathrm{CH}_4(g)\)
[Methane]

⇒ \(\mathrm{Mg}_3 \mathrm{~N}_2(s)+6 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 3 \mathrm{Mg}(\mathrm{OH})_2(s)+2 \mathrm{NH}_3(g)\)

7. Hydration reaction: Water is able to combine with some metal salts to form compounds known as hydrates.

There are three types of hydrates:

  1. Water molecules may combine with metal ions through coordinate bonds to form complexions.
    For example:
    ⇒ \(\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\left(\mathrm{NO}_3^{-}\right)_2 ;\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3 ;\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right] \mathrm{Cl}_3\)
  2. Water molecules may remain hydrogen-bonded to certain oxygen-containing anions. For example, in CuSO4-5H2O, the four water molecules are coordinated to the central Cu2+ ion while the fifth water molecule is hydrogen bonded to the sulfate group. Thus, CuSO4 . 5H2O be represented as [C(H2O)4]SO4.H2O.
  3. Water molecules may occupy the interstitial sites in the may crystal lattice. In barium chloride dihydrate, (BaCl2-2H2O), for example, the two H2O molecules occupy the voids of the crystal lattice.

8. Water absorbents: Many substances like concentrated H2SO4, P2O5, fused CaCl2, CaO, magnesium perchlorate [Mg(CO4)2, anhydrous], dehydrated silica gel (SiO2 . xH2O); anhydrous Na2SO4, etc., has the capacity to absorb a certain amount of water. These are known as dehydrating or desiccating agents. Although such absorption of water by the desiccating agents is often a physical process, in some cases chemical changes may also occur. For example, P2O5 and CaO absorb water to form H3PO4 and Ca(OH)2 respectively.

Desiccating agents are usually employed to dry moist substances and also to remove water from the sphere of the ‘ reaction.

  1. Concentrated sulphuric acid is used as a dehydrating agent in the esterification of an organic acid with 1 alcohol, where it absorbs the water formed in the reaction and thus helps to increase the yield of the ester.
  2. A moist gas or a moist liquid may be dried with the help of a desiccating agent. But, in such cases, the desiccating agent should be properly chosen so that it does not react chemically with the substance to be dried. Thus, moist NH3 gas cannot be dried by P2O5 or concentrated H2SO4 because they react with NH3 to form ammonium phosphate and ammonium sulfate respectively. Also, it cannot be dried by fused CaCl2 1 because it forms an additional compound CaCl2 • 8NH3.
  3. Similarly, moist H2S cannot be dried by the cone. H2SO4 or CaO and this is because cone. H2SO4 oxidizes H2S to sulfur and CaO reacts with acidic H2S to form calcium sulphide.
  4. Organic liquids are generally dried by anhydrous Na2SO4, fused CaCl2, or anhydrous K2CO3.

Unusual Properties Of Water

  1. The three states of water (solid, liquid, and gas) can easily be interconverted.
  2. Despite having low molecular mass water is a liquid at room temperature because its molecules remain associated through hitermolcciilar hydrogen bonding.
  3. Water is an excellent solvent for many Ionic as well as covalent compounds because of its high dielectric constant, dipole moment, and ability to form 11 bonds.
  4. The density of water is the highest in cm-3 at 4C, Ice has a larger volume for a given mass of water (11 cm3 of water freezes to yield 12cm, of ice), Thus, the density of ice is less than that of water and it floats over water.
  5. Conversion of ice into water and vaporization of water involves cleavage of numerous H-bonds and because of this, the melting point of ice, the latent heat of fusion of ice, the specific heat of water, the boiling point of water, the latent heat of vaporization of water, etc., have remarkably high values. Due to much higher values of the specific heat of the water and the latent heat of its vaporization, water plays a significant role in controlling the atmospheric and body temperatures.
  6. Water is a very stable compound and its dissociation temperature is extremely high. So the production of superheated steam by application of heat under pressure has been feasible and with its help, the generation of electricity through a turbine has been a very common commercially available process.
  7. Pure water is a conductor of heat and electricity.

Identification Of Water

  1. When a drop of water is added to anhydrous copper sulfate (CuSO4), its color changes from white to blue due to its conversion into hydrated copper sulfate (CUSO4-5H2O).
  2. Blue-colored silica gel (SiO2-xH2O) containing Co (2) salt becomes reddish-pink in the presence of water.
  3. Pure water can be identified from its melting point (0°C) and boiling point (100°C) at atmospheric pressure.

Heavy Water Or Deuterium Oxide (D2O)

Chemically, heavy water is deuterium oxide (D2O). It is called heavy water because it is obtained when oxygen combines with deuterium \(\left({ }_1^2 \mathrm{H}\right)\), the heavy isotope of hydrogen. It was discovered by Harold C. Urey, an American chemist in 1932. He showed that 6000 parts of ordinary water contain 1 part of heavy water.

Preparation

The main source of heavy water is ordinary water from which it is prepared by

The following methods:

By prolonged electrolysis of ordinary water:

The electrolysis of H2O occurs at a faster rate as compared to D2O and as the electrolysis continues, the concentration of heavy water in ordinary water gradually increases. When the amount of the liquid reduces to a small volume, almost pure D2O is obtained.

Electrolyte: Alkaline solution of water [-0.5 (N) NaOH].

Anode: Cylindrical sheet of nickel.

Cathode: Cylindrical steel cell.

Hydrogen Electrolysis of ordinary water

Procedure: In this method, electrolysis of ordinary water containing NaOH [nearly 0.5(N) NaOH solution] is carried out in a cylindrical steel cell. The cell itself acts as a cathode. A perforated cylindrical sheet of nickel acts as an anode. The electrolysis is carried out in different stages. The concentration of D2O in the residual liquid obtained after the 7th stage is about 99%. Almost 29000L of ordinary water is to be electrolyzed to obtain 1L of 99% pure D2O.

When ordinary water is electrolyzed, diprotium (H2) is liberated much more rapidly than deuterium (D2) because:

  1. Relatively smaller H+ ions have greater mobility than that of D+ ions.
  2. H+ ions having lower discharge potential are discharged at the cathode more easily than D+ ions.
  3. H-atoms combine more rapidly to form H, than D-; atoms to form D2.
  4. The O —H bond is weaker than the O—D bond.

By fractional distillation of ordinary water: The boiling points of ordinary water (H2O) and heavy water (D2O) are 100°C and 101.42°C respectively. Because of the small difference in boiling points, they cannot be separated by ordinary distillation but they can be separated by fractional distillation.

The fractional distillation of ordinary water is carried out in a very long (about 13m) fractionating column and the process is repeated several times. The lighter fraction (H2O) is distilled first while the heavier fraction (D2O) is left behind in the vessel. This residual liquid becomes rich in D2O.

Properties Of Heavy Water

Physical properties: Like ordinary water, heavy water is a colorless, tasteless, and odorless liquid. Because of higher molecular mass, all the physical constants of heavy water are higher than the corresponding values of ordinary water.

Chemical properties: Although heavy water is chemically similar to ordinary water, its reactions are slower than those of ordinary water and this is because the O —D bond is stronger than the O —H bond.

Some of the important reactions are given below:

1. Reaction with alkali and alkaline earth metals:

2Na+2D2O→2NaOD+D2; Ca+2D2O→Ca(OD)2+D2

2. Reaction with metal oxides:

Na2O+D2O→2NaOD [Sodium deuteroxide]

CaO+D2O→Ca(OD)2 [Calcium deuteroxide]

3. Reaction with non-metallic oxides:

N2O5+D2O→2DNO3 [Deuteronitric acid]

SO3+D2O→D2SO4 [Deuterosulphuric acid]

CO2+D2O→D2CO3 [Deuterocarbonic acid]

P2O5+3D2O→2D3PO4 [Deuterophosphoric acid]

4. Reactions with metallic carbides, nitrides, phosphides and arsenides:

CaC2+2D2O→Ca(OD)2+C2D2 [Deuteroacetylene]

Al4C3+12D2O→4A1(OD)3+3CD4 [Deuteromethane]

Hydrogen Reactions with metallic carbides, nitrides, phosphides and arsenides

Na3As+3D2O→3NaOD+AsD3 [Deuteroarsine]

5. Electrolysis: 2D2O→2D2 [at cathode]+O2 [at anode]

4. Exchange reactions: When compounds having active hydrogen are treated with D2O, hydrogen is exchanged by deuterium partially or completely. For example:

⇒ \(\mathrm{NH}_4 \mathrm{Cl}+4 \mathrm{D}_2 \mathrm{O} \rightleftharpoons \mathrm{ND}_4 \mathrm{Cl}+4 \mathrm{HDO}\)

⇒ \(\mathrm{NaOH}+\mathrm{D}_2 \mathrm{O}\rightleftharpoons\mathrm{NaOD}+\mathrm{HDO}\)

⇒ \(\mathrm{HCl}+\mathrm{D}_2 \mathrm{O} \rightleftharpoons \mathrm{DCl}+\mathrm{HDO}\)

⇒ \(\mathrm{CHCl}_3+\mathrm{D}_2 \mathrm{O} \rightleftharpoons \mathrm{CDCl}_3+\mathrm{HDO}\)

Formation of deuterates: Like ordinary water heavy water combines with many salts as heavy water of crystallization. The heavy hydrates thus obtained are called deuterates. Some examples are Na2SO4-10D2, CuSO4-5D2O, MgSO4-7D2O etc.

Physiological effect: Heavy water (D2O) is injurious to men, animals, and plants because it slows down the reactions occurring in them. It has also been established that heavy water has germicide and bactericide properties.

Uses Of Heavy Water

  1. Heavy water is extensively used as a moderator (the substance used for slowing down the speed of neutrons) in 4 nuclear reactions.
  2. It is used as a tracer compound for studying various mechanisms or organic reactions and various physiological processes occurring in the body.
  3. It is used for the preparation of deuterium (D2).
  4. It is used for the preparation of various deuterium-containing compounds.
  5. It is used as a solvent in NMR spectroscopic studies.

Soft Water And Hard Water

Water may be classified into two categories depending on its behavior towards soap.

These are as follows:

Soft water: Water that readily forms lather with soap is called soft water. Some examples of soft water are distilled water, demineralized water, and rainwater.

Hard water: Water that does not form lather with soap readily is called hard water. Hard water forms insoluble scum with soap. Some examples of hard water are river water, seawater, spring water, lake water, well water, and tap water.

Causes Of Hardness Of Water

  1. The hardness of natural water is due to the presence of soluble salts like bicarbonates, chlorides, and sulfates of calcium and magnesium. Water gets contaminated by these salts when it passes through the soil, mountains, and rocks.
  2. Ordinary soap is sodium or potassium salt of certain higher fatty acids such as stearic acid, palmitic acid, oleic acid, etc. These salts are soluble in water and dissolve in water to form a lather.
  3. But the calcium and magnesium salts of these acids, being insoluble in water, do not produce lather. If the water contains calcium or magnesium salts, then they react with soap to form scum or curdy white precipitates of calcium or magnesium salts of the higher fatty acids.

⇒ \(\begin{aligned} & 2 \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa}+\mathrm{M}^{2+} \rightarrow\left(\mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}\right)_2 \mathrm{M} \downarrow+2 \mathrm{Na}^{+} \\ & \text {Sodium stearate } \quad \text { (from } \quad \text { Metal stearate } \quad(\mathrm{M}=\mathrm{Ca}, \mathrm{Mg}) \\ & \text { (soap) hard water) (white precipitate) } \end{aligned}\)

As the soap water now contains no sodium salt or fatty acids, lather is not produced. After the complete removal of Ca2+ and Mg2+ ions as precipitate by using a sufficient amount of soap, lather is again produced. For these reasons, the use of soap in hard water leads to the wastage of soap. Hard water is, therefore, not suitable for washing purposes.

  • With the knowledge of the cause of the hardness of water, it becomes quite clear that the presence of Na and K-salts in water does not make it hard. But, if some soluble salts of heavy metals like Zn, Al, Ag, Pb, etc. whose stearates, palmitates, and oleates are insoluble in water, are added to a sample of soft water (For example distilled water), will behave as hard water.
  • If some acid (For example HCl, H2SO4, etc.) which may react with soap to precipitate the fatty acids, is added to soft water, it will also behave as hard water

For example:

⇒ \(\begin{aligned} & \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa}+\mathrm{HCl} \rightarrow \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COOH} \downarrow+\mathrm{NaCl} \\ & \text { Sodium stearate } \quad \text { Stearic acid } \\ & \end{aligned}\)

Types Of Hardness Of Water

Depending on the nature of the salt present, the hardness of water may be divided into two types Temporary hardness and Permanent hardness.

Temporary hardness: The hardness of water which is caused by the presence of bicarbonates of calcium and magnesium and can easily be removed by simply boiling the water is known as temporary hardness and water possessing such hardness is called temporary hard water. It is also termed as carbonate hardness.

Rainwater dissolves small quantities of atmospheric carbon dioxide forming a very dilute solution of carbonic acid. This water reacts with the calcium and magnesium carbonates present mainly in mountains and rocks over which it flows and as a result, soluble bicarbonates are formed. Thus soft rain water becomes hard.

⇒ \(\begin{aligned} & \mathrm{MCO}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{M}\left(\mathrm{HCO}_3\right)_2[\mathrm{M}=\mathrm{Ca} \text { or } \mathrm{Mg}] \\ & \text { [insoluble] } \quad \text { [soluble] } \\ & \end{aligned}\)

Permanent hardness: The hardness of water which is caused by the presence of chlorides and sulfates of calcium and magnesium and cannot be removed by simply boiling is known as permanent hardness and water possessing such hardness is called permanent hard water. It is also called non-carbonate hardness.

Disadvantages of using hard water

In domestic use: Hard water is not suitable for cooking because pulses and vegetables are not cooked well in it. Moreover, water with excessive hardness is not suitable for drinking and is harmful to health.

In laundry use: Hard water is not suitable for laundry purposes because its use results in considerable wastage of soap. Yellow stains may appear on clothsifiron salts are present in hard water.

This disadvantage can be overcome If detergent is used In hard water instead of soap. Calcium and magnesium salts of higher fatty acids are insoluble in water while calcium or magnesium salts of detergent are soluble in water. So, the use of hard water does not involve any wastage of detergent. Moreover, it gives lather more easily than soap.

In boiler use: Hard water cannot be used to produce steam in boilers.

The reasons are as follows:

  1. Hard water containing Mg(HCO3)2 and Ca(HCO3)2, on boiling, forms a hard heat-insulating thick layer or scale of MgCO3 and CaCO3 on the inner surface of the boiler. As a result of this, much heat is required to raise the temperature of the boiler, and thus, fuel economy is adversely affected.
    Again at much higher temperatures, the boiler scales and the metal of the boiler expand unequally. Due to such uneven expansion, cracks are formed on the scales.
    Through these cracks, when hot water comes in contact with the hot metal surface of the boiler, it is suddenly converted into steam. Due to the high pressure thus developed, the boiler may burst leading to serious accidents.
  2. MgCl2, MgSO4, CaCl2, etc., likely to be present in hard water, may undergo hydrolysis at high temperatures, producing strong acids like HCl or H2SO4. These acids slowly corrode the inner surface of the boiler and thus, reduce the longevity of the boiler.

⇒ \(\mathrm{MgCl}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Mg}(\mathrm{OH})_2+2 \mathrm{HCl}\)

⇒ \(\mathrm{MgSO}_4+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Mg}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{SO}_4\)

4. In industrial use: Hard water cannot be used in industry for cooling. This is because if it is used, the inner surface of the cooling coil may be coated with a layer or scale having poor thermal conductivity. Thus, cooling efficiency is affected by the consequent wastage of energy.

Removal Of Hardness Of Water Or Softening Of Hard Water

The process of removal of Ca2+ and Mg2+ ions responsible for the hardness of water is known as softening. water. Depending upon the nature of dissolved salts, many methods are available to soften hard water.

Removal Of Temporary Hardness

The temporary hardness of water can be removed by the following methods:

Boiling process: When temporary hard water is boiled, the bicarbonate salts of calcium and magnesium decompose to form Insoluble calcium and magnesium carbonates respectively which is filtration.

⇒ \(\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \rightarrow \mathrm{CaCO}_3 \downarrow+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{Mg}_{\left(\mathrm{HCO}_3\right)_2} \rightarrow \mathrm{MgCO}_3 \downarrow+\mathrm{CO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}\)

As MgCO3 has significant solubility in water, the temporary hardness caused by Mg(HCO3)2 cannot be removed completely.

Clark’s process: In this process, a calculated amount of slaked lime, Ca(OH)2, is added to the temporary hard water. The soluble bicarbonates are converted into insoluble carbonates and get precipitated. The precipitate is removed by filtration.

⇒ \(\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2+\mathrm{Ca}(\mathrm{OH})_2 \rightarrow 2 \mathrm{CaCO}_3 \downarrow+2 \mathrm{H}_2 \mathrm{O}\)

Because of the appreciable solubility of MgCO3, it further reacts with Ca(OH)2 to give a precipitate of Mg(OH)2.

⇒ \(\mathrm{MgCO}_3+\mathrm{Ca}(\mathrm{OH})_2 \rightarrow \mathrm{Mg}(\mathrm{OH})_2 \downarrow+\mathrm{CaCO}_3 \downarrow\)

If the quantity of Ca(OH)2 is less than the requisite amount, hardness due to magnesium still persists in water and this is because 1 mol of Mg(HCO3)2 requires 2 mol of Ca(OH)2.

Again, if the quantity of Ca(OH)2 is more than required, artificial hardness is created due to the absorption of atmospheric CO2 by Ca(OH)2 leading to the formation of Ca(HCO3)2. Therefore, in this process, the calculated quantity of slaked lime should be used.

Removal Of Permanent Hardness

The permanent hardness of water can be removed by the following methods:

Washing soda process: in this process, hard water is treated with a calculated amount of washing soda (Na2CO3.10H2O) when Ca2+ and’ Mg2+ ions are precipitated as their insoluble carbonates which can be easily filtered off.

⇒ \(\mathrm{MCl}_2+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{MCO}_3 \downarrow+2 \mathrm{NaCl} ;\)

⇒ \(\mathrm{MSO}_4+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{MCO}_3 \downarrow+\mathrm{Na}_2 \mathrm{SO}_4[\mathrm{M}=\mathrm{Mg} \text { or Ca}]\)

Calgon process: In the Ihls method, Ca2+ and Mg2+ Ions are rendered Ineffective (masked) by the addition of sodium Na2[Na4(PO3)6] commercially called ‘Calgon’ (meaning calcium gone) which forms soluble complexes with soap and so, facilitates the formation of lather.

⇒ \(\underset{\text { Calgon }}{2 \mathrm{Ca}^{2+}+\mathrm{Na}_2\left[\mathrm{Na}_4\left(\mathrm{PO}_3\right)_6\right]} \underset{\text { Soluble complex }}{\mathrm{Na}_2\left[\mathrm{Ca}_2\left(\mathrm{PO}_3\right)_6\right]+4 \mathrm{Na}^{+}}\)

⇒ \(\begin{array}{cc} 2 \mathrm{Mg}^{2+}+\mathrm{Na}_2\left[\mathrm{Na}_4\left(\mathrm{PO}_3\right)_6\right] & \mathrm{Na}_2\left[\mathrm{Mg}_2\left(\mathrm{PO}_3\right)_6\right]+4 \mathrm{Na}^{+} \\ \text {Calgon } & \text { Soluble complex } \end{array}\)

Ion Exchange Process: This is the most modern method for softening hard water. In this method, Ca2+ and Mg2+ ions present in hard water are exchanged by those present in ion, exchangers (complex inorganic and organic compounds) which are mainly of two types:

Inorganic cation exchangers (Pcrmutit): These are complex inorganic salts like hydrated sodium-aluminum silicates represented by the general formula Na2Z, where Z = Al2Si2O8. xH2O, which possess interestingproperty of exchanging Ca2+ and Mg2+ ions presentin hard water with their Na+ ions. These complex salts are known as zeolites (naturally occurring) or permit (synthetic).

⇒ \(\underset{\substack{\text { Sodium } \\ \text { zeolite }}}{\mathrm{Na}_2 \mathrm{Z}}+\underset{\text { from hard }}{\mathrm{MCl}_2} \rightarrow \underset{\substack{\text { Calcium or } \\ \text { Magnesium } \\ \text { zeolite }}}{\mathrm{MZ}}+2 \mathrm{NaCl}[\mathrm{M}=\mathrm{Ca} \text { or } \mathrm{Mg}]\)

Regeneration of permit: As the process continues, the whole of the permit gets exhausted because of its conversion into calcium and magnesium zeolite. The exhausted resin can, however, be regenerated by passing a 10% solution of NaCl (called brine) through it.

⇒ \(\underset{\substack{\text { Exhausted } \\ \text { resin }}}{\mathrm{MZ}+2 \mathrm{NaCl}} \underset{\substack{\text { Regenerated } \\ \text { resin }}}{\mathrm{Na}_2 \mathrm{Z}}+\mathrm{MCl}_2[\mathrm{M}=\mathrm{Ca} \text { or } \mathrm{Mg}]\)

Advantages of the permit process: It is an efficient and cheap process (only NaCl is consumed) that can be used to remove both the temporary and permanent hardness of water completely.

Organic ion exchangers (Resins): These synthetic ion exchangers (also called ion exchange resins) are superior to zeolites because these can remove all types of cations [Na+, Ca2+, Mg2+, etc.] and anions [Cl-, SO4, HCO3, etc.] presentin’ hard water. Thus, water obtained by this method is free from all types of cations and anions and is as good as distilled water. This is called deionized or demineralized water. Ion exchange resins (giant organic molecules of high molecular mass) are of two types.

Cation exchange resins: These are complex organic molecules consisting of a giant hydrocarbon framework attached to acidic groups such as —COOH (carboxyl) or —SO3H (sulphonic acid) groups and are represented by the general formula R —COOH or R —SO3H.

Since these resins are capable of exchanging the H+ ions of their acidic groups with cations such as Ca2+, Mg2+, etc. present in hard water, these are called cation exchange resins or simply cation exchangers.

Anion exchange resins: These are also complex organic molecules consisting of giant hydrocarbon frameworks attached to basic groups, such as OH ions derived from amines usually in the form of substituted ammonium hydroxide.

These may be represented by the general formula \(\mathrm{R}-\stackrel{\oplus}{\mathrm{N}} \mathrm{H}_3 \stackrel{\ominus}{\mathrm{O}} \mathrm{H}\). Since these resins are capable of exchanging their OH ions with anions such as Cl, \(\mathrm{SO}_4^{2-}\), \(\mathrm{HCO}_3^{-}\) etc. present in hard water, these are called anion exchange resins or simply anion exchangers.

Function of resin: Hard water is first passed through cation exchange resins when all the cations (Ca2+, Mg2+, Na+, etc.) present in water are exchanged with H+ ions of the resin.

Hydrogen Preparation of deionised water

⇒ \(\mathrm{M}^{2+}+2 \mathrm{RSO}_3 \mathrm{H} \rightarrow \mathrm{M}\left(\mathrm{RSO}_3\right)_2+2 \mathrm{H}^{+}[\mathrm{M}=\mathrm{Ca} \text { or } \mathrm{Mg}]\)
From hard Cation exchange Exhausted water resin resin

Due to the release of the proton, the resulting water becomes acidic. This water is then passed through anion exchange resin when all the anions (Cl-, \(\mathrm{SO}_4^{2-}\), etc.) present in water are exchanged with OH- ions of the resin.

⇒ \(\underset{\substack{\text { From hard } \\ \text { water }}}{\mathrm{CI}^{-}}+\underset{\substack{\text { Anion exchange } \\ \text { resin }}}{\stackrel{+}{\mathrm{NH}_3 \mathrm{OH}^{-}}} \rightarrow \underset{\substack{\text { Exhausted } \\ \text { resin }}}{\stackrel{+}{\mathrm{NH}_3 \mathrm{Cl}^{-}}+\mathrm{OH}^{-}}\)

⇒ \(\underset{\substack{\text { From hard } \\ \text { water }}}{\mathrm{SO}_4^{2-}}+\underset{\substack{\text { Anion exchange } \\ \text { resin }}}{2 \mathrm{RN}_3^{+} \mathrm{OH}^{-}} \rightarrow \underset{\substack{\text { Exhausted } \\ \text { resin }}}{\left(\mathrm{RN}_3\right)_2 \mathrm{SO}_4^{2-}}+2 \mathrm{OH}^{-}\)

The liberated OH ions neutralize the H+ ions set free in cation exchange resin (H++OH→H2O). Therefore, the collected water from anion exchange resin is free from all types of cations as well as anions. This is what is called deionized or demineralized water.

This is as pure as distilled water and can be used instead of distilled water in industry and in laboratories. However, this water may contain some organic impurities, some dissolved gases, or pyrogen. Deionized water is prepared by a machine known as a deioniser.

Regeneration of resins: The exhausted cation exchange resin is regenerated by treating it with moderately concentrated HCl or H2SO4 and the exhausted anion exchange resin is regenerated by treating it with moderately concentrated NaOH solution.

⇒ \(\begin{array}{lc} \mathrm{M}\left(\mathrm{RSO}_3\right)_2+2 \mathrm{HCl} \rightarrow \mathrm{MCl}_2 & +2 \mathrm{RSO}_3 \mathrm{H} \\ \quad \text { Exhausted } & \text { Regenerated } \\ \text { resin } & \text { resin } \end{array}\)

⇒ \(\underset{\substack{\text { Exhausted } \\ \text { resin }}}{\stackrel{+}{\mathrm{N}} \mathrm{H}_3 \mathrm{Cl}^{-}}+\mathrm{NaOH} \rightarrow \underset{\substack{\text { Regenerated } \\ \text { resin }}}{\stackrel{+}{\mathrm{N}} \mathrm{H}_3 \mathrm{OH}^{-}}+\mathrm{NaCl}\)

The ion exchange resins are not capable of removing non¬ electrolytes like sugar, urea, etc. from water. Therefore, the deionized water obtained by this process may contain these types of impurities. Distilled water does not contain similar impurities.

Manufacturing Of Drinking Water

Drinking water should necessarily be free from suspended impurities, organic matter, and germs. Water from rivers or lakes, properly purified, is supplied in towns and cities. Sometimes underground water, lifted by a pump, is also supplied for domestic use because it is not generally contaminated with microbes.

Drinking water is prepared from the river or, Jake water through the steps as follows:

  1. Precipitation of colloidal particles and some bacteria by the process of coagulation using alum [K2SO4-Al2(SO4)3-24H2O],
  2. Removal of these impurities by filtration,
  3. Removal of different ions causing hardness by ion exchange method and
  4. Sterilization by passing Cl2 or O3 gas or by exposing it to UV rays.

Hydrogen Preparation of highlypure water

Degree Of Hardness Of Water

Degree Of Hardness Of Water Definition: The degree of hardness of water is defined as the number of parts by mass of CaCO3 (calcium carbonate) equivalent to various calcium and magnesium salts present in one million parts by mass of water.

Unit: It is expressed in ppm (parts per million).

Example: A million parts by mass of a sample of water contain salts causing its hardness which are equivalent to x parts by mass of calcium carbonate, then the hardness of this sample of water is x ppm.

Hydrogen Equivalent mass ofvarious salts causing hardness

Now, the equivalent mass of CaCO3 \(=\frac{100}{2}=50\)

Therefore, 1 gram-equivalent or 1 /2 mol or 50 g of CaCO3 = 1

Gram-equivalent of any salt causing hardness =\(\equiv \frac{1}{2} \mathrm{~mol} \mathrm{or} 47.5 \mathrm{~g}\) \(\equiv \frac{1}{2} \mathrm{~mol} \text { or } 55.5 \mathrm{~g}\) of \(\mathrm{CaCl}_2 \equiv \frac{1}{2} \mathrm{~mol} \text { or } 60 \mathrm{~g} \text { of }\) \(\mathrm{MgSO}_4 \equiv \frac{1}{6} \mathrm{~mol} \text { or } 57.9 \mathrm{~g} \text { of } \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\) etc.

Hydrogen Hardness scale table

Numerical Examples

Question 1. Calculate the degree of hardness of a sample of hard water that is found to contain 72mg of MgSO4 per kg of water.
[Here our objective is to find out the mass of CaCO3 equivalent of MgSO4 present in one million parts of water.]
Answer: Now, 1 kg or 103 g of water contains 72 mg MgSO4

106 g of water contains 72 X 103 mg = 72g MgSO4

Now, 60 g MgSO4 = 50g CaCO3

or, 72g MgSO4 \(\equiv \frac{50 \times 72}{60} \mathrm{~g}\) CaCO3 = 60 g CaCO3

Thus, the degree of hardness of that sample = 60 ppm.

Question 2. Estimate the hardness of a sample of water 1L which contains 0.001 mol of dissolved MgCl2.
Answer: We know, 1 mol MgCl2= 1 mol CaCO3

or, 0.001 mol MgCl2 = 0.001 mol CaCO3

Now, 0.001 mol CaCO3 = 100 x 0.001 for O.lgof CaCO3

So, in 1L 1000 g, or 103g water, the amount of CaCO3 equivalent to 0.001 mol MgCl2 is 0.1 g.

In 106 g water, the amount of CaCO3 is\(\frac{0.1 \times 10^6}{10^3} \mathrm{~g}=100\)

Hence, the degree of hardness of that sample of water = 100 ppm

Question 3. IL of river water contains 6 mg Mg2+ and 20 mg Ca2+ ions as chloride salts. Determine the degree of hardness of that sample of river water.
Answer: 6mg Mg2+ = 0.006 g Mg2+

and20mg Ca2+ = 0.02 g Ca2+ ion.

Now, 24 g Mg2+ = 95 g MgCl2 = 100 g CaCO3.

[∴ atomic mass of. Mg = 24 and molecular mass of MgCl2 = 95 and CaCO3 = 40]

∴ 0.006 g Mg2+\(=\frac{100 \times 0.006}{24} \mathrm{~g}=0.025 \mathrm{~g} \mathrm{CaCO}_3 .\)

Again, 40 g Ca2+ = 111 g CaCl2 = 100 g CaCO3

[∴ atomic mass of Ca = 40 & molecular mass of CaCl2 = 111]

∴ \(0.02 \mathrm{~g} \mathrm{Ca}^{2+}=\frac{100 \times 0.02}{40} \mathrm{~g}=0.05 \mathrm{~g} \mathrm{CaCO}_3\)

So, the quantity of CaCO3 equivalent to MgCl2 and CaCl2 presentin 1Lor 1 O3 g water = (0.025 + 0.05) g =0.075 g.

∴ \(10^6 \mathrm{~g} \text { of water contains } \frac{0.075 \times 10^6}{10^3}=75 \mathrm{~g}^3 \mathrm{CaCO}_3 .\)

Hence, the degree of hardness of the sample is 75 ppm.

Question 4. The degree of hardness of a sample of water is 40 ppm. If the hardness is only due to the presence of MgSO4, then determine the amount of MgSO4 in 1 kg of that water.
Answer: The hardness of the sample of water is 40 ppm.

Therefore, 106 g of that sample contains 40 g CaCO3.

\(1 \mathrm{~kg} \text { or } 10^3 \mathrm{~g} \text { of water contains } \frac{40 \times 10^3}{10^6}=0.04 \mathrm{~g} \mathrm{CaCO}_3\)

Now,1 mol CaCO3 = 1 mol MgSO4 or, 100 g CaCO3= 120 g MgSO4

[molecular mass of CaCO3 = 100 and MgSO4 = 120]

\(\text { or, } \quad 0.04 \mathrm{~g} \mathrm{CaCO}_3 \equiv \frac{120 \times 0.04}{100} \mathrm{~g} \quad \text { or, } 0.048 \mathrm{~g} \mathrm{MgSO}_4\)

Hence, the amount of MgSO4 present per kg of that water is 0.048 g or 48 mg.

Question 5. 10 mL of 0.01 (N) HCl is required for titrating 100mL of a sample of cold water usingmethyl orange as an indicator. Determine the temporary hardness of that sample of water.
Answer: 10mL O.Ol(N) HCl solution =1 mL 0.1(N) HCl

1 g equivalentHCI=1 g equivalent CaCO3

\(1000 \mathrm{~mL} 1(\mathrm{~N}) \mathrm{HCl} \equiv \frac{100}{2} \mathrm{~g} \mathrm{CaCO}_3\) \(1 \mathrm{~mol} 0.1(\mathrm{~N}) \mathrm{HCl} \equiv 50 \times \frac{1}{1000} \times 0.1 \mathrm{~g} \mathrm{CaCO}_3\)

= 0.005g CaCO3

Therefore, in lOOmL or lOOg of that sample of water contains some hardness-producing substance which is equivalent to 0.005g CaCO3

106g of water contains the hardness-producing substance equivalent to \(\frac{0.005 \times 10^6}{10^2}=50 \mathrm{~g} \mathrm{CaCO}_3\)

Hence, the hardness of that sample of water is = 50 ppm.

Question 6. Determine the weight of CaO required to remove the hardness of a sample of 105L water, 1L of which contains 1.62gof Ca(HCO3)2.
Answer: When CaO is added to a sample of hard water containing Ca(HCO3)2, dissolved bicarbonate is precipitated as CaC3 and as a result, the hardness is removed.

\(\begin{gathered} \mathrm{CaO}+\mathrm{Ca}_{\left(\mathrm{HCO}_3\right)_2}=2 \mathrm{CaCO}_3 \downarrow+\mathrm{H}_2 \mathrm{O} \\ \text { molecular mass }=56 \text { molecular mass }=162 \end{gathered}\)

I L of water contains 1.62 g of Ca(HCO3)2

∴ 105L of water contains 1.62 x 105 g of Ca(HCO3)2.

According to the above equation,

For removing 162 g Ca(HCO3)2, 56 g CaO is required

∴ \(\text { Amount of } \mathrm{CaO} \text { required }=\frac{56 \times 1.62 \times 10^5}{162} \mathrm{~g}=56 \times 10^3 \mathrm{~g}\)

Structure Of H2O2 Molecule

The structure of the hydrogen peroxide molecule is H—O—O—H. There is a peroxide linkage (—0—0— ) in the molecule. The molecule is non-planar and it has an open-book-like shape. The two 0 —H bonds lie in different places and this is due to repulsion between the —OH groups. In the gas phase, the dihedral angle and the H —O —O bond angle are 111.5° and 94.8° respectively. However, in the crystalline state, these are 90.2° and 101.9° due to intermolecular hydrogen bonding, Because of such shape, H2O2 is polar.

Hydrogen Structure of H202 molecule

Properties Of Hydrogen Peroxide

Physical properties

  1. Pure hydrogen peroxide is an almost colorless syrupy liquid (with a tinge of pale blue) that has a bitter taste and smell similar to HNO3.
  2. It is denser (1.44 g-cm-3) and more viscous than water because the molecules of H2O2 (with two different —OH groups) are even more highly associated through intermolecular H-bonding than H20 molecules. Also due to this reason, its boiling point is higher than that of water.
  3. It is soluble in water, ether, and alcohol in all proportions.
  4. It has both polar and non-polar bonds.

⇒ \(\begin{array}{r} \text { polar bond } \\ \mathrm{H}-\mathrm{O}-\mathrm{O} \downarrow-\mathrm{H} \\ \text { non-polar bond } \end{array}\)

Hydrogen Values of some physical properties of H2O2

Chemical Properties

Decomposition: Hydrogen peroxide is an unstable liquid that decomposes into water and oxygen on long-standing or heating.

⇒ \(2 \mathrm{H}_2^{-1} \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2^{-2} \mathrm{O}+\stackrel{0}{\mathrm{O}_2} ; \Delta H=-196.0 \mathrm{~kJ}\)

It is an example of a disproportionation reaction because H2O2 undergoes both oxidation and reduction. Since it is a very unstable compound and decomposes readily on heating, it is impossible to determine its boiling point at atmospheric pressure. However, it can be determined by the extrapolation method.

  • The presence composition of metal of powders H2O2is[e.g,furtherCo, Au, accelerated, Pt, etc.), by some metal ions [Example Fe2+, Co2+, Ni2+, etc.), metal oxides (for example MnO2), charcoal, basic substances, dust particles, rough surfaces or even sunlight Because of such properties, the solution of H2O2 must be handled with care as uncontrolled rapid decomposition may resultin an explosion.
  • Its decomposition may, however, be suppressed by the addition of glycerol, acetanilide, phosphoric acid, or urea (all acting as negative catalysts).
  • H2O2, because of its very unstable nature, is preserved with a small amount of stabilizers like H3PO4, glycerol, or acetanilide in an opaque polythene bottle or a wax-lined colored glass bottle away from light and at low temperature.

Acidic nature:

Pure H2O2 turns blue litmus red but its dilute solution is neutral to litmus. Thus, it behaves as a very weak acid. In fact, it is a slightly stronger acid (Ka = 1.55 X 1012 at 25°C) (Ka = 1.0 x 10-14 at 25°C). Since it contains at than water two ionizable or replaceable H-atoms, it reacts with bases like NaOH to form two types of salts, for example, hydroperoxides (acidic salt) and peroxides (normal salts). This property of H2O2 is known as peroxidizing property.

⇒ \(\mathrm{H}_2 \mathrm{O}_2 \rightleftharpoons \mathrm{H}^{+}+\mathrm{HO}_2^{-} \text {(Hydroperoxide ion) }\)

⇒ \(\mathrm{H}_2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{O}_2^{2-} \text { (Peroxide ion) }\)

⇒ \( \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{NaHO}_2 \text { (acidic salt) }+\mathrm{H}_2 \mathrm{O}\)
Sodium hydroperoxide

⇒ \( 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{Na}_2 \mathrm{O}_2+\mathrm{H}_2 \mathrm{O}\)
Sodium peroxide

H2O2 cannot decompose carbonates to CO2. H2O2 is weaker add than carbonic add (H2CO3) so it cannot decompose carbonate or bicarbonate salts to liberate CO2

Oxidizing and reducing properties: H2O2 is an oxidizing agent. But in the presence of other oxidizing agents, It behaves as a reducing agent. These oxidising and reducing properties are exhibited both In acidic and alkaline solutions. The oxidation state of oxygen in H2O2 is. It lies in between the tire oxidation state of oxygen in H2O or OH (-2) and that of O2 (zero). So the oxidation state of oxygen in H2O2 may decrease to\(-2\left(\mathrm{H}_2^{-1} \mathrm{O}_2 \rightarrow \mathrm{H}_2^{-2} \mathrm{O}^{\mathrm{O}} \text { or } \stackrel{-2}{\left.\mathrm{OH}^{-}\right)}\right.\) or increase to zero \(\left(\mathrm{H}_2 \mathrm{O}_2^{-1} \rightarrow \stackrel{0}{\mathrm{O}_2}\right)\) and bacause of this,h2o2 is formed to exhibit both oxidising and reducing properties.

Oxidizing properties: H2O2 can act as an oxidizing agent in both acidic and basic mediums.

In acidic medium: \(\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{H}^{+}+2 e \rightarrow 2 \mathrm{H}_2 \mathrm{O}\)

In basic medium: \(\mathrm{H}_2 \mathrm{O}_2+2 e \rightarrow 2 \mathrm{OH}^{-}\)

Hydrogen Oxidationin acidic medium
Hydrogen Oxidationin acidic medium.

Restoration of the color of oil paintings: H2O2 is used to restore the original color of old oil paintings.

  1. In oil paintings, lead white, a mixture of basic lead acetate, lead carbonate, lead sulfate, etc. is used.
  2. Lead white, if left exposed to open air for a long time, the compounds present in it react with H2S of air to form insoluble black lead sulfide and as a result, the oil painting gets blackened.
  3. If the oil painting is washed with an H2O2 solution, lead sulfide init is oxidized to white lead sulfate, and the brightness of the oil painting is regained.

⇒ \(\mathrm{PbCO}_3+\mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{PbS} \downarrow \text { (black) }+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

⇒ \(\mathrm{PbSO}_4+\mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{PbS}(\text { black }) \downarrow+\mathrm{H}_2 \mathrm{SO}_4 ;\)

⇒ \(\mathrm{Pb}\left(\mathrm{CH}_3 \mathrm{COO}\right)_2+\mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{PbS}(\text { black }) \downarrow+2 \mathrm{CH}_3 \mathrm{COOH}\)

⇒ \(\mathrm{PbS}(\text { black })+4 \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{PbSO}_4(\text { white })+4 \mathrm{H}_2 \mathrm{O}\)

Reducing properties: H2O2 can act as a reducing agent in both acidic and alkaline mediums.

In acidic medium: \(\mathrm{H}_2 \mathrm{O}_2 \rightarrow 2\mathrm{H}^{+}+\mathrm{O}_2+2 e\)

In alkaline medium: \(\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2+2 e\)

Hydrogen Reductionin acidic medium
Hydrogen Reductionin acidic medium.

Antichlor property of H2O2: Since H2O2 destroys Cl2 by reducing it to HCl, it is used to remove excess chlorine after bleaching operations in the textile industry. It is known as the chlorine-destroying or antichlor property of H2O2.

Bleaching action: Hydrogen peroxide is used for bleaching delicate articles like ivory, feathers, silk, wool, etc. The bleaching action of H2O2 is due to oxidation of the coloring matter by nascent oxygen liberating on decomposition: H2O2 H2O+[O]

Colouring matter + [O]→ colorless matter

Formation of additional compounds: H2O2 combines with some salts to form additional compounds.

For example, Na2SO4 -H2O2 9H2O; NaBO2-H2O2-3H2O; (NH4)2SO4-H2O2 etc.

These additional compounds are known as perhydrates (hydrates where the water molecules have been replaced by H2O2). Also, it reacts with alkenes to form glycols.

CH2=CH2 + H2O2 → HOCH2CH2OH [ethylene glycol]

Identification And Uses Of H2O2

Identification of H2O2:

1. Perchromic acid test: H2O2 is added to K2Cr2O? solution acidified with dilute H2SO4 and the mixture is shaken with ether. Perchromic acid or chromium pentoxide (CrO5), produced by the reaction of H2O2 with acidified K2Cr2O7 solution, dissolves in ether through the formation of an addition compound that turns the ether layer blue.

K2Cr2O7 + H2SO4 + 4H2O2K2SO4 + 2CrO5 + 5H2O

2. When H2O2 is treated with an acidified solution of titanium salt, the orange color is produced due to the formation of pertitanic acid (H2TiO4).

Ti(SO4)2 + H2O2 + 2H2O → H2TiO4 + 2H2SO4

3. H2O2 liberates iodine from an acidified solution of KI. The liberated iodine turns the starch solution blue

2KI + 2HCl + H2O2 → I2 + 2KCl + 2H2O

Starch + I2 → Deep blue colored complex

Uses of hydrogen peroxide:

  1. The most important industrial application of hydrogen peroxide acts as a bleaching agent for fine and delicate materials like silk, wool, ivory, paper pulp, leather, oils fats, etc.
  2. Its dilute solution is used to impart a golden color to hair.
  3. It is used as an antichlor for removing chlorine from articles bleached by chlorine.
  4. The dilute solution of H2O2 is used as an antiseptic for washing wounds. This solution is known as perhydrol.
  5. It is used for restoring the color of old oil paintings.
  6. It is largely used as an oxidizing agent in the laboratory. A mixture of H2O2 and FeSO4 is called Fenton’s reagent. It is used to oxidize many organic compounds.
  7. A mixture of H2O2 and hydrazine hydrate with copper (II) catalyst is used as a rocket propellant.
  8. It is used for preparing chemicals like sodium peroxoborate and peroxocarbonate which are largely used as brighteners in high-quality detergents.
  9. It is used in the synthesis of hydroquinone, tartaric acid, and certain food products and pharmaceuticals [e.g, cephalosporin) etc.
  10. H2O2 is being increasingly used to control pollution by—treatment of domestic and industrial effluents, oxidation of cyanides, restoration of aerobic conditions to sewage wastes, etc.

Volume Strength: The volume strength of an H2O2 solution indicates the volume in mL of oxygen that will be evolved at STP on the complete decomposition of lmL of the H2O2 solution. Thus, ’20 volume’ H2O2 solution means that lmL of that solution yields 20mL of oxygen at STP as a result of its complete decomposition.

Percentage Strength: The percentage strength of an H2O2 solution indicates the amount of H2O2 in gram present in 100 mL of this solution. Thus, 30% H2O2 solution means that 100 mL of that solution contains 30 g of H2O2.

Relation between volume strength & percentage strength:

H2O2( 68 g) → 2H2O + O2 [22400 mL at STP]

The equation states that 68 g H2O2 yields 22400 mL O2 at STP.

∴ lg of H2O2 yields 22400/68 mL or 329.4 mL of O2 at STP. Now, if 100 mL of H2O2 solution contains lg of H2O2, then the strength of the solution is 1%.

Hence 100 mL 1% H2O2 solution on being completely decomposed liberates 329.4 mL of O2 at STP.

∴ 1 mL of 1% H2O2 solution on being completely decomerate 329.4/100 mL or 3.249 mL of O2 at STP.

∴ The strength of 1% of H2O2 solution is ‘3.294 volume

Thus, ‘x volume’ H2O2 solution = \(\frac{x}{3.294} \%\) or 0.3036x

H2O2 solution, i.e., if the volume strength of any H2O2 solution is known, then its percentage strength can readily be calculated by multiplying its volume strength by 0.3036. It thus follows that the x% H2O2 solution is stronger than the ‘ x volume’ H2O2 solution.

Relation between volume strength and normal strength:

The equivalent mass of H2O2 \(=\frac{68}{32} \times 8=17.0\) [From the dissociation equation of H2O2 (2H2O2 -> 2HaO + 02) it becomes clear that 8g of O2 is obtained from 17g of H2O2]

Now, 68g of H2O2 produces 22400 mL of O2 at STP.

or, 17g H2O2 produces\(\frac{22400}{68} \times 17=5600 \mathrm{~mL}\) of O2 at STP.

The strength of an H2O2 solution will be (N) if 1000 mL of that solution contains 17g of H2O2.

Thus 1000 mL of (N) H2O2 solution will produce 5600 mL O2 at STP.

or, lmL(N) H2O2 solution will produce 5.6mL O2 at STP.

∴ The volume strength of (N) H2O2 solution =5.6 or, the volume strength of x (N) H2O2 solution = 5.6 x. The volume strength of the H2O2 solution of any normality can be obtained by multiplying its normal strength by 5.6.

1. Volume strength

= 5.6 x normality

⇒ \(=5.6 \times \frac{\text { percentage strength }}{17} \times 10\)

⇒ \(=5.6 \times \frac{\text { strength in gram per litre }}{17}\)

2. Volume strength

= 11.2 x molarity

⇒ \(=11.2 \times \frac{\text { percentage strength }}{34} \times 10\)

Numerical Examples

Question 1. Determine the strength of’10 volume H2O2 solution in

  1. Gram per liter,
  2. Normality and
  3. Percentage strength.

Answer: 2H2O2(68 g) 2H2O + O2 [22.4L at STP]

1. Now, 10 volume H2O2 solution means that at STP 10 mL O2 is obtained from 1 mL of this solution.

∴ 22400 mL O2 at STP is obtained from

\(\frac{22400}{10}=2240 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}_2\)

∴ 2240 mL of H2O2 solution contains 68g of H2O2

∴ 1000 mL H2O2 solution\(=\frac{68 \times 1000}{2240}=30.36 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2\)

So, strength of10 volume H2O2 solution = 30.36 g-L-1.

2. Amount of H2O2 present in 1000 mL solution = 30.36 g.

∴ In normality the strength of10 volume H2O2solution

= 30.36/17 = 1.7858 (N)

3. The amount of H2O2in 1000 mL solution = 30.36 g

∴ Amount of H2O2 in 100 mL of the solution

= 30.36 x 100/1000 =3.036 g.

∴ Percentage strength of H2O2 solution = 3.036

2. Determine the volume strength of 1.5 (N) H2O2.
Answer: 1L of 1.5 (N)H2O2 solution contains 1.5 x 17 =25.5 g

1 mL of l.5 (N) H2O2 solution contains 25.5/1000=0.0255g.

Now 68g H2O2 liberates 22400 ml, O2 STP.

∴ 0.0255g I-I202 liberates\(22400 \times \frac{0.0255}{68}=8.4 \mathrm{~mL} \mathrm{O}_2 \text { at STP. }\)

Therefore, the volume strength of 1.5 (N) H2O2 solution =8.4.

Question 3. Determine the volume strength of a 6.07% H2O2 solution.
Answer: 6.07% H2O2 solution means 100 mL of the solution contains 6.07g of H2O2.

∴ \(1 \mathrm{~mL} \text { solution contains } \frac{6.07}{100}=0.0607 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2 \text {. }\)

2H2O2(68 g) -> 2H2O + O2 [22400 mL at STP]

Now, 68 g H2O2 liberates 22400 mL O2 at STP.

∴ 0.0607 g H2O2 liberates\(\frac{22400 \times 0.0607}{68} \approx 20 \mathrm{~mL} .\)

Thus, 1 mL H2O2 solution produces 20 mL O2 at STP.

Hence, the volume strength of 6.07% H2O2 solution is ’20 volume.

Question 4. The strengths of the three H2O2 solutions are 10, 15, and 20 volumes respectively. 0.5L of each of these solutions is mixed and an equal amount of water is added to it. Determine the volume strength of the mixed solution.
Answer: Volume strength = 5.6 x normality

∴ The normality of the first solution, N1 = 10/5.6, the normality of the second solution, N2 – 15/5.6, and the normality of the third solution, N3 = 20/5.6.

Now, if the volumes of the first, second, third, and mixed solutions are V1, V2, V3, and VR respectively, and if the normality of the mixed solution is NR, then

V1V1 + N2V2 + N3V3 = NRVP

\(\text { or, } \quad \frac{10}{5.6} \times \frac{1}{2}+\frac{15}{5.6} \times \frac{1}{2}+\frac{20}{5.6} \times \frac{1}{2}=N_R \times 3\)

⇒ \(\text { or, } \quad N_R=\frac{(5+7.5+10)}{5.6 \times 3}=1.339\)

Therefore, the volume strength of the mixed solution

=NR X 5.6 = 1.339 X 5.6 = 7.5 volume.

Question 5. 20mL of a H2O2 solution after acidification required 20 mL of N/10 KMn04 solution for complete oxidation. Calculate the percentage and volume strength of the H2O2 solution.
Answer: From die given data, for H2O2 solution, V1 = 20mL and for KMnO4 solutions V2 = 20mL,\(N_2=\frac{N}{10}\)

Applying the normality equation, N1V1 = N2V2

⇒ \(\text { or, } \quad 20 \times N_1=20 \times \frac{1}{10} \quad \therefore N_1=0.1(\mathrm{~N})\)

Thus, the normality of the H2O2 solution = 0.1(N).

Now, amount of H2O2 in 1 L solution = 0.1 x 17 = 1.7g

∴ The amount of H2O2 in 100 mL of the solution \(=\frac{1.7 \times 100}{1000}=0.17 \mathrm{~g}\)

∴ The percentage strength of the solution = 0.17 %.

Now, 68g of H2O2 produces 22400 mL of O2 at STP.

∴ \(\text { 1.7g of } \mathrm{H}_2 \mathrm{O}_2 \text { produces } \frac{22400}{68} \times 1.7=560 \mathrm{~mL} \mathrm{O}_2 \text { at STP. }\)

This 1.7g of H2O2 is present in 1000 mL of H2O2 solution.

Hence, 1000 mL of H2O2 solution gives 560 mL of O2 at STP.

∴ \(1 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}_2 \text { solution gives } \frac{560}{1000}=0.56 \mathrm{~mL} \text { of } \mathrm{O}_2 \text { at STP. }\)

Question 14. Arrange the following: CaH2, BeH2, and TiH2 in order of increasing electrical conductance. LiH, NaH, and CH in order of increasing ionic character.If —D, D—D, and F—F in order of increasing bond dissociation enthalpy. NaH, MgH2, and H2O in order of Increasing reducing properties.
Answer: Being a covalent hydride BeH2 does not conduct electricity at all. Being an Ionic hydride CaH2 conducts electricity in the fused state while TiH2, being a metallic hydride, conducts electricity at room temperature, Thus, the order of increasing electrical conductance is: BeH2 < CaH2 < TiH2.

The electronegativity of the alkali metals decreases down the group from Li to Cs. Therefore, the ionic character of their hydrides also increases in the same order, l.e., LIH < NaH < CsH.

The bond dissociation enthalpy of the: F—F: bond is the lowest (242.6 kj. mol-1) and this is due to the high concentration of electron density around each F atom in the form of three unshared pairs which have significant repulsive interactions. Again, because of the marginally smaller size of D as compared to H, the bond dissociation enthalpy of the D—D bond (443.35 kj-mol-1) is slightly higher than that of the H —H bond (435.88 kj-mol-1). Hence, the bond dissociation enthalpy increases in the order: of F —F < H —H < D —D.

NaH, being an ionic hydride, is a more powerful reducing agent than the covalent hydrides MgH2 and H2O. MgH2 is a stronger reducing agent than H2O because the bond dissociation enthalpy of the Mg—H bond is much lower than that of the O —H bond. Therefore, the reducing property increases in the order: H2O < MgH2 < NaH

Question 15. Compare the structures of H2O and H2O2.
Answer: The oxygen atom in water is sp3 -sp3-hybridized. The two O —H bonds are sp3-s sigma bonds. The H —O —H bond angle is 104.5°. This value is a little less than the tetrahedral angle (109°28/) because of stronger lone pair-lone pair and lone pair-bond pair repulsions than bond pair-bond pair repulsion.

Thus, water is a bent molecule. Each oxygen atom in H2O2 is also sp3 hybridized. The O —0 bond is a sp3-sp3 sigma bond and the two O —H bonds are sp3-s sigma bonds. The two O —H bonds are, however, present in different planes. In the gas phase, the dihedral angle between the two planes (i.e., the planes containing H —O —O system) is 111.5°. So, the molecule has an open-booklike structure.

Hydrogen Hydrogen Peroxide

Question 16. What do you understand by the term ‘auto-protolysis’ of water? What is its significance?
Answer: Self-ionization of water is called auto-protolysis. Self-ionization of water can be expressed by the given equation—

⇒ \(\begin{aligned}
& \mathrm{H}_2 \mathrm{O}(l)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \\
& \text { Acid-1 Base-2 Acid-2 Base-1 } \\
&
\end{aligned}\)

Water exhibits amphoteric properties because of protolysis. Thus water reacts with both acids and bases. It usually acts as a base in the presence of an acid stronger than it and acts as an acid in the presence of a base stronger than it. For example,

\(\underset{\text { Acid-1 }}{\mathrm{H}_2 \mathrm{O}(l)}+\underset{\text { Base-2 }}{\mathrm{NH}_3(a q)} \longrightarrow \underset{\text { Acid-2 }}{\mathrm{NH}_4^{+}(a q)}+\underset{\text { Base-1 }}{\mathrm{OH}^{-}(a q)}\) \(\underset{\text { Base-1 }}{\mathrm{H}_2 \mathrm{O}(l)}+\underset{\text { Acid-2 }}{\mathrm{HCl}(a q)} \longrightarrow \underset{\text { Acid-1 }}{\mathrm{H}_3 \mathrm{O}^{+}(a q)}+\underset{\text { Base-2 }}{\mathrm{Cl}^{-}(a q)}\)

Question 17. Consider the reaction of water with F2 and suggest, In terms of oxidation and reduction, which species are oxidized/reduced.
Answer: \(
2 \mathrm{~F}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{~F}^{-}(a q)
Oxidant Reductant\)

⇒ \(
3 \mathrm{~F}_2(g)+3 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{O}_3(\mathrm{~g})+6 \mathrm{H}^{+}(a q)+6 \mathrm{~F}^{-}(a q)
Oxidant Reductant\)

In these reactions, water acts as a reductant and itself gets oxidized to oxygen or ozone. In this case, highly electronegative fluorine acts as an oxidant and gets reduced to F-

Question 18. Complete the following chemical reactions. Classify the above into [a] hydrolysis, [b] redox and [c] hydration reactions
Answer: \(\mathrm{PbS}(s)+4 \mathrm{H}_2 \mathrm{O}_2(a q) \longrightarrow \mathrm{PbSO}_4(s)+4 \mathrm{H}_2 \mathrm{O}(l)\)

\(\begin{aligned}
2 \mathrm{MnO}_4^{-}(a q)+5 \mathrm{H}_2 \mathrm{O}_2(l)+6 \mathrm{H}^{+}(a q) \\
2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_2 \mathrm{O}(l)+5 \mathrm{O}_2(g)
\end{aligned}\) \(\mathrm{CaO}(s)+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_2(a q)\) \(\begin{aligned}
& \mathrm{AlCl}_3(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(l) \\
& {\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)+3 \mathrm{Cl}^{-}(a q)} \\
&
\end{aligned}\)

⇒ \(\begin{array}{r}
{\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow} \\
{\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{OH})\right]^{2+}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)}
\end{array}\)

Reactions 1 and 2 are redox reactions. Reactions 3 and 5 are hydrolysis reactions. The reaction is a hydration reaction.

Hydrogen Warm-Up Type Questions

Question 1. Name the isotopes of hydrogen and state their mass ratio.
Answer: The three isotopes of hydrogen are— Protium \({ }_1^1 \mathrm{H}\) or H, Deuterium \({ }_2^1 \mathrm{H}\) or D, Tritium \({ }_3^1 \mathrm{H}\) or T Their mass ratio is protium: deuterium: tritium =1:2:3.

Question 2. What is the source of solar energy?
Answer: The main source of solar energy is the given nuclear fusion reaction
\(4{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+2{ }_{+1} e^0 \text { (positron) + Energy }\)

Question 3. Although Fe is placed above hydrogen in the electrochemical series, dihydrogen is not obtained by its reaction with nitric acid. Explain with reasons.
Answer: HNO3 being a strong oxidizing agent oxidizes
dihydrogen into the water and itself gets reduced to nitrogen dioxide \(\mathrm{Fe}+6 \mathrm{HNO}_3 \rightarrow \mathrm{Fe}\left(\mathrm{NO}_3\right)_3+3 \mathrm{NO}_2+3 \mathrm{H}_2 \mathrm{O}\)

Question 4. How can one prepare H2 gas from water by using a reducing agent?
Answer: Reaction between metals such as Na or. JC (strong reducing agents) and water produce hydrogen gas \(2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \uparrow\)

Question 5. Name two compounds, in one ofwhich hydrogen is in +1 and the other in -1 oxidation state.
Answer: In HC1, hydrogen is in a +1 oxidation state and in NaH it is in a -1 oxidation state

Question 6. Holli dihydrogen and carbon monoxide burn in the air with a blue flame. How will you distinguish between them?
Answer: Dihydrogen burns with a blue flame in the air to form water vapor, turning white anhydrous CuSO4 into hydrated copper sulfate (CuSO4-5H2O). However, carbon monoxide on combustion forms CO2 which does not bring about any change in CuSO4

Question 7. What characteristics do you expect from an electron-deficient and an electron-rich hydride to their structures?
Answer: Electron-deficient hydrides are electron acceptors, i.e., they act as Lewis acids whereas electron-rich hydrides are electron donors, i.e., Lewis bases. For example, B2H6 is a Lewis acid while NH3 is a Lewis base.

Question 8. Why the boiling point of HF is higher than that of other hydrogen halides?
Answer: Due to the formation of strong intermolecular hydrogen bonding, the boiling point of HF is higher than that of other hydrogen halides.

Question 9. How can you separate H2 or D2 from He?
Answer: Red hot palladium is cooled in an atmosphere of He mixed with H2 or D2. Consequently, a large amount of H2 or D2 gets adsorbed by palladium but not He. When palladium is heated, occluded H2 or D2 gels are liberated as free hydrogen or deuterium.

Question 10. Why ionic or salt-like hydrides are used to dry organic solvents?
Answer: Ionic or salt-like hydrides are used to dry organic solvents because they readily react with water to form the corresponding metal hydroxide along with the evolution of H2 8as- The solvent is then separated from the metallic hydroxide by distillation.

Question 11. Why concentration of D2O increase when electrolysis of water is carried out for a long time?
Answer: Electrolysis of H2O occurs at a faster rate than D2O because the bond dissociation energy of the O—H bond is greater than that of the O—D bond. So, electrolysis of ordinary water for a prolonged time results in an increase in concentration of D2O

Question 12. How would you prepare deuterium peroxide (D2O2)?
Answer: Deuterium peroxide (D2O2) can be prepared by the reaction between barium peroxide (BaO2) and deuterosulphuric acid (D2SO4) BaO2 + D2SO4 → BaSO4 + D2O2

Question 13. How will you prepare deuteroammonia (ND3) from N2?
Answer: Magnesium burns in nitrogen to produce magnesium nitride which further reacts with D2O to produce ND3 (deuteroammonia)

⇒ \(\begin{gathered}
3 \mathrm{Mg}+\mathrm{N}_2 \rightarrow \mathrm{Mg}_3 \mathrm{~N}_2 \\
\mathrm{Mg}_3 \mathrm{~N}_2+6 \mathrm{D}_2 \mathrm{O} \rightarrow 3 \mathrm{Mg}(\mathrm{OD})_2+2 \mathrm{ND}_3
\end{gathered}\)

Question 14. How will you prove thathypophosphorusacid (H3PO2) is a monobasic acid?
Answer: When H3PO2 is treated with D2O, only one of its hydrogen atoms is replaced by D. So, it can be said that only one H-atom remains attached to O-atom in hypophosphorous acid (H3PO2). Therefore, it is a monobasic acid

Question 15. Sodium chloride is less soluble in heavy water than ordinary water—why?
Answer: As the dielectric constant of D2O is less than that of H2O, NaCl (sodium chloride) is less soluble in heavy water than ordinary water.

Question 16. Explain why the water obtained after passing hard water through cation exchange resins is acidic.
Answer: When hard water is passed through an organic ion exchange resin, the water obtained is acidic because all the metal ions present in water are exchanged with H+ ions of the resin. As a result, the resulting water is free of cations and has a high concentration of H+ ions. So, the water turns blue litmus paper red.

Question 17. A sugar solution prepared in distilled water is passed successively through cation and anion exchange resins. What will be the taste ofthe collected water and why?
Answer: ion- exchange cannot remove sugar(non-electrolyte) from water. therefore, when a sugar solution is passed successively through cation and anion exchange resins after being collected tastes sweet.

Question 18. The hardness of water in a tube well is 300 ppm. What do you mean by this statement?
Answer: The statement means that in million parts by mass of the sample of water from the tube, the well contains salts causing its hardness which are equivalent to 300 parts by mass of calcium carbonate.

Question 19. Will the water obtained by passing hard water through anion exchange resin, form lather with soap? Why?
Answer: As the sample of water is not free from Ca2+ and Mg2+ ions, it will not form a lather with soap easily.

Question 20. A sample of water contains MgS04 and urea. How can they be eliminated easily?
Answer: They can be eliminated by a simple distillation method.

Question 21. It is better to preserve H2O2 in a polythene bottle than in a
glass bottle—why?
Answer: The decomposition of H2O2 is accelerated by the presence of glass, sunlight, and basic substances. So, H2O2 is preserved in a polythene bottle rather than a glass bottle.

Question 22. What do you understand by the expression ’30 volume H2O2 solution?
Answer: ’30 volume H2O2 solution’ means that 1 mL of that solution yields 30 mL of oxygen at STP as a result of its complete decomposition.

Question 23. what do you mean by 20% H2O2 solution?
Answer: 20% H2O2 solution means that momT. of that solution contains 20g of H2O2

Question 24. Calculate the percentage strength of6.588 volume H2O2
Answer: Percentage strength of solution= \(\frac{\text { volume strength } \times 34}{11.2 \times 10}\)

⇒ \(=\frac{6.588 \times 34}{11.2 \times 10}=1.99=\frac{6.588 \times 34}{11.2 \times 10}=1.99\)

Hydrogen Very Short Answer Type Questions

Question 1. Explain why concentrated HCI is not used in the laboratory preparation of H2 gas.
Answer: Concentrated HCI is not used for the laboratory preparation of dihydrogen because HCI, being highly volatile, gets mixed with dihydrogen.

Question 2. Write down the name and formula of a compound that on electrolysis produces dihydrogen at the anode.
Answer: Sodium hydride (NaH)

Question 3. What is syngas?
Answer: All mixtures of CO and H2 irrespective of their composition are called synthesis gas or syngas

Question 4. Which isotope of hydrogen is used as a tracer in organic reactions?
Answer: Deuterium 21D is usually used as a tracer in determining the mechanism of organic reactions.

Question 5. Explain why dihydrogen is not suitable for balloons.
Answer: As dihydrogen, the lightest substance known, is a highly inflammable gas, it is not suitable for balloons.

Question 6. Which bond between two atoms has the highest bond dissociation enthalpy?
Answer: Bond dissociation enthalpy ofthe H —H bond is highest.

Question 7. Explain why H2 is more reactive than D2.
Answer: This is because the H —H bond dissociation enthalpy is less than the D —D bond dissociation enthalpy.

Question 8. What change is expected to take place when vegetable oils are hydrogenated?
Answer: Carbon-carbon double bonds are converted to carbon- V carbon single bonds.

Question 9. Which isotope of hydrogen is used in nuclear rectors?
Answer: Deuterium (21H or D )

Question 10. Why are ionic hydrides used as solid fuels?
Answer: When heated, ionic hydrides decompose to evolve dihydrogen gas which ignites readily.

Question 11. The densities of ionic hydrides are greater than that of the metal from which they are formed—why?
Answer: This is because hydride ions (H) occupy the holes in the lattice ofthe metal without distorting the metal lattice.

Question 12. Give examples of two interstitial hydrides.
Answer: CuH and FeH.

Question 13. Which gaseous compound on treatment with dihydrogen produces methanol?
Answer: Carbon monoxide (CO).

Question 14. Give the chemical reaction that occurs when hydrogen is used as a rocket fuel.
Answer: \(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l)+286 \mathrm{~kJ}\)

Question 15. A sample of water containing KC1 does not behave as hard water, but a sample of water containing CaCI2 or MgCl2 behaves as hard water—why?
Answer: The potassium salt of soap is soluble in water and forms a lather while calcium or magnesium salt of soap is insoluble in water and does not form a lather.

Question 16. What is EDTA, a compound used to determine the hardness of water?
Answer: EDTA is the disodium salt of ethylenediamine tetraacetic acid[NaO2C(COOH)NCH2CH2N(C00H)COONa].

Question 17. Can distilled water be called deionized water?
Answer: Distilled water can be called deionized water because it does not contain any cations and anions.

Question 18. What is the difference between the water softened by the permit process and the water softened by the organic ion exchangers?
Answer: Although the water softened by the permit process contains no cation, it contains various anions (e.g., Cl-, SO- etc.). However, the water softened by the organic ion exchangers contains no cations and anions.

Question 19. What will be the hardness of a sample of water, 106 g of which contains i mol A12(SO4)3?
Answer: 50ppm

Question 20. What is Calgon?
Answer: Sodiumhexametaphosphate, Na2[Na4(PO3)g]

Question 21. Give the chemical formula of the permit.
Answer: Na2Al2Si2O8 x H2O.

Question 22. What is the main source of heavy water?
Answer: Ordinary water is the main source of heavy water.

Question 23. Can sea animals survive in distilled water?
Answer: Sea animals cannot survive in distilled water because distilled water contains no salt and dissolved oxygen.

Question 24. Although D2O resembles H2O chemically, it is a toxic substance—why?
Answer: Dilute solution of H2O2,

Question 25. What is the trade name of hydrogen peroxide used as an antiseptic?
Answer: Perhydrol.

Question 26. What is the strength in the normality of an ‘11.2 volume’ H2O2 solution?
Answer: 2(N) H2O2 solution.

Question 28. Name a compound that suppresses the decomposition of H2O2.
Answer: Acetanilide (PhNHCOCH3).

Question 29. H2O2 molecule has an open-book-like structure. What is the angle between the two pages of the book in the gas phase?
Answer: 111.50.

Question 30. Name an organic compound without peroxo bond that is used to manufacture H2O2.
Answer: 2-ethylanthraquinol

Question 25. why H2O2 is a better oxidant than water?
Answer: H2O2 is a better oxidant than water because H2O2 being unstable readily dissociates to form stable water molecules along with the evolution of O2 gas. 2H2O2 2H2O + O2 + Heat

Question 1. What is heavy water? Why is it so-called?
Answer: Deuterium oxide is commonly known as heavy water because its density is higher than that of ordinary water, i.e., it is heavier than ordinary water.

Question 2. A water sample contains 1 millimole of Mg2+ ion per liter. Calculate the hardness of the water sample in ppm units.
Answer: 1 millimole Mg2+=1 millimole of MgCl2 s 0.095 g of MgCL,.

1 1. or 1000 g or 103 g of water contains 0.1 g of MgCl2 106g of water contains 0.095 x 103 = 95g of MgCl2 The degree of the hardness of water is 95 ppm

Question 3. BaO2 is a peroxide but MnO2 is not a peroxide— explain?
Answer: There is peroxide linkage in the BaO2 molecule ( —O —O—). But there is no such bonding present in MnO, molecule (O —Mn=0). Thus MnO2 is not a peroxide.

Hydrogen Short Answer Type Questions

Question 1. Why do most of the reactions of H2 occur at much higher temperatures?
Answer: Due to the much higher bond dissociation enthalpy of the H—H bond, dihydrogen is quite stable and relatively inert at temperature. It dissociates into atoms at about 5000K. For this reason, most of the reactions of dihydrogen occur at much higher temperatures.

Question 2. What characteristics do you expect from electron-deficient hydrides to their structure and chemical reactivity?
Answer: Electron-deficient hydrides have less number of electrons in the valence shell of the central atom and so, their mononuclear units do not satisfy the usual Lewis octet rule. Due to a deficiency of electrons, these hydrides act as Lewis acids and form complex entities with Lewis bases such as NH3, H- ion, etc.

⇒ \(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NH}_3 \rightarrow\left[\mathrm{BH}_2\left(\mathrm{NH}_3\right)_2\right]^{+}\left[\mathrm{BH}_4\right]^{-} \text {; }\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{LiH} \rightarrow 2 \mathrm{Li}^{+}\left[\mathrm{BH}_4\right]^{-} \text {(Lithium borohydride) }\)

Question 3. Explain why it is harmful to bathe in heavy water and use it for drinking purposes.
Answer: Being a very hygroscopic substance, heavy water (D2O) absorbs water from the body and thereby damages body cells. Also, it retards some cellular processes such as mitosis, cell division, and various enzyme-catalyzed reactions. For these reasons, it is harmful to bathe in heavy water and use it for drinking purposes

Question 4. Explain why the thermal stability of H2O2 is verylow.
Answer: The bond dissociation enthalpy of the O —O bond presenting H2O2 molecule is very low (35kcal. mol-1) i.e., the bond is very weak. For this reason, the thermal stability of H2O2 is extremely low.

Question 5. How the presence of H- ions be confirmed in ionic hydrides?
Answer: In the molten state, ionic hydrides conduct electricity with the liberation of dihydrogen at the anode. This confirms the presence of hydride (H ) ions in them.

Question 6. How do you separate 2 allotropic forms of hydrogen?
Answer: Ordinary hydrogen is a mixture of 75% ofortho and 25% of para-isomer at room temperature. On passing through activated charcoal kept at 20K, the para-isomer is adsorbed leaving behind the ortho-isomer. From the charcoal surface, para-hydrogen can be released by reducing pressure.

Question 7. Mention the difference in chemical characteristics of the two hydrides obtained when hydrogen combines with two elements having atomic numbers 17 and 20.
Answer: The highly electronegative Cl atomhaving atomic number 17 combines with hydrogen to form the covalent hydride H —Cl. On the other hand, the highly electropositive Ca atom having atomic number 20 combines with hydrogen to form the ionic hydride CaH2

Question 8. Two samples of hard water contain the same cations, Ca2+ & Mg2+. One is marked as temporary and the other as permanent. In which respect do they differ?
Answer: The two samples of magnesium of water salts are different present. with The sample of watermarked as temporary hard water containing bicarbonates of Ca2+ and Mg2+ ions while the sample marked as permanent hard water containing chlorides and sulfates of Ca2+ and Mg2+ ions.

Question 9. Tube-well water, if left for some time, assumes a brownish turbidity—explain.
Answer: Tube-well water sometimes contains soluble ferrous bicarbonate [Fe(HCO3)2]. This compound, on aerial oxidation, is converted into brown ferric hydroxide, Fe(OH)3, which remains in water as colloidal suspension, and because of this, water assumes a brownish turbidity.

⇒\(4 \mathrm{Fe}\left(\mathrm{HCO}_3\right)_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \rightarrow \underset{\text { (brown) }}{4 \mathrm{Fe}(\mathrm{OH})_3}+8 \mathrm{CO}_2\)

Question 10. Write the reactions for the Preparation of H2O2 from two sodium salts and preparation of D2O2 from potassium persulphate
Answer: \(\begin{aligned}
& \mathrm{Na}_2 \mathrm{O}_2+2 \mathrm{NaH}_2 \mathrm{PO}_4 \rightarrow 2 \mathrm{Na}_2 \mathrm{HPO}_4+\mathrm{H}_2 \mathrm{O}_2 \\
& \text { Sodium dihydrogen Disodium hydrogen } \\
& \text { phosphate phosphate } \\
&
\end{aligned}\)

Question 11. Between deionized water and distilled water which one is more pure and why?
Answer: Both deionized and distilled water are free from ions. Yet distilled water is superior to deionized water in terms of purity. This is because deionized water contains a small amount of dissolved silica and CO2 along with some germs and organic impurities. However, distilled water prepared in a glass apparatus does not contain any impurities other than trace amounts of silica and CO2.

Question 12. Why is Na2O2 used for purifying air in submarines and crowded places?
Answer: Na2O2 reacts with CO2 of air to evolve O2 (2Na2O2 + 2CO2 →2Na2CO3+ O2). For this reason, Na2O2 is used for the purification of air in submarines and in crowded places.

Question 13. Comment on the reactions of dihydrogen with Chlorine Sodium and Copper (2) oxide.
Answer: Dihydrogen reduces chlorine (Cl) to chloride ion (Cl-) and itself gets oxidized to form H+ ions. These two ions (H+ and Cl-) share an electron pair between themselves to form a covalent molecule of hydrogen chloride (HC1)

⇒ \(\mathrm{H}_2(g)+\mathrm{Cl}_2(g) \rightarrow 2 \mathrm{HCl}(g)\)

Sodium reduces dihydrogen to hydride ion (H-) and itself gets oxidized to sodium ion (Na+). In this reaction, an electron gets completely transferred from Na toH thereby forming ionic sodium hydride (NaH).

⇒\(2 \mathrm{Na}(s)+\mathrm{H}_2(s) \xrightarrow{\Delta} 2 \mathrm{Na}^{+} \mathrm{H}^{-}(s)\)

Dihydrogen reduces copper(2) oxide to metallic copper while itself gets oxidized to form a covalent molecule of H2O

⇒ \(\stackrel{+2}{\mathrm{CuO}}(\mathrm{s})+\stackrel{0}{\mathrm{H}}{ }_2(s) \xrightarrow{\Delta} \stackrel{0}{\mathrm{Cu}}(\mathrm{s})+\stackrel{+1}{\mathrm{H}}_2 \stackrel{-2}{\mathrm{O}}(l)\)

Question 14. An ionic alkali metal hydride has a covalent character to some extent and it does not react with oxygen and chlorine. This hydride is used in the synthesis of another hydride. Write the formula of the hydride and what happens when it reacts with A12CI6.
Answer: Since the alkali metal hydride possesses sufficient covalent character, the hydride is of the smallest alkali metal, Li, i.e. the hydride is LiH. As LiH is quite stable, it does not react with oxygen and chlorine. Lithium hydride reacts with A12C16 to form lithium aluminum hydride (LiAH4) which is extensively used as a reagent in the synthesis of different organic compounds.

⇒ \(8 \mathrm{LiH}+\mathrm{Al}_2 \mathrm{Cl}_6 \longrightarrow 2 \mathrm{LiAlH}_4+6 \mathrm{LiCl}\)

Question 15. Sodium reacts with dlliydrogcn to form a crystalline ionic solid. It is non-volatile and a non-conductor of electricity. It also reacts vigorously with water to liberate H2 gas. Write the formula of the Ionic solid and give a reaction between the lids solid & water. What happens when the ionic solid in its molten stater Is electrolysed?
Answer: Sodium reacts with dihydrogen to form sodium hydride which is a crystalline ionic solid \(2 \mathrm{Na}+\mathrm{H}_2 \xrightarrow{\Delta} 2 \mathrm{Na}^{+} \mathrm{H}^{-}\)

Sodium hydride reacts with water as follows—

⇒ \(2 \mathrm{NaH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+2 \mathrm{H}_2\)

Sodium hydride, in its solid state, does not undergo electrolysis. However, in its molten state, undergoes electrolysis. However, in its molten state, NaH undergoes electrolysis to liberate dihydrogen (H2) at the anode and metallic sodium at the cathode

⇒ \(\mathrm{Na}^{+} \mathrm{H}^{-}(\text {molten }) \xrightarrow{\text { Electrolysis }} \underset{\text { Cathode }}{2 \mathrm{Na}(\text { molten })}+\underset{\text { anode }}{\mathrm{H}_2(g)}\)

Question 16. Why is seawater not used in boilers?
Answer: Sea water is hard water. It contains Mg(HCO3)2 and Ca(HCO3)2, which on boiling, form a hard heat-insulating thick layer or scale of MgCO3 and CaCO3 on the inner surface ofthe boiler. Consequently, much heat is required to raise the temperature of the boiler, and thus, fuel economy is adversely affected.

Again at much higher temperatures, the boiler scales and the metal of the boiler expand unequally. Due to such uneven expansion, cracks are formed on the scales. When hot water comes in contact with the hot metal surface of the boiler through these cracks, it is suddenly converted into steam. Due to the high pressure, thus developed, the boiler may burst leading to accidents.

Again, MgCl2 and MgSO4 present in seawater undergo hydrolysis to form HCI and H2SO4 which degrade the metallic component of the boiler. So, seawater is not used in boilers.

Question 17. The values of melting point, enthalpy of vaporization, and viscosity of H2O and D2O are given below: From the given data, determine which liquid has a greater magnitude of intermolecular forces of attraction
Answer: The magnitude of intermolecular forces of attraction depends on the magnitudes of melting point, enthalpy of vaporization, and viscosity of the liquid. As, these parameters have a higher value in the case of D2O, the magnitude of intermolecular forces of attraction is greater for D2O than H2O.

Question 18. How will you prepare heavy water from ordinary water? Explain its principle
Answer: Heavy water (D2O) is prepared by prolonged electrolysis of ordinary water. As water is not a good conductor of electricity, an alkaline solution of water [~0.5(N) NaOH] is used for electrolysis. The bond dissociation energy of the O —H bond is less than that of the O —D bond. So, electrolysis of H2O occurs at a faster rate and more easily than D2O. Consequently, the amount of D2O in ordinary water increases. Pure D20 is obtained when the amount of residual liquid decreases

Question 19. Can phosphorus form PHg with its outer electronic configuration of 3s23p3?
Answer: Phosphorus cannot form PH5 although it shows +3 and +5 oxidation states. Dihydrogen acts as a weak oxidizing agent due to the high bond dissociation enthalpy of H—Hbond (435.88 kJ-moH) and slightly negative electron-gain enthalpy (-73 kj mol-1). So, dihydrogen can oxidize phosphorus to a +3 oxidation state but not to its highest oxidation state of +5. Therefore, phosphorous can form only PH3 and not PH5.

Question 20. How will you prepare dinitrogen from HNO3?
Answer: Magnesium and manganese react with a very dilute solution of HN03(2%) to form dihydrogen.

⇒ \(\begin{aligned}
& \mathrm{Mg}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow \\
& \mathrm{Mn}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Mn}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow
\end{aligned}\)

Question 21. Do you think the hydrides of N, O, and F will have lower boiling points than the hydrides of their corresponding group members? State reasons?
Answer: The hydrides of the elements N, O, and F are NH3, H2O, and HF respectively. These hydrides are expected to have lower boiling points than that of their corresponding group members (PH3, H2S & HC1) when their masses are considered. However, because of the high electronegativity of N, O, and F, their hydrides undergo extensive hydrogen bonding (intermolecular). As a result, boiling points of NH3, H2O, and HF are much higher than the hydrides of their corresponding group members, i.e., PH3, H2S, and HC1 respectively.

Question 22. KF reacts with HF to form the compound, KF-2HF. Discuss the probable structure of the compound.
Answer: The h-bond in the HF molecule is very strong. When KF gets added to HF, one F” ion forms H — bond with two HF molecules to form H2F3 ion [F—-H —F—H —F —]“

Question 23. Calculate the amount of energy liberated due to combustion of 4g dihydrogen.
Answer: Amount of energy liberated due to combustion of 1 mol, i.e., 2g dihydrogen = 242 kj .mol-1.

⇒ \(\left[\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \Delta \mathrm{H}=-242 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\right]\) Amount of energy liberated due to 4g dihydrogen = 242×2 = 484 kj-mol-1

Question 24. Under what conditions, water reacts with calcium cyanamide and what are the products formed due to this reaction?
Answer: Superheated steam reacts with calcium cyanamide under high pressure to form ammonia gas and calcium carbonate as a result of hydrolysis

⇒ \(\mathrm{CaNCN}\left(\text { Calcium cyanamide) }+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CaCO}_3+2 \mathrm{NH}_3\right.\)

Question 2. Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Answer: In nature, there are three isotopes of hydrogen. These are
protium \(\left({ }_1^1 \mathrm{H}\right)\) deuterium \(\left[{ }_1^2 \mathrm{H} \text { or } \mathrm{D}\right]\) and tritium \(\left[{ }_1^3 \mathrm{H} \text { or } \mathrm{T}\right]\).
The mass ratio of \({ }_1^1 \mathrm{H},{ }_1^2 \mathrm{H} \text { and }{ }_1^3 \mathrm{H}\) is 1: 2: 3.

Question 3. Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Answer: A hydrogen atom has only one electron and thus, it has one electron less than the stable configuration of the nearest noble gas helium. Thus, to achieve a stable configuration it shares its single electron with the electron of other H -atoms to form a stable diatomic molecule (H2 ).

Question 6. Complete the following reactions

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{M}_m \mathrm{O}_0(\mathrm{~s}) \xrightarrow{\Delta}\)

⇒ \(\mathrm{CO}(g)+\mathrm{H}_2(g) \xrightarrow[\text { catalyst }]{\Delta}\)\(\mathrm{C}_3 \mathrm{H}_8(g)+3 \mathrm{H}_2 \mathrm{O}(g) \xrightarrow[\text { catalyst }]{\Delta}\) \(\mathrm{Zn}(s)+\mathrm{NaOH}(a q) \xrightarrow{\text { heat }}\)
Answer: \(o \mathrm{H}_2(\mathrm{~g})+\mathrm{M}_m \mathrm{O}_o(s) \xrightarrow{\Delta} m \mathrm{M}(s)+o \mathrm{H}_2 \mathrm{O}(l)\)

\(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2(\mathrm{~g}) \xrightarrow[\text { catalyst }]{\Delta} \mathrm{CH}_3 \mathrm{OH}(l)\) \(\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \xrightarrow{\mathrm{Nl}, 1270 \mathrm{~K}} 3 \mathrm{CO}(\mathrm{g})+7 \mathrm{H}_2(\mathrm{~g})\)

⇒ \(\mathrm{Zn}(s)+2 \mathrm{NaOH}(a q) \xrightarrow{\Delta} \underset{\text { Sodium zincate }}{\mathrm{Na}_2 \mathrm{ZnO}_2(a q)}+\mathrm{H}_2(g)\)

Question 7. Discuss the consequences of high enthalpy of the H —H bond in terms of the chemical reactivity of dihydrogen.
Answer: The 2 —2 bond length is very small (74 pm) because the H-atom has a very small atomic size. Consequently, the H—H bond dissociation endialpyis very high (435.9 kj-mol-1) which makes hydrogen completely inert at ordinary temperature. However, at higher temperatures, the H—H bond undergoes dissociation in the presence of a catalyst to form hydrides with metals & non-metals.

Hydrogen Solved WBCHSE Scanner

 

Question 8. What characteristics do you expect from an electron-deficient hydride concerning its structure and chemical reactions?
Answer: There is an insufficient number of electrons in the valence shell of the central atom. So, in this type of hydride, the valence shell of the central atom does not have a complete octet configuration. For example, in BH3, the valence shell of B has six electrons and the hydrides are trigonal planarian shape.

Due to electron deficiency, this type of hydrides act as Lewis acids, i.e., they accept electron pairs. For example, H3B + NH3→ H3B→NH3

To compensate for the electron deficiency, the hydrides form dimers, trimers, and polymers and attain stability.

For example, B2H6, B4H10, (A1H3)b

Electron-deficient hydrides are extremely reactive. They easily react with both metals and non-metals.

Question 9. Do you expect the carbon hydrides of the (CMH2n + 2) to act as ‘Lewis’ acid or base? Justify your answer.
Answer: The hydrides of carbon of the type CnH2n + 2 are electron-precise hydrides, i.e., they have an exact number of electrons in the valence shell of the central atom to write their conventional Lewis structures. Therefore, they do not tend to either gain or lose electrons and consequently, they do not act as Lewis acids or bases.

Question 10. What do you understand by the term “nonstoichiometric hydrides”? Do you expect this type of hydride to be formed by alkali metals? Justify your answer.
Answer: Hydrides of d and f-block elements that are deficient in hydrogen and in which the ratio of the metal to hydrogen is fractional are called non-stoichiometric hydrides.

The alkali metals, being highly reducing in nature, transfer their lone pair of electrons to the H-atom forming Hions. Since a hydride ion H- is formed by the complete transfer of an electron, the ratio of metal to hydrogen is always fixed in these hydrides and their compositions correspond to a simple whole-number ratio. Hence, the alkali metals form only stoichiometric hydrides.

Question 11. How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Answer: Some transition metals like palladium (Pd), Platinum (Pt), etc., adsorb a large volume of hydrogen on their surface (as Hatoms) forming hydrides. Due to the inclusion of H-atoms, the metal lattice expands and thus becomes less stable.

Therefore, when such metal hydrides are heated, they decompose to release dihydrogen and change back to a metallic state in finely divided form. The dihydrogen evolved in this manner and can be used as a fuel. Thus, the transition metals or their alloys can be used to store as well as for transportation of hydrogen used as a fuel.

Question 12. Among NH3 H2O and HF, which would you expect to have the highest magnitude hydrogen bonding & why?
Answer: The strength of a hydrogen bond depends upon the atomic size mid the electronegativity of the atom to which the 2-atom Is covalently linked. Smaller size and higher electronegativity of the other atom favor the formation of stronger H -bonding and that Is due to an increase In the magnitude of bond polarity. Among N, I’, and O atoms, V has the lowest atomic size and the highest electronegativity. Hence the 2 — F bond is maximum polar and as a result, it will have the highest magnitude of H-bonding.

Question 13. Saline hydrides are known to react with water violently producing fire. Can CO2, a well-known fire extinguisher, be used in this case? Explain.
Answer: The saline hydrides (Example; NaH, CaH2, etc.), react with water violently to yield the corresponding metal hydroxides with the evolution of H2 gas. The liberated H2 gas undergoes spontaneous combustion causing fire and this is because of the highly exothermic nature of the combustion reaction.

⇒ \(\begin{gathered}
\mathrm{NaH}(s)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{NaOH}(a q)+\mathrm{H}_2(\mathrm{~g}) ; \\
2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{gathered}\)

In this case, CO2 cannot used as a fire extinguisher because its gels are reduced by the hot metal hydride to form formate ions.

⇒ \(\mathrm{NaH}+\stackrel{+1}{\mathrm{CO}_2} \rightarrow \stackrel{+1+2}{\mathrm{HCOONa}}\)

 

Question 19. Is demineralized or distilled water useful for drinking purposes? If not, how can it be made useful?
Answer: Although very pure, demineralized or distilled water is not useful for drinking purposes and this is because it does not contain even useful minerals. To make it useful for drinking purposes, useful minerals in proper amounts should be added to demineralized or distilled water.

Question 20. Describe the usefulness of water in biosphere and biological systems
Answer: Water is extremely necessary for the existence of different forms of life on Earth. It constitutes about 65-70% of the body weight of plants and animals. Since water has a high specific heat, thermal conductivity, surface tension, dipole moment, and dielectric constant compared to other liquids, it plays a vital role in the biosphere.

Due to the high heat of vaporization of water, it can moderate the body temperature. Water also indirectly helps in climate control.

In the living body, water helps to transport minerals, nutrients, and enzymes to the different parts of the body and also affects the process of metabolism. Water is a vital component in the process of photosynthesis. Therefore, waterways a significant role in the biosphere.

Question 21. Knowing the properties of H2O and D2O, do you think that D2O can be used for drinking purposes?
Answer: Drinking heavy water (D2O) is harmful to the human beings. Heavy water being highly hygroscopic, absorbs water from different parts of the body. As a result, the body cells may get destroyed. Apart from this, heavy water slows down different biochemical reactions such as mitosis, cell division, etc.

Question 22. What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Answer: Hydrolysis refers to the reaction of salt with water to form acidic or basic solutions. For example:

⇒ \(
\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 2\left[\mathrm{Na}^{+}+\mathrm{OH}^{-}\right]+\mathrm{H}_2 \mathrm{CO}_3 \text {; }
(basic solution)\)

⇒ \(
\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons\left[\mathrm{H}^{+}+\mathrm{Cl}^{-}\right]+\mathrm{NH}_4 \mathrm{OH}
(acidic solution)\)

Hydration, on the other hand, refers to the addition of H2O to ions or molecules to form hydrated ions or hydrated salts. For example

⇒ \(\underset{\text { Salt }}{\mathrm{NaCl}(s)+\mathrm{H}_2 \mathrm{O}} \rightarrow \underbrace{\mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)}_{\text {Hydrated ions }}\)

⇒ \(\underset{\text { Anhydrous salt (colourless) }}{\mathrm{CuSO}_4(s)+5 \mathrm{H}_2 \mathrm{O}(l)} \rightarrow \underset{\text { (Hydrated salt blue) }}{\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(s)}\)

Question 23. What do you expect the nature of hydrides to be if formed by elements of atomic numbers 15, 19, 23, and 44 with dihydrogen? Compare their behavior towards water.
Answer: The element with Z = 15 is non-metal phosphorus and hence it forms the covalent hydride PH3.

The element with Z = 19 is the alkali metal potassium and hence it forms the saline or ionic hydride K+HT.

The element with Z = 23 is the transition metal vanadium of group-3 and hence it forms the interstitial hydride (VH1.6).

The element with Z = 44 is the transition metal ruthenium (Ru) belonging to group 8. It does not form anhydride (hydride gap).

Only the ionic hydride, KH reacts with water evolving dihydrogen.

Question 24. Do you expect different products in solution when aluminum (3) chloride and potassium chloride are treated separately with Normal water Acidified water and Alkaline water? Write equations wherever necessary.
Answer: In normal water: KC1, being a salt of a strong acid and strong base, does not undergo hydrolysis in normal water. It simply dissociates to form K+(aq) and Cl-(aq) ions.

⇒ \(\mathrm{KCl}(s) \xrightarrow{\text { water }} \mathrm{K}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

On the other hand, A1C13, being a salt of a weak base Al(OH)3 and a strong acid HC1, undergoes hydrolysis giving an acidic solution

⇒ \(\mathrm{AlCl}_3(s)+3 \mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{Al}(\mathrm{OH})_3(s)+3 \mathrm{H}^{+}(a q)+3 \mathrm{Cl}^{-}(a q)\)

In acidified water: The H+ ions react with Al(OH)3 to form Al3+(aq) ions and H20. Therefore, in acidic water, A1C13 exists as A13+(aq) and Cl-(aq) ions.

⇒ \(\mathrm{AlCl}_3(s) \xrightarrow{\text { acidified } \mathrm{H}_2 \mathrm{O}} \mathrm{Al}^{3+}(a q)+3 \mathrm{Cl}^{-}(a q)\)

Potassium chloride, KC1 simply dissociates to give K+(aq) and Cl-(aq) ions

⇒ \(\mathrm{KCl}(s) \xrightarrow{\text { acidified water }} \mathrm{K}^{+}(a q)+\mathrm{Cl}^{-}(a q)\).

In alkaline water: A1(OH)3 reacts with OH ions to form soluble tetra hydroxo aluminate complex or meta-aluminateion (A1O-2).

Question 25. What do you understand by the terms: hydrogen economy hydrogenation ‘syngas’?
Answer: Hydrogenation is the process of addition of hydrogen to unsaturated organic compounds to form saturated organic compounds. For hydrogenation of oils.

Hydrogen Higher Order Thinking Skill Questions

Question 1. What is ‘hydrogenite’? Mention its use.
Answer: A mixture of silicon, caustic soda, and slaked lime is called hydrogenite. Dihydrogen can be prepared by heating hydrogenate with water.

⇒ \(\begin{aligned}
& \mathrm{Si}+2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Na}_2 \mathrm{SiO}_3+\mathrm{H}_2 \uparrow \\
& \mathrm{Si}+\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CaSiO}_3+2 \mathrm{H}_2 \uparrow
\end{aligned}\)

Question 2. What is denoted by [HgO4]+ ion? Explain
Answer: High charge density and high hydration energy of the H+ ion (proton) cause the H+ ion to remain solvated to form a hydroxonium ion or hydronium ion (H3O+). Hydronium ion again forms hydrogen bonds with three water molecules and remains solvated. Therefore, in water, a proton forms [H3O(H2O)3]+ or [H9O4]+ ion.

Hydrogen High Charge Density

Question 3. Pure para-hydrogen is available but not pure ortho¬ hydrogen. Explain.
Answer: At ordinary temperature, ordinary hydrogen is a mixture of 75% of ortho and 25% of para-isomer. With the decrease in temperature, the amount of ortho-hydrogen decreases while that of para-hydrogen increases. At 20K, pure para-hydrogen is obtained. As para-hydrogen is more stable, it is found in the pure form. However, if the temperature is increased, the amount of ortho-isomer of dihydrogen increases but at 400K or above, the ratio of ortho-and para-isomer is fixed (3: 1). Therefore, pure para-hydrogen is available but not pure ortho-hydrogen.

Question 4. How many hydrogen-bonded water molecule(s) are associated with CuSO4-5H2O?
Answer: Only one molecule of water, which remains outside the brackets (coordination sphere) is linked by a hydrogen bond to SO2-4 as shown below. The remaining four water molecules are associated with Cu2+ ions by coordination bonds.

Hydrogen Hydrogen Bonded water molecule

Question 5. Write down the reaction between H2O2 and hydrazine (NH2NH2) in the presence of a Cu(2) catalyst. Mention the use of this reaction.
Answer: A highly concentrated solution (about 40%) of H2O2 (called high test peroxide) oxidizes hydrazine (N2H4) in the presence of Cu(II) into nitrogen gas, itself being oxidized to water (steam)

⇒ \(\stackrel{-2}{\mathrm{~N}_2} \mathrm{H}_4(l)+2 \mathrm{H}_2{ }^{-1} \mathrm{O}_2(l) \xrightarrow[\text { catalyst }]{\mathrm{Cu}(\mathrm{II})} \stackrel{0}{\mathrm{~N}}(\mathrm{~g})+4 \mathrm{H}_2 \stackrel{-2}{\mathrm{O}}(\mathrm{g})+\text { heat }\)

The reaction is highly exothermic and is accompanied by a large increase in volume which in turn generates high pressure. Due to this, the reaction is employed for propelling rockets.

Hydrogen Multiple Choice Questions

Question 1. At absolute zero—

  1. Only para-hydrogen exists
  2. Only ortho-hydrogen exists
  3. Both ortho- and para-hydrogen exist
  4. Neither para- nor ortho-hydrogen exists

Answer: 1. Only para-hydrogen exists

Question 2. In which of the following reaction dihydrogen acts as an oxidizing agent—

  1. F2 +H2→2H
  2. Cl2 + H2→2hc1
  3. N2 + 3H2→2nH3
  4. 2Na + H2→2naH

Answer: 4. 2Na + h2→2naH

Question 3. Which of the following halogens has the least affinity towards hydrogen—

  1. H
  2. Cl2
  3. Br2
  4. F2

Answer: 1. H

Question 4. Which of the following compounds on electrolysis produces hydrogen—

  1. Dil. H2sO4
  2. Dil. Solution of NaOH
  3. Ba(oh)2 solution
  4. Koh solution

Answer: 3. Ba(oh)2 solution

Question 5. Thermal stability of gr.-15 Hydrides follows the order—

  1. Ash3 > ph3 > nh3 > sbh3 > bih3
  2. Nh3 > ph3 > ash3 > sbh3 > bih3
  3. Nh3 > ash3 > ph3 > sbh3 > bih3
  4. Bih3 > sbh3 > ash3 > ph3 > nh3

Answer: 1. Ash3 > ph3 > nh3 > sbh3 > bih3

Question 6. The correct order of vaporization enthalpy of the following hydride is—

  1. Nh3<ph3<ash3
  2. Ash3<ph3<nh3
  3. Ph3<ash3<nh3
  4. Nh3<ash3<ph3

Answer: 3. Ph3<ash3<nh3

Question 7. Interstitial hydrides are formed by—

  1. 5-Block elements
  2. P-block elements
  3. R f-block elements
  4. Intert gas elements

Answer: 3. Rf-block elements

Question 8. The correct descending order of thermal stability of alkali metals hydrides is—

  1. Lih > nah > kh > rbh > csh
  2. Csh > rbh > kh > nah > lih
  3. Nah > kh > lih > csh > rbh
  4. Csh > lih > kh > nah > rbh

Answer: 1. Lih > nah > kh > rbh > csh

Question 9. Solubility of nacl in the solvents h3O and d2O is—

  1. Equal in both
  2. More in d2O
  3. More in H2O
  4. Only in H2O

Answer: 3. More in D20

Question 10. Degree of hardness of 1 sample water containing 0.002 mol mgsO4 is—

  1. 20 Ppm
  2. 200 Ppm
  3. 2000 Ppm
  4. 120 Ppm

Answer: 2. 200 Ppm

Question 11. Which of the following reacts with water to produce electron-precise hydrides—

  1. Ca3p2
  2. A14c3
  3. Mg3n2
  4. None of these

Answer: 2. A14c3

Question 12. Which of the following couples reacts with water to produce the same gaseous product—

  1. K and kO2
  2. Ca and cah2
  3. Na and na2O2
  4. Ba and baO2

Answer: 2. Ca and cah2

Question 13. Which of the following compounds contain free hydrogen—

  1. Water
  2. Marsh gas
  3. Water gas
  4. Acid

Answer: 3. Water gas

Question 14. Which of the following reacts with metallic sodium to produce hydrogen—

  1. CH4
  2. C2H6
  3. C2H4
  4. C2H2

Answer: 4. C2h2

Question 18. Semi-water gas is—

  1. Co + H2 + N2
  2. H2 + ch4
  3. Co + H2 + O2
  4. Co + H2

Answer: 1. Co + H2 + N2

Question 10. Which of the following metals does not react with cold water but liberates H2 with boiling water—.

  1. Na
  2. K
  3. Pt
  4. Fe

Answer: 2. K

Question 17-. Volume of ’10 volume’ h2O2 required to convert 0.01 mol PBS into pbsO4 is—

  1. 11.2 ml
  2. 22.4 ml
  3. 33.6 ml
  4. 44.8 ml

Answer: 4. 44.8 ml

Question 18. On dilution of h2O2, the value of dielectric constant—

  1. Increases
  2. Remains same
  3. Decreases
  4. None of these

Answer: 1. Increases

Question 19. By which of the following water gets oxidized to oxygen—

  1. C1O2
  2. KmnO4
  3. H2O2
  4. F2

Answer: 4. F2

Question 20. Which of the following does not get oxidized by h2O2 —

  1. Na2sO3
  2. Pbs
  3. Ki
  4. O3

Answer: 4. O3

Question 21. The temperature at which the density of d2O is maximum is—

  1. 9°C
  2. 11.5°c
  3. 15.9°c
  4. 20°C

Answer: 2. 11.5°c

Question 22. Which of the following undergoes disproportionation reaction with water—

  1. SO3
  2. F2
  3. Cl2
  4. N2

Answer: 3. Cl2

Question 23. The non-inflammable hydrides—

  1. Nh3
  2. Ph3
  3. Ash3
  4. Sbh3

Answer: 1. Nh3

Question 24. The Triplepoint of water is—

  1. 203K
  2. 193K
  3. 273K
  4. 373K

Answer: 3. 273K

Question 25. The process by which hydrogen is prepared by the reaction of silicon, iron alloy, and NaOH is—

  1. Wood process
  2. Haber’s process
  3. Silicol process
  4. Bosch process

Answer: 3. Silicol process

Question 26. An element reacts with hydrogen to form a compound a, which on reaction with water liberates hydrogen again. The element is—

  1. Cl
  2. Cs
  3. Se
  4. N2

Answer: 2. Cs

Question 27. Only one element of which of the following groups forms metal hydride—

  1. Gr-6
  2. Gr-7
  3. Gr-8
  4. Gr-9

Answer: 1. Gr-6

Question 28. An acidic solution of which of the following turns orange in the presence of h2O2

  1. BaO2
  2. Na2O2
  3. TiO2
  4. PbO2

Answer: 3. TiO2

Question 29. In the following reaction the isotopic oxygens—

2MnO4 + 3h2O8 2mnO2 + 3O2 + 2H2O + 2Oh-

  1. Both get converted into O2
  2. Both get converted into Oh-
  3. Both get converted into mnO2
  4. One of them gets converted to O2, another to mnO2

Answer: 1. Both get converted into O2

Question 30. X on electrolysis produces y which on vacuum distillation produces h202. The numbers of peroxo linkage present in x and y are—

  1. 1,1
  2. 1, 2
  3. 0 > 1
  4. 0, 0

Answer: 3. 0 > 1

Question 31. The compound which on electrolysis in its molten or liquid state liberates hydrogen at anode is—

  1. Noah
  2. Cah2
  3. Hc1
  4. H2O

Answer: 2. Cah2

Question 32. Which of the following couples exhibit the maximum isotope effect—

  1. \({ }_1^1 \mathrm{H},{ }_1^2 \mathrm{D}\)
  2. \({ }_8^{16} \mathrm{O},{ }_8^{18} \mathrm{O}\)
  3. \({ }_{17}^{35} \mathrm{Cl},{ }_{17}^{37} \mathrm{Cl}\)
  4. \({ }_6^{12} \mathrm{C},{ }_6^{14} \mathrm{C}\)

Answer: 1. \({ }_1^1 \mathrm{H},{ }_1^2 \mathrm{D}\)

Question 33. Which of the following emits by tritium—

  1. Neutron
  2. Y-ray
  3. 3-Particle
  4. O-particle

Answer: 3. 3-Particle

Question 34. Oxidation of benzene by H2O9 in the presence of ferrous sulfate produces—

  1. Phenol
  2. Cyclohexane
  3. Anisole
  4. Benzaldehyde

Answer: 1. Phenol

Question 35. The oxidation state of cr in the product obtained by the reduction of k2cr2O7 by atomic hydrogen is—

  1. +6
  2. +2
  3. 0
  4. +3

Answer: 4. +3

Question 38. Which of the following does not get reduced by h2 in its aqueous solution—

  1. Cu2+
  2. Fe3+
  3. Zn2+
  4. Ag+

Answer: 1. Cu2+

Question 37. Which of the following compounds has a similar odor as that of h2O2

  1. Caustic soda
  2. Chloroform
  3. Alcohol
  4. Nitric acid

Answer: 4. Nitric acid

Question 38. Which of the following compounds reacts with atomic hydrogen to form formaldehyde—

  1. Co
  2. CO2
  3. Ch4
  4. C2H2

Answer: 1. Co

Question 39. Which of the following isotopes of hydrogen is the most reactive—

  1. \({ }_1^1 \mathrm{H}\)
  2. \({ }_2^1 \mathrm{H}\)
  3. \({ }_3^1 \mathrm{H}\)
  4. All the isotopes are equally reactive

Answer: 1. \({ }_1^1 \mathrm{H}\)

Question 40. When equal amounts of zn are allowed to react separately with excess h2SO4 and excess NaOH, then the ratio of the volumes of hydrogen produced for the first and the second case respectively is—

  1. 1:2
  2. 2:1
  3. 4:9
  4. 1:1

Answer: 4. 1:1

Question 41. Which of the following hydrides of s-block elements have a polymeric structure—

  1. Lih
  2. Beh2
  3. Nah
  4. Mgh2

Answer: 2. Beh2

Question 42. Which of the following statements is true—

  1. If z= 15, the element forms a covalent hydride
  2. If z = 23, the element forms an ionic hydride
  3. If z= 19, the element forms an ionic hydride
  4. If z = 44, the element forms a metalic hydride

Answer: 1. If z= 15, the element forms a covalent hydride

Question 43. Which of the following hydrides are polynuclear hydrides—

  1. Nah
  2. C3h8
  3. N2h4
  4. H2

Answer: 2. C2H8

Question 44. Which of the following statements is correct—

  1. Metallic hydrides are hydrogen-deficient
  2. Metallic hydrides are conductors of heat and electricity
  3. Ionic hydrides in their solid state do not conduct electricity
  4. Ionic hydrides on electrolysis in their molten state produce h2 at the cathode.

Answer: 1. Metalic hydrides are hydrogen deficient

Question 45. Which of the following ions get exchanged with the na+ ion of zeolite when the zeolite is added to the hard water—

  1. H+ ion
  2. Ca2+ ion
  3. SO4 ion
  4. Mg2+ ion

Answer: 2. Ca2+ ion

Question 46. Which of the following reactions are neurolysis—

  1. 2Na + 2D2O → 2NaOD + D2
  2. AIcI3 + 3D2O →AI(OD)3 + 3DCI
  3. Ca + 2D2O→Ca(OD)2 + D2
  4. Fe2(SO4)3 + 6D2O→2Fe(OD)3 + 3D2SO4

Answer: 2. AIcI3 + 3D2O →AI(OD)3 + 3DCI

Question 47. Which of the following reactions are redox reactions —

  1. H2O+SO2→H2SO
  2. CaO + H2O → Ca(OH)2
  3. 2Na + 2H2O → 2NaOH + H2
  4. 2F2 + 2H20 →O2 + 4HF

Answer: 3. 2Na + 2H2O → 2NaOH + H2

Question 40. In which of the following reactions H2O2 acts as a reductant—

  1. CgHg + H2O2 C6H5OH + H2O
  2. PbS + 4H2O2→PbSO4 + 4H2O
  3. NaOBr +H2O2→NaBr + H2O +O2
  4. 2MnO4+6H+→+ 5H2O2→2Mn2+ + 8H2O + 5O2

Answer: 3. NaOBr +H2O2→NaBr + H2O +O2

Question 49. Which of the following properties are the same for metal and its hydride—

  1. Hardness
  2. Metallic lustre
  3. Electrical conductance
  4. Magnetic property

Answer: 1. Hardness

Question 50. The correct orders are—

  1. H2 < D2 < T2 : boiling point
  2. H2 < D2 < T2 : freezing point
  3. H2 < D2 < T2 : latent heat of vaporisation
  4. T20 > H2O > D20 : dissociation constant

Answer: 1. H2 < D2<T2 : boiling point

Question 51. Which of the following reacts with zinc to produce hydrogen gas—

  1. Dil. Hc1
  2. Hot naoh solution
  3. Cold water
  4. Cone. H2SO4

Answer: 1. Dil. Hc1

Question 52. Which of the following properties have a greater magnitude in D2O than that in h2O —

  1. Viscosity
  2. Surface tension
  3. Dielectric constant
  4. Latent heat of vaporization

Answer: 1. Viscosity

Question 53. Which of the following metal hydrides get reduced by hydrogen

  1. Cuo
  2. Pb3O4
  3. Na2O2
  4. MgO

Answer: 1. Cuo

Question 54. Multimolecular covalent hydrides of 5-block are—

  1. Lih
  2. Beh2
  3. Nah
  4. Mgh2

Answer: 2. Beh2

Question 55. The oxidation numbers of the most electronegative element in the product were obtained due to the reaction between baO2 and dil. H2sO4 are—

  1. -1
  2. -2
  3. 0
  4. +1

Answer: 1. -1

Question 55. Which of the following compounds decreases the rate of decomposition of h2O2

  1. Co(nh2)2
  2. PbNHCOCH3
  3. MnO2
  4. (Cooh)2

Answer: 1. Co(nh2)2

Question 57. Which of the following produces h2O2 on hydrolysis—

  1. Pernitric acid
  2. Perchloric acid
  3. Perdisulphuric acid
  4. Caro’s acid

Answer: 1. Pernitric acid

Question 58. Choose the correct statements—

  1. The concentration of 20 volume h2O2 solution is 60.7g
  2. Volume strength of 2(n)h2O2 solution is 15
  3. Volume strength of 2(n)h2O2 solutions 11.2
  4. The concentration of 20 volume h2O2 solution is 50.7g

Answer: 1. Concentration of 20-volume h2O2 solution is 60.7g

Question 59. Choose the correct alternative—

  1. A mixture of HCI and hair is formed when chlorine reacts with cold water
  2. Arrange color of the k2cr2O7 solution turns blue when it reacts with H2O2
  3. Under low pressure, isopropyl alcohol reacts with a small amount of H2O2 to produce formaldehyde
  4. Hydrolith produces black coloured product when it reacts with pbs04

Answer: 1. Mixture of hc1 and hair is formed when chlorine reacts with cold water

Question 60. Which of the following alternatives is not true—

  1. Correct order of reactivity of H2 towards the halogens is: cl2 > br2 > i2 > f2
  2. The concentration of H2O2 used in rockets is 90%
  3. H2 gets more readily absorbed on the surface of pt-metal than D2
  4. Conversion of atomic hydrogen into molecular hydrogen is an exothermic process

Answer: 1. Correct order of reactivity of H2 towards the halogens is: cl2 > br2 > i2> f2

Question 61. The normality of ’30 volume’ of H2O2 is—

  1. 2.678 (N)
  2. 5.336 (N)
  3. 8.034 (N)
  4. 6.685 (N)

Answer: 2. 5.336 (N)

Volume strength = 5.6 x normality or, 30= 5.6 x normality or, noemality \(=\frac{30}{5.6} \mathrm{~N}=5.357 \mathrm{~N}\) The normality of 30 volumes of H2O2 is 5.357N.

Question 62. When H2O2 is shaken with an acidified solution of K2Cr2O7 in the presence of ether, the ether layer turns blue due to the formation of— 2-

  1. Cr2O3
  2. Cr2(SO4)3
  3. CrO4
  4. CrO5

Answer: 4. CrO5

Question 63. A commercial sample of H2O2 is labeled as 10V. Its % strength is nearly—

  1. 3
  2. 6
  3. 9
  4. 12

Answer: 1. 6

Question 64. In an aqueous alkaline solution, two-electron reduction of HO¯² given—

  1. HOθ
  2. H2O
  3. O2
  4. 2

Answer: 1. HOθ

Hydrogen Reduction

Question 65. Which statement is not correct for ortho- and para-hydrogen—

  1. They have different boiling points
  2. Ortho-form is more stable than para-form
  3. They differ in their nuclear spin
  4. The ratio of ortho to para-hydrogen changes with a change in temperature

Answer: 2. Ortho-form is more stable than para-form

At normal or high temperatures, ortho-hydrogen is more stable than para-hydrogen but at very low temperatures para-hydrogen is more stable than orthohydrogen.

Question 66. At room temperature, the reaction between water and fluorine produces

  1. HF and H2O2
  2. HF,O2 and F2O2
  3. F, O2 and H+⊕
  4. HOF and HF

Answer: 3. F, O2 and H+⊕

⇒ \(\mathrm{F}_2+\mathrm{H}_2 \mathrm{O} \xrightarrow{\text { room temperature }} \underset{\substack{\downarrow \\ \mathrm{H}^{+}+\mathrm{F}^{-}}}{\mathrm{HF}}+\mathrm{O}_2\)

Question 67. Very pure hydrogen (99.9%) can be made by which of the following processes—

  1. Mixing natural hydrocarbons of high molecular weight
  2. Electrolysis of water
  3. Reaction of salt-like hydrides with water
  4. Reaction of methane with steam

Answer: 3. Reaction of salt-like hydrides with water

Question 68. In which of the following reactions, H2O2 acts as a reducing agent—

  1. \(\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{H}^{+}+2 e \rightarrow 2 \mathrm{H}_2 \mathrm{O}\)
  2. \(\mathrm{H}_2 \mathrm{O}_2-2 e \rightarrow \mathrm{O}_2+2 \mathrm{H}^{+}\)
  3. \(\mathrm{H}_2 \mathrm{O}_2+2 e \rightarrow 2 \mathrm{OH}^{-}\)
  4. \(\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-}-2 e \rightarrow \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}\)

Choose The Correct Option

  1. 2,4
  2. 1,2
  3. 3,4
  4. 1,3

Answer: 1. 2,4

Question 69. From the following statements regarding H202, choose the incorrect statement—

  1. It has to be stored in plastic or wax-lined glass bottles in dark
  2. It has to be kept away from dust
  3. It can act only as an oxidizing agent
  4. It decomposes on exposure to light

Answer: 3. It can act only as an oxidizing agent

Question 70. Which one of the following statements about water is false—

  1. Water can act both as an acid and as a base
  2. There is extensive intramolecular hydrogen bonding in the condensed phase
  3. Ice formed by heavy water sinks in normal water
  4. Water is oxidized to oxygen during photosynthesis

Answer: 2. There is extensive intramolecular hydrogen bonding in the condensed phase

Water molecules are associated with the formation of intermolecular hydrogen bonding

Question 71. Hydrogen peroxide oxidises [Fe(CN)6]4- to [Fe(CN)6]3- in acidic medium but reduces [Fe(CN)6]3- to [Fe(CN)6]4- in alkaline medium. The other products formed are, respectively—

  1. H2O and (H2O + O2)
  2. H2O and (H2O + OH-)
  3. (H2O + O2) and H2O
  4. (H22O + O2) and (H2O + OH-)

Answer: 1. H2O and (H2O + O2)

⇒ \(\left.2\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}+\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{H}^{+} \longrightarrow \longrightarrow \mathrm{Fe}(\mathrm{CN})_6\right]^{3-}+2 \mathrm{H}_2 \mathrm{O}\)

Alkaline medium:

⇒ \(\begin{aligned}
2\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}+2 \mathrm{OH}^{-}+ & \mathrm{H}_2 \mathrm{O}_2 \\
& {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 }
\end{aligned}\)

Question 72. Which of the following electron-deficient-

  1. (BH3)2
  2. PH3
  3. (Ch3)2
  4. (SiH3)2

Answer: 1. (BH3)2

Question 73. The reaction of aqueous KMnO4 with H2O2 conditions gives—

  1. Mn4+ and O2
  2. Mn2+ and O2
  3. Mn2+ and O3
  4. Mn2+ and MnO2

Answer: 2. Mn2+ and O2

Question 74.

  1. H2O2 + O3→H2O + 2O2
  2. H2O2 + Ag2O→2Ag + H2O + O2

The role of hydrogen Peroxide in the above reactions is respectively—

  1. Oxidsing in 1 and reducing in 2
  2. Reducing in 1 and oxidising in 2
  3. Reducing in 1 and 2
  4. Oxidizing in 1 and 2

Answer: 3. Reducing in 1 and 2

Question 75. Which of the following statements about hydrogen is incorrect—

  1. Hydrogen has three isotopes of which tritium is the most common
  2. Hydrogen never acts as cationic ionic salts
  3. Hydronium ion, H3O exists freely in solution
  4. Dihydrogen does not act as a reducing agent

Answer: 1. Hydrogen has three isotopes of which tritium is the most common

Question 76. In ice, the oxygen atom is surrounded—

  1. Tetrahedrally by 4 hydrogen atoms
  2. Octahedrally by 2 oxygen and 4 hydrogen atoms
  3. Tetrahedrally by 2 hydrogen 2 oxygen atoms
  4. Octahedrally by 6 hydrogen atoms

Answer: 1. Tetrahedrally by 4 hydrogen atoms

X-ray studies have shown that in ice, four hydrogen atoms tetrahedrally surround each oxygen atom.

Hydrogen Tetrahedral Arrangement Of Water

Question 77. Predict the product of the reaction of I2 with H2O2 in basic mL of H2O2
medium—

  1. I‾
  2. i2O3
  3. IO2
  4. 3

Answer: 1. I-

⇒ \(\begin{array}{r}
\mathrm{I}_2(s)+\mathrm{H}_2 \mathrm{O}_2(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \downarrow \\
2 \mathrm{I}^{-}(a q)+2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(g)
\end{array}\)

Question 78. The strength of H2O2 is 15.18 g. L-1, then it is equal to—

  1. 1 volume
  2. 10 volume
  3. 5 volume
  4. 7 volume

Answer: 3.05 volume

Question 79. Which of the following reactions increases the production of dihydrogen from synthesis gas-

  1. \(\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \xrightarrow[\mathrm{Ni}]{\mathrm{1270K}} \mathrm{CO}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g})\)
  2. \(\mathrm{C}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \xrightarrow{1270 \mathrm{~K}} \mathrm{CO}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g})\)
  3. \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \xrightarrow[\text { Catalyst }]{673 \mathrm{~K}} \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})}\)
  4. \(\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \xrightarrow[\mathrm{Ni}]{\frac{1270 \mathrm{~K}}{\longrightarrow}} 2 \mathrm{CO}+5 \mathrm{H}_2\)

Answer: 3. \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \xrightarrow[\text { Catalyst }]{673 \mathrm{~K}} \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})}\)

The production of dihydrogen can be increased by reacting the carbon monoxide of syngas with steam in the presence of iron chromate as a catalyst.

⇒ \(\mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \xrightarrow[\text { Catalyst }]{673 \mathrm{~K}} \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})}\)

This is called the water-gas shift reaction.

Question 80. Which ofthe following produces hydrogen

  1. Mg + H2O
  2. H2S2O8 + H2O
  3. BaO2 + HCl
  4. Na2O2 + 2HC1

Answer: 1. Alkali and alkaline earth metals react with water to produce hydrogen gas and metal hydroxides. This occurs due to high electropositive character ofthe metals

⇒ \(\mathrm{Mg}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Mg}(\mathrm{OH})_2+\mathrm{H}_2\)

Question 81. H2O2 can be obtained when the following reacts with H2SO4 except with-

  1. BaSO2
  2. PbO2
  3. Na2O2
  4. SrO2

Answer: 2. PbO2

H2O2 is prepared by the reaction of peroxide with H2SO4.PbO2 is a dioxide. Hence, it does not give H2O2 with dilute H2SO4

Hydrogen Very Short Answer Type Questions

Question 1. Which is the lightest gas known?
Answer: Dihydrogen;

Question 2. Which isotope of hydrogen is radioactive?
Answer: Tritium \(\left({ }_1^3 \mathrm{H}\right)\)

Question 3. Give examples of an ionic, a covalent, and a metallic hydride.
Answer: CaH2 (ionic); NH3 (covalent) & CrH (metallic)

Question 4. What is hydrolith?
Answer: Calcium hydride (CaH2)

Question 5. Name the two nuclear spin isomers of dihydrogen.
Answer: Ortho-hydrogen and para-hydrogen;

Question 6. Give an example of an electron-deficient hydride in which three centre-two
electron bonds are present.
Answer: B2H6

Question 7. Which gaseous compound on treatment with dihydrogen produces methanol?
Answer: Carbon monoxide

Question 8. How will you prove that a colorless liquid is water?
Answer: White anhydrous CuSO4 becomes blue in contact with water

Question 9. What is the unit for expressing the degree of hardness of water?
Answer: ppm (parts per million);

Question 10. write the names of two chemical substances which are used for removing
dissolved oxygen from water meant for the boiler.
Answer: Hydrazine (NH2NH2) and sodium sulphite (Na2SO3);

Question 11. Why is heavy water used in atomic reactors?
Answer: It is used as a moderator

Question 12. Name a solid and a liquid absorbent of water.
Answer: Cone. H2SO4 and P2O5

Question 13. Which chemical is commercially known as ‘perhydrol’?
Answer: H2O2 solution

Question 14. What is called ‘hyper lol or horizon’?
Answer: A compound of hydrogen peroxide and urea is called hyper lol (NH2CONH2-H2O2);

Question 15. What is the die volume strength of a molar solution of H2O2?
Answer: 11.2 volume

Question 16. Which organic reagent is used for the manufacture of H2O2?
Answer: 2-ethylanthraquinol

Question 17. 10 volume of H2O2 = x(N)H2O2 .What is the value of
Answer: X = 56

Question 18. What are how water molecules are bonded to the anhydrous salt
to form hydrates?
Answer: Coordinate bond and H bond.

Hydrogen Fill In The Blanks

Question 1. The radioactive isotope of hydrogen is____________.
Answer: Tritium

Question 2. When NaH is electrolysed,____________ is obtained at the anode.
Answer: Dihydrogen

Question 3. Syngas is a mixture of hydrogen and____________.
Answer: Carbon monoxide

Question 4. para-hydrogen is____________ stable than ortho-hydrogen.
Answer: More

Question 5. The oxygen atom in the water molecule is____________ hybridized.
Answer: SP3

Question 6. H2O undergoes electrolysis____________than D20.
Answer: Faster

Question 7. Heavy water is used as a____________in nuclear reactors.
Answer: Moderate

Question 8. Temporary hardness is also known as ____________hardness.
Answer: Carbonate

Question 9. Rainwater is____________water but sea water is water.
Answer: Soft, Hard

Question 10. The reaction between CaC2 and D2O forms____________.
Answer: C2D2

Question 11. D2O2 can be prepared by electrolysing____________by d2o.
Answer: K2S2O8

Question 12. H2O2____________ isslighdy acid than water.
Answer: Stronger

Question 13. Decomposition of H2O2 is suppressed by ____________
Answer: Urea

Question 14. Boiling point of H2O2 is____________ than water.
Answer: Higher

Question 15. The mixture of & H2O2 is known as Fenton’s____________ reagent.
Answer: FeSO4

Question 16. Volume strength of 1.5(N) H2O2 is ____________
Answer: 8.4

Hydrogen Numerical Examples

Question 1. 1L of a sample of hard water contains 1 mg CaCl2 and 1 mg MgCl2. Estimate the degree of hardness of this sample of water
Answer: Molecular mass of CaCl2 =111

Now lllg of CaCl2=100g of CaCO3

∴ 1 mg CaCl2 \(\equiv \frac{100 \times 0.001}{111} \mathrm{~g}\) of CACO3

= 9×10-4 g of CACO3 = 0.9 mg of CaCO3

The molecular mass of MgCl2 = 95

Now, 95g of MgCl2=100g of CaCO3

⇒ \(1 \mathrm{mg} \text { of } \mathrm{MgCl}_2 \equiv \frac{100}{95} \mathrm{mg}\)

CaCO3 = 1.05mg of CaCO3

An equivalent amount of CaCO3 corresponding to CaCl2 and MgCl2 present l Lor 103g of hard water

= (0.90 + 1.05)mg = 1.95mg [1L water = 103g = 106mg water]

The mass of the equivalent amount of CaC03 corresponding to CaCl2 and MgCl2 presenting 106mg of water is 1.95 mg.

Hence, the degree of hardness of the given sample is 1.95 ppm.

Hence, the hardness of that sample of water is 75 ppm.

Question 2. Determine the strength of ’30 volume’ H2O2 in normality
Answer: Volume strength = 5.6 x normality

or, 30 = 5.6 x normality or, normality \(=\frac{30}{5.6} \mathrm{~N}=5.35 \mathrm{~N}\)

∴ The normality of 30 volumes of H2O2 is 5.35N

Question 3. Determine the volume (in liter) of O2 obtained at STP when 0.1 liter of 2(M)H2O2 solution is decomposed.
Answer: \(\begin{array}{cc}
2 \mathrm{H}_2 \mathrm{O}_2 & 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\
(2 \times 34) \mathrm{g}=68 \mathrm{~g} & 22.4 \mathrm{~L}(\text { at STP) }
\end{array}\)

Now, 1L1(M) H2O2 = 34g of H2O2

1L2(M) H2O2 →(2 X 34)g of H2O2

0.1L 2(M) H2O22 = (2 X 34 X 0.1 )g of H2O2 →Kg of H2O2

Again at STP, 68g of M2O2 produces 22.4L of O2

∴ 6.8 g of H2O2 produces \(\frac{22.4}{68} \times 6.8 \mathrm{~L} \text { of } \mathrm{O}_2\)

Question 4. When 100 ml of tube-well water is titrated using methyl orange as an indicator, it requires 15 ml of 0.01 (N) HCl. Estimate the hardness of that sample of water.
Answer: 15 mL 0.01 (N) HCI = 1 mL 0.15(N) HCI 1000 ml HCL=\(\mathrm{HCl} \equiv \frac{100}{2} \mathrm{~g} \mathrm{CaCO}_3\)

1 mL 0.15 (N) HCI \(\begin{aligned}
& =50 \times \frac{1}{1000} \times 0.15 \mathrm{~g}^{-} \text {of } \mathrm{CaCO}_3 \\
& =7.5 \times 10^{-3} \mathrm{~g} \text { of } \mathrm{CaCO}_3
\end{aligned}\)

Therefore, 100, or 100g of that sample of water contains some hardness-producing substance which is equivalent to 7.5 x 10-3g of CaCO3. 106g of water contains the hardness-producing substance equivalent to \(\frac{7.5 \times 10^{-3} \times 10^6}{100}=75 \mathrm{~g} \text { of } \mathrm{CaCO}_3\)

Question 5. Calculate the amount of H2O2 present in 600 mL of 10-volume H2O2 solution.
Answer: 10 volume H2O2 means that 1 mL of H2O2 solution will produce 10 mL of O2 at STP

\(\begin{gathered}
2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\
\left(2^{\prime} 34\right) \mathrm{g}=68 \mathrm{~g} \quad 22400 \mathrm{~mL} \text { (at STP) }
\end{gathered}\)

At STP, 10 mL O2 is obtained from 1 mL of 10vol H2O2 solution

22400 mL of O2 is obtained from \(\frac{22400}{10}\) of 10 ml of 10 vol; H2O2 solution =2240 mL of H2O2 2240 mL of H2O2 solution contains 68g of H2O2

100 mL of H2O2 solution contains \(\frac{68 \times 100}{2240}\) =3.03g of H2O2.

So, 600 mL of H2O2 solution contains \(=\frac{3.03 \times 600}{100} \mathrm{~g}\) of H2O2 = 18.18g of H2O2 =18.2g of H2O2

Question 6. An excess of acidic KI solution is added to 25 mL of a H2O2 solution when iodine is liberated. 20 mL of 0.1(N) sodium thiosulfate solution is required to titrate the liberated iodine. Calculate the percentage strength, volume strength, and strength in normality of the H2O2 solution.
Answer: 25 x (N) = 20 x 0.1(N)

⇒ \(x=\frac{20 \times 0.1}{25}=\frac{2}{25}=0.08(\mathrm{~N})\)

Amount of H2O2 in 25mL0.08(N) H2O2 solution

1000 mL 1(N) H2O2 solution = 17g H2O2 25mL0.08(N) H2O2 solution \(\equiv \frac{17 \times 25 \times 0.08}{1000}=0.034 \mathrm{~g}\) H2O2

100ml 0.08(N) H2O2 solution contains \(=\frac{0.034 \times 100}{25}\)
= 0.136gH2O2

% strength of H2O2 solution = 0.136

⇒ \(\begin{array}{cr}
2 \mathrm{H}_2 \mathrm{O}_2 \\
2 \times 34=68 \mathrm{~g} & 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\
& 22.4 \mathrm{~L} \text { at STP }
\end{array}\)

68g H2O2 gives→ 22.4 L O2 at STP

0.034g H2O2 solution \(\begin{aligned}
\rightarrow & \frac{22.4 \times 0.034}{68} \text { Litre } \mathrm{O}_2 \text { at STP } \\
& =0.0112 \mathrm{Litre}_2 \text { at STP } \\
& =11.2 \mathrm{~mL} \mathrm{O}_2 \text { at STP }
\end{aligned}\)

25mLH2O2 solution gives 11.2mLO2 atSTP

lmL H2O2 solution given \(\frac{11.2}{25}=0.448 \mathrm{~mL} \mathrm{O}_2\) at stp.

Volume strength ofthe given H2O2 solution = 0.448.

WBCHSE Class 11 Chemistry Notes For Periodic Trends of Elemental Properties

Periodic Trends In Properties Of The Elements

The properties of elements can be divided into two categories:

Properties of individual atoms: The properties such as atomic and ionic radii, ionisation energy, electron affinity, electro negativity and valency are the properties of the individual atoms & are directly related to their electronic configurations.

Properties of groups of atoms: The properties such as melting point, boiling point, the heat of fusion, heat of vaporisation, atomic volume, density etc. are the bulk properties i.e., the properties of a collection of atoms and are related to their electronic configurations indirectly.

All these properties which are directly or indirectly related to the electronic configurations of the elements are called atomic properties.

Since electronic configurations of the elements are periodic functions of their atomic numbers, these atomic properties are also periodic functions of atomic numbers of the elements. Thus, atomic properties are also called periodic properties.

Remember that, when we descend a group, the chemical properties ofthe elements remain more or less the same due to the same valence shell configuration, but there is a gradual change in physical properties due to a gradual change in the size ofthe atoms owing to the addition of new electronic shells.

Atomic size or atomic radius

If an atom is assumed to be a sphere, the atomic size is given by the radius of the sphere, called atomic radius.

Atomic size or atomic radius Definition: The distance from the centre of the nucleus to the outermost shell containing the electrons is called the atomic radius

Difficulties in precise measurement of atomic radius:

It is not possible to isolate a single atom for the measurement of its radius.

The electron cloud surrounding the nucleus does not have a sharp boundary as the probability of finding an electron can never be zero even at a large distance from the nucleus.

The magnitude of atomic radius changes from one bonded state to another.

Types of atomic radius: As already mentioned, the size of an atom varies from one environment to another.

Therefore, several kinds of atomic radii have been defined. These are—

  1. Covalent radius
  2. Metallic radius
  3. van der Waals radius.

Covalent radius: It is defined as one-half of the distance between the centres of two atoms of the same element bonded by a single covalent bond.

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Covakent radius

Thus for homonuclear diatomic molecules, covalentradius\((r)=\frac{1}{2} x\)
internuclear distance.

Example: In the H2 molecule, the internuclear distance is = 0.74 A =74pm. So, covalent radius ofhydrogen atom=\(=\frac{1}{2} \times 0.74\) = 0.37 A= 37pm. lA = 10-10m, 1 pm = 10-12m.

For the heteronuclear diatomic molecule A—B, internuclear distance, rAB = rA + rB.

Therefore rA = rAB-rB rB = rAB-rA

[rA and rB represent radii ofthe atoms, A & B respectively]

Example: For HC1 molecule, internuclear distance, rHCl = L36A & covalent radius of hydrogen atom (rH) = 0.37 A

Therefore Covalent radius ofchlorine atom (rcl) = rHC1- rH = 1.36-0.37 = 0.99 A

Metallic radius: It is defined as one-half of the internuclear distance between two adjacent atoms in a metallic lattice.

Example: The distance between two adjacent copper atoms in solid copper is 2.56 A (determined by X-ray diffraction). Hence, the metallic radius ofcopperis1.28

A. Similarly, the metallic radii of sodium and potassium have been determined as 1.86 A and 2.31 respectively.

Note—covalent radii of Na and K are 1.54 A and 2.03 A.

Metallic radius is always greater than the covalent radius van der Waals radius: It is defined as one-half of the distance between the nuclei of two non-bonded neighbouring atoms belonging to two adjacent molecules of an element in the solid state.

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Van der waals radius

Example: The distance between nuclei of two adjacent Cl -atoms of two adjacent chlorine molecules in a solid state is 3.6 A.

So, the van der Waals radius of Cl-atom is \(\frac{3.6}{2}=1.8 \AA\) Since the 2 van der Waals force of attraction is weak even in the solid state, the internuclear distances between the atoms of two adjacent molecules held by van der Waals forces are much larger than those between covalently bonded atoms (which involve mutual overlap of atomic orbitals). Van der Waals radii are always greater than covalent radii. For example, the van der Waals radius ofchlorine is1.80 A while its covalent radius is 0.99 A.

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Sequences of three type of atomic radii

The sequence of three types of atomic radii: van der Waals radius >Metallic radius > Covalentradius.

Variation of atomic radii or sizes in the periodic table: While moving from left to right across a period in the periodic table, atomic radii or atomic sizes progressively decrease.

Explanation: The principal quantum shell remains unchanged in the same period. So, the differentiating electrons enter the same shell but due to an increase in the number ofprotons, the positive charge of nucleus also increases.

So, the attractive force of the nucleus for electrons in the outermost shell also increases.

Hence, the atomic sizes, rather than the atomic radii of elements in the same period, gradually decrease with an increase in atomic number. In any period, the atomic size of the element of group LA is maximum and that of the halogen of group VILA is minimum.

Variation of atomic [covalent] radii of the elements of the third period (n = 3]

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Variatioon of atmoic [Covalent] radii of the elements of thirs period [n=3]

Variation of atomic radii or sizes down a group: On moving down along any group of the periodic table, the atomic sizes rather the atomic radii of the elements increase remarkably.

Explanation: On moving down a group, a new electronic shell is added to each succeeding element, though the number of electrons in the outermost shell remains the same. This tends to increase the atomic size.

At the same time, there is an increase in nuclear charge with an increase in atomic number. This tends to decrease the size.

However, the effect of the increased nuclear charge is partly reduced by the shielding effect of the inner electronic shells. In practice, it is found that the effect of the addition of a new electronic shell is so large that it outweighs the contractive effect of the increased nuclear charge.

Hence, there occurs a gradual increase in atomic radii on moving down a group in the periodic table.

Variation of size in a group for heavier elements: On moving down a group, the relative rate of increase of atomic radii slow for heavier elements.

So, the differences in size for heavier species such as Cs (6s1) and Fr(7s1) of group-1 or Ba(6s2) and Ra(7s2) of group-2 are very small.

This is due to the presence of electrons in the inner d and f-orbitals having poor screening effect.

D and f-electrons do not screen the outer electrons effectively from the pull of the nucleus. There is only a small increase in size despite addition of a new electronic shell. Example— Na(l.54 A), K(2.03 A), Rb(2.16 A), Cs(2.35 A) etc.

In the case of transition elements, such a decrease in the atomic size is relatively less. Here the differentiating electron instead of entering the outermost orbit, goes to the penultimate (n- 1 )d -subshell which is closer to the nucleus.

However, due to the poor shielding effect of the additional d -electrons, there is a small increase in effective nuclear charge and hence, the decrease in the size with an increase in the atomic number is relatively small.

In the case of lanthanide elements, the differentiating electrons enter the (n – 2)/-orbital having a very poor shielding effect.

So, In such cases, the increase in the effective nuclear charge is greater than that in the case of transition elements.

Thus, the decrease in atomic sizes of the lanthanides is much more regular and distinct as compared to the transition elements.

So, the difference in the extent of the decreased atomic sizes ofthese elements is due to their relative inability to screen the outermost electrons from attraction ofthe nucleus.

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Covalent radiii[pm] of representative element

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Variation Of atomic radii with atomic numbers across the second period

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Variation of atomic radii with atomic numbers for group -1 metals

Screening effect or shielding effect: in multi-electron atoms, the electrons present in the inner shells shield the electrons in the valence shell from the pull of the nucleus.

It means that the electrons of the inner shells act as a screen between the nucleus and the electrons in the valence shell. This is known as the screening effect shielding effect.

Screening effect or shielding effect Definition: The ability of the inner electronic shells to shield the outer electrons from the attraction of the nucleus is called the screening effect or shielding effect.

The magnitude of the screening effect of electrons belonging to different subshells follows the sequence: s>p> d>f.

Important points regarding atomic (covalent) radius:

The alkali metals occupying positions at the extreme left side ofthe periodic table have the largest size in each period.

The halogens occupying positions at the right side of the periodic table have the smallest size in each period.

The noble gases present at the extreme right of the periodic table have larger atomic radii than those of the preceding halogens because van der Waals radii (but not covalent radii) are taken into consideration for noble gases.

In transition series {d -block elements), there is only a small decrease in size with successive increases in atomic number because the differentiating electrons enter into (n-l)d subshell, which partially shields the increased nuclear charge acting on the valence electrons. This is known as d contraction.

For the inner-transition series ( f-block elements), the decrease in atomic radii and wide increase in atomic number is relatively greater and more regular as compared to the transition series elements.

This is so because the differentiating electrons enter into (n- 2)f -subshell havingverypoor shielding effect.

In a group of representative elements, there is a continuous increase in atomic radii with an increase in atomic number.

On going down a group of transition elements, there is an increase in size from first member to second member as expected, but for higher members, the increase in size is quite small. This is due to lanthanide contraction. For example, Cu(1.28A), Ag(1.44A), Au (1.44A)etc.

Ionic radii

Ionic radii refers to the radii of the ions in the ionic crystals. Ionic radius may be defined as the effective distance from the centre of the nucleus of an ion to the point upto which it exerts its influence on the electron cloud.

Theoretically, the electron cloud may extend itself to a very large distance from the nucleus. So, it is not possible to measure the ionic radius directly.

It is measured indirectly as follows. The equilibrium distance between nuclei of adjacent cation and anion in an ionic crystal can be determined by X-ray analysis.

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Ionic radii

Regarding ions as spheres, the internuclear distance may be taken as the sum of the radii of the adjacent ions. Knowing the radius of one, that ofthe other can be calculated. Several methods have been developed to fix the absolute value of at least one ion.

Pauling’s method is the most widely accepted and the values given here are based on this method.

For example, based on Pauling’s method, the radius ofthe Na+ion has been determined to be 0.95 A.

Again, the intemuclear distance between Na+ and Cl- ions in NaCl crystal is 2.76 A (from X-ray studies). So, (2.76-0.95) = 1.81 A.

Some characteristics of ionic radii:

The radius ofthe cation is always smaller than that ofthe
neutral atom: A cation is formed loss of one or more electrons from the neutral atom (e.g., Na-e→Na+; Mg- 2e→Mg2+ ).

With the removal of electron (s) from an atom, the magnitude of nuclear charge remains the same while the number of electrons decreases. That means the same nuclear charge now acts on a decreased number of electrons.

As a result, effect of nuclear charge per electron increases and the electrons are more strongly attracted towards the nucleus. This causes a contraction in size ofthe cation.

In most cases, all the electrons from the outermost shell of the atom are completely removed so that the penultimate shell of the atom now becomes the outermost shell ofthe cation.

As a result, the size ofthe cation becomes smaller than that of the parent atom from which it is formed.

\(\begin{array}{cc}
\mathrm{Na}\left(1 s^2 2 s^2 2 p^6 3 s^1\right) \stackrel{-e}{\longrightarrow} & \mathrm{Na}^{+}\left(1 s^2 2 s^2 2 p^6\right) \\
r_{\mathrm{Na}}=1.54 \AA & r_{\mathrm{Na}^{+}}=0.95 \AA
\end{array}\)

When an element forms more than one type of cation, then the cationwithhigher chargewillbe smallerin size: The cations (e.g., Fe2+ and Fe3+) derived from a specific element contain same number of protons but different number of electrons.

The cation with a higher charge (e.g., Fe3+) contains fewer electrons and the effective nuclear charge per electron is greater, thereby causing a contraction in size.

Example: Ionic radius of Fe2+ = 0.75 A but ionic radius of Fe3+ = 0.60 A

The radius ofthe anionic is always greater than that ofthe neutral atom: An anion is formed by the gain of one or more electrons by a neutral atom (e.g., Cl+e→Cl-; O+2e→O2-)

This increases the number of electrons in the anion but the nuclear charge remains the same as that in the parent atom.

That means the same nuclear charge now acts on an increased number of electrons. As a result, effective nuclear charge per electron decreases and the electrons are less tightly held by the nucleus. This causes an increase in the size of the anion.

Furthermore, the addition of one or more electrons would result in increased repulsion among the electrons, thereby causing the expansion of the outermost electronic shell. This also causes an increase in the size of the anion

Example-1

\(\begin{array}{cc}
\mathrm{Cl}(17 p+17 e) \stackrel{+e}{\longrightarrow} \mathrm{Cl}^{-}(17 p+18 e) \\
r_{\mathrm{Cl}}=0.99 \AA & r_{\mathrm{Cl}^{-}}=1.81 \AA
\end{array}\)

Variation of ionic radii within a group:

For cations: Like covalent radii of atoms, the radii of cations also increase on moving down a group. This is primarily due to the addition of a new electronic shell at each succeeding member on moving down a group. Thus,

\(r_{\mathrm{Li}^{+}}<r_{\mathrm{Na}^{+}}<r_{\mathrm{K}^{+}}<r_{\mathrm{Rb}^{+}}\)

For anions: Like cationic radii, the radii of anions also increase on moving down a group. This is primarily due to the addition of a new electronic shell at each succeeding member on moving down a group. Thus, \(r_{\mathrm{F}^{-}}<r_{\mathrm{Cl}^{-}}<r_{\mathrm{Br}^{-}}<r_{1^{-}}\)

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Variation of ionic radii of cations of Gr-1[1A] and anions of group- 17

Variation of ionic radii along a period:

For cations: On moving from left to right along a period, the radii of isoelectronic cations (cations having the same number of electrons) decrease progressively due to an increase in the magnitude of nuclear charge (i.e., an increase in the number of protons).

Thus, rNa+ > rMg2+ > rAl3+.

For anions: Onmovingfromlefttorightalong period, the radii of isoelectronic anions (anions having the same number of electrons) decrease progressively due to increase in the magnitude of nuclear charge [i.e., an increase in the number of protons). Thus, rN3¯ > rQ2‾ > rp¯

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Variation Of ionic Radii Of Isoelecronic Ions

Variation of ionic radii of isoelectronic ions (cations or anions) belonging to the same or different periods:

Isoelectronic species Cations or anions or neutral atoms having the same number of electrons but different magnitude of nuclear charge [i.e., different number of protons) are called isoelectronic species.

A set of isoelectronic ions (belonging to periods 2 and 3) are shown in the following table. All the ions have an equal number (10) of extranuclear electrons but a different number of protons.

As we move from one member to another, the nuclear charge increases. Consequently, the electrons are pulled more and more strongly, thereby causing a gradual decrease in size.

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Isoelectronic anions having same number of electrons

Note that the cation with a greater positive charge has a smaller radius while the anion with a greater negative charge has a greater radius.

For any atom or ion, (Z/e) oc(1f): where Z- atomic number or nuclear charge, e = number of electrons and r = atomic or ionic radius.

Since the quantity, Z -e varies inversely with ionic radius (r), so increase in the magnitude of Z/e brings about a decrease in the value of ionic radius, and a decrease in the magnitude of Zfe results in an increase in ionic radius.

Forinstance,in case of O -atoms,(Z/e) = 1 and for O2¯ ion, (Z/e) = (8/10) = 0.8. So, the ionic radius of O2¯ ion (1.40 A) is greater than the atomic radius of O-atom (0.66A).

Similarly for K-atom, (Z/e) = 1 and for K+ ion, (Z/e) = (19/18) = 1.05. Hence, the ionic radius of the K+ ion (1.33 A) is less than the atomic radius of the K -atom (2.27 A)

Important points to remember about ionic radii:

  1. The radius of the cationic is always smaller than that of the parent atom.
  2. Theradius of anion is always greater than that of the parent atom.
  3. The ionic radii (both cationic and anionic) in any group increase on moving down the group.
  4. The radii of isoelectronic cations decrease with increasing atomicnumber across a period.
  5. The radii of isoelectronic anions decrease with increasing atomicnumber across a period.
  6. The radii of isoelectronic cations or anions (of the same or different period) decrease with increasing atomic number.
  7. The sizes of cations (of the same element) decrease with increasing positive charge (e.g., rpe2+ > rFe3).

Atomic volume Definition: It is the volume occupied by the gram-atom (or mole atom) of an element at its melting point in the solid state.

Measurement: The gram-atomic mass of an element, when divided by its density, gives the atomic volume ofthe element.

Atomic masses. He plotted atomic volumes of different elements against the corresponding atomic masses and the curve thus obtained looked like a wave with several sharp peaks and broad minima.

The curve reveals that the atomic volumes of elements having similar properties are periodic functions of their atomic masses. For instance, l] Each peak ofthe curve is occupied by the first member of a period.

That means the reactive and light alkali metals, Li, Na, K, Rb, Cs and Fr occursuccessivelyat at the peaks. They have the largest atomic volumes.

The less reactive transition elements (e.g., Cr, Mn, Fe, Co, Ni )which are heavy and possess high melting points occupy the bottom portions ofthe curve.

Electronegative elements (e.g., F, Cl, Br, I) occupy the ascending portions of the curve (i.e., the left side of each peak) and their non-metallic character increases while moving towards the peak.

Each of the descending portions (i.e., the right side of each peak) of the curve is occupied by electropositive metals (e.g., Mg, Ca, Sr, Ba). v] Elements occupying similar positions on the curve are placed in the same group of the periodic table.

\(\text { Atomic volume }=\frac{\text { Gram-atomic mass }}{\text { Density }\left(\text { in } \mathrm{g} \cdot \mathrm{cm}^{-3}\right)}\)

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Variation Of atomic volume with the change in atomic masses of different elements

Variation of atomic volume along s period: On moving from left to right along a period, the atomic volume of the elements gradually decreases to a minimum value and then increases successively and attains a maximum value the first member of the next period.

Variation of atomic volumes of elements belonging to the second and third periods

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Variation of atomic volumes of elemnets belonging to second nad third periods

Variation of atomic volume down a group: On moving down a group, (i.e., with increasing atomic mass) the atomic volume of elements increases almost regularly due to successive addition of new principal quantum shells.

Metallic and non-metallic character

The metallic property or electropositive character of an element is measured as the tendency of its atom to form a cation with loss of electron (s). The more the tendency ofthe atom of an element to lose electrons, the more will be its metallic property.

Variation of metallic and non-metallic character across a period: In any period, as we move from left to right, the atomic size ofthe elements progressively decreases with a consequent steady increase in nuclear attractive force for the valence electrons; i.e., the valence electrons lose their freedom.

As a result, metallic property gradually decreases which is reflected in the fall of thermal and electrical conductivity of the elements. On the other hand, the non-metallic character ofthe elements increases along a period from left to right.

Explanation: Because of the large size of the metallic atoms, electrons in their valence shell are loosely held. So, they exhibit both thermal and electrical conductivity.

On the contrary, atoms of non-metals are smaller in size and hence their valence shell electrons are relatively strongly held therefore, they are thermally and electrically non-conductive. Moreover, non-metals have a strong tendency to form anions by gaining electrons. So, they are electronegative.

Metalloids like the nonmetals, are characterised by their small atomic sizes and their valence-electrons are more or less strongly confined.

However, at high temperatures, the confined electrons become mobile and contribute to electrical conductivity.

Example: Elements belonging to groups 1 and., Na and Mg are good conductors of of heat and electricity, and A1 in group 13 has moderate conductivity. Other elements of this period are either or conductors on-conductors.

Variation of the metallic and non-metallic character down a group: In the periodic table, as we move down a group, the metallic character gradually increases and consequently, the non-metallic character gradually decreases.

Explanation: Due to the introduction of new electronic shells, the hold of the nucleus on the valence electrons gradually decreases thereby making these electrons more and more mobile on moving down a group.

Consequently, metallic character increases with increasing atomic number. Thus in group 15, N and P are typical non-metals, As and Sb are metalloids and Bi is a typical metal. C in group-14 is a typical non-metal, Ge possesses metallic character while Pb and Sn are typical metals.

All transition elements as well as the elements oflanthanoid and actinoid series are typical metals. Transition elements are good conductors of heat and electricity. Coinage or noble metals (Cu, Ag, Au and Pt) have maximum thermal and electrical conductivity.

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Positions of mentals, metalloids and non-mentals in periodic table

Variation of the metallic and non-metallic character of elements from the nature of their oxides: On moving across a period from left to right, metallic character decreases and within a group, the metallic character increases from top to bottom.

Generally, oxides of strongly electropositive elements are basic while oxides of strongly negative elements exhibit acidic character.

Example:

Oxides of the representative elements ofthe third period show that their basic character i.e., the metallic property gradually decreases while their acidic character i.e., non-metallic character gradually increases from left to right across the period.

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Nature of oxides of elements of third period

Oxides of elements of group VA reveal that the acidic character progressively decreases while the basic character increases down a group. Hence, the non-metallic character gradually decreases and the metallic character gradually increases down a group.

Class 11 Chemistry Classification Of Elements And Periodicity in Properties Nature Of Oxides Of elements Belonging to group-5A

Oxidising and reducing properties

An oxidising agent can gain electrons while a reducing agent can donate electrons. So strong oxidising agents have a pronounced tendency to accept electrons and strong reducing agents give up electrons easily.

Variation of oxidising & reducing properties across a period:

On moving from left to right a period, nuclear charge increases by one unit and at the same time one electron is added to the same outermost shell in each succeeding element.

So nuclear pull on the electrons of the outermost shell increases with increasing atomic number.

Consequently, the tendency of an atom to give up electrons decreases and the reverse tendency i.e., the tendency to accept electrons increases on increasing atomic numbers over a period.

In other words, the elements on the right side of a period [Le., gr.-VA(15), VIA(16) and VIIA(17) elements] are good oxidising agents because they are good acceptors of electrons.

Their oxidising power increases as: VA < VIA < VIIA. On the other hand, the elements on the left side of a period [le., gr-IA(l), IIA(2)] are good-reducing agents because they are good donors of electron. Their reducing power decreases as 1A > 2A > 3A

Example: Oxidisingpower: Na < Mg < A1 < Si < P < S < Cl

Reducing power: Na > Mg > Al > Si > P > S > Cl

Variation of oxidising and reducing properties down a group:

On moving down a group, a new electronic shell is added at each succeeding element. This tends to decrease the nuclear pull on the outer electrons.

But at the same time, there is an increase in nuclear charge with an increase in atomicnumber This tends to increase the nuclear pull on the outer electrons. In practice, it is found that the effect of the addition of a new electronic shell is greater than the effect of increased nuclear charge.

This means that, on moving down a group, nuclear pull on the outer electrons gradually decreases.

Consequently, reducing power gradually increases while the oxidising power gradually decreases on moving down a group in the periodic table.

Example: The oxidising power of halogens follows the order: of F(+2.87) > Cl(+1.36) > Br(+1.06) >I(+0.53). The reducing power of group-IIA metals follows the order:

Be(-1.85) < Mg(—2.37) < Ca(-2.87) < Sr(-2.89) < Ba(-2.90). where the quantities within the bracket indicate the standard reduction potential, £°ed(volt) at 25°C for the indicated (12X2 or M2+/M ).

Class 11 Chemistry Classification of Elements and Periodicity in Properties Question and Answers

Class 11 Chemistry Classification of Elements and Periodicity in Properties Question and Answers

Question 1. What is the basic theme of organization in the periodic table?
Answer: The basic theme of organization in the periodic table is to study different physical and chemical properties of all the elements and their compounds simply and systematically.

Elements belonging to the same group have similar physical and chemical properties.

So, if the physical and chemical property of any one element of a group is known, then it is possible to predict the physical and chemical properties of the remaining elements of that group.

Therefore, it is not important to keep in mind the physical and chemical properties of all elements in the periodic table.

Question 2. Which property did Medeleev use to classify the elements in his periodic table? Did he stick to that?
Answer: Mendeleev classified the elements based on their increasing atomic weights in the periodic table. He arranged almost 63 elements in order of their increasing atomic weights placing together elements with similar properties in a vertical column

He observed that while classifying elements in the periodic table according to increasing atomic weight, certain elements had different properties than those elements belonging to the same group.

For such cases, Mendeleev prioritized the properties of the element over its atomic weight.

So, he placed an element with a higher atomic weight before an element with a lower atomic weight.

For example, iodine [I (126.91)] with a lower atomic weight than tellurium [Te (127.61)] is placed after tellurium in group VII along with elements like fluorine, chlorine, etc., due to similarities in properties with these elements.

Thus, Mendeleev did not stick to his idea of classifying elements only according to the increasing atomic weights.

Question 3. What is the basic difference in approach between Mendeleev’s Periodic Law & the Modern Periodic Law?
Answer: According to Mendeleev’s periodic law, the physical and chemical properties of elements are a periodic function of their atomic weights.

On the other hand, the modern periodic table states that the physical and chemical properties of the elements are a periodic function of their atomic numbers.

Thus the basic difference in approach between Medeleev’s periodic law and modern periodic law is the change in the basis of classification of elements from atomic weight to atomic number.

Based on quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

In the modern periodic table, each period begins with the filling of a new principle energy level. Therefore, the sixth period starts with the filling of the principal quantum number, n = 6.

In the sixthperiod elements, the electron first enters the 6s -orbital, and then from left to right across a period the electrons enter the 4/, 5d, and 6p orbitals of the elements.

Filling of electrons in orbitals case of6th period continues till new principal energy level of quantum number, n = 7 begins, i.e., for elements of the sixth period, electrons fill up the 6s, 4/, 5d, and 6p orbitals total number of orbitals in this case =1 + 7 + 5 + 3 = 16. Since each orbital can accommodate a maximum of two electrons, there can be 16 X 2 or 32 elements in the sixth period.

Question 5. In terms of period and group where would you locate the element with Z = 114?
Answer: it is known that the difference between atomic numbers of the successive members of any group is 8,8,18,18 and 32 (from top to bottom). So, the element with atomic number 114 will lie just below the element with atomic number (114- 32) = 82.

The element with atomic number 82 is lead (Pb), which is a member of the 6th period belonging to group number 14 (p -block element). Thus the element with atomic number 114 takes its position in the 7th period and group- 14 (p -block element) ofthe periodic table.

Question 6. Write the atomic number of the element present in the third period & seventeenth group of the periodic table.
Answer: The general electronic configuration of the valence shell of the elements of group-17 (halogens) is ns2np5.

For the third period, n = 3. Therefore, the electronic configuration of the valence shell of the element ofthird period and group-17 Is 3ia3p1 and the complete electronic configuration of this element Is ls22s22p63s23p5.

There are a total of 17 electrons in this element. Thus, the element In the third period and seventeenth group of the periodic table has atomic number = 17.

Question 7. Which clement do you think would have been named by—Lawrence Berkeley Laboratory and Seaborg’s group?
Answer: Lawrencium, Lr and Berkellum, UK. Seaborgium, Sg.

Question 8. Why do elements In the same group have similar physical and chemical properties?
Answer: Elements belonging to the same group have similar valence shell electronic configurations so they have similar physical and chemical properties.

Question 9. What do you mean by isoelectronic species? Name a species that will be isoelectronic with each of the given atoms or ions: F- Ar, Mg2+, Rb+.
Answer: Isoelectronic ions are ions of different elements that have the same number of electrons but different magnitudes of nuclear charge.

There are (9 + 1) or 10 electrons in F-. Isoelectronic species of F- are nitride (N3-) ion [7 + 3], oxide (O2-) ion [8 + 2], neon (Ne) atom [10], sodium (Na+) ion [11-1], magnesium (Mg2+) ion [12-2], aluminum (Al3+) ion [13-3].

There are 18 electrons in Ar. Isoelectronic species of Ar phosphide (P3-) ion [15 + 3], sulfide (S2-) ion [16 + 2]. chloride (Cl ion [17 + 1], potassium (K+) ion [19 -1], calcium (Ca2+) ion [20-2].

There are (12-2) = 10 electrons in Mg2t. Isoelectronic species of Mg2+ are nitride (N3-) ton [7 + 3], oxide (O2-) ion [8 + 2], fluoride (F-)ion [9+1] sodium (Na+) ion [11-1].

There are (37-1)- 36 electrons In Kb 1 . Isoelectronic species of Rb+ are bromide (Br-) Ion [35 + 1], krypton (Kr) atom [36], strontium (Sr2′) Ion [30-2].

Question 12. Consider the given species: N3-, 02-, I1-, Nn+, Mg2+ and Al3+. What Is Common In them? Arrange them in the order of increasing ionic radii.
Answer: Each of the given ions has 10 electrons. Hence, they are all isoelectronic species. Ionic radii isoelectronic ions decrease with an increase in the magnitude of the nuclear charge.

The order of increasing nuclear charge of the given isoelectronic ions is N3- < O2- <F- < Na+ < Mg2+ < Al3+.

Therefore, the order of increasing ionic radii is: Al3+ < Mg2+ < Na+ < F- < O2- < N3-.

Question 13. What is the significance of the terms—’ isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron-gain enthalpy?
[Hint: Requirements for comparison purposes]
Answer: The ionization energy of an element is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of an isolated gaseous atom existing in its ground state to form a cation in the gaseous state.

Electron-gain enthalpy is defined as the enthalpy change involved when an electron is added to an isolated gaseous atom in its lowest energy state (ground state) to form a gaseous ion carrying a unit negative charge.

The force with which an electron gets attracted by the nucleus of an atom is influenced by the presence of other atoms in the molecule or the neighborhood. Thus, to determine the ionization enthalpy, the interatomic forces should be minimal.

Interatomic forces are minimal in the case of the gaseous state as the atoms are far apart from each other.

Consequently, the value of ionization enthalpy is less affected by the surroundings. Similarly, for electron affinity, the interatomic forces of attraction should be minimal for the corresponding atom. Tor Tills reason, the term ‘Isolated gaseous atom’ Is used while defining Ionisation enthalpy and electron-gain enthalpy.

Ground state means the state at which the atom exists In Its most stable state. If the atom Is In the excited state, then the amount of heat applied to remove an electron or the amount of heat liberated due to the addition of an electron is low.

So, for comparison, the term ‘ground state’ Is used while defining ionization enthalpy and electron-gain enthalpy.

Question 15. The energy of an electron in the ground state of the Hatom is -2.18 X 10-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J . mol-1. [Hint: Apply the idea of the mole concept.]
Answer: Amount of energy required to remove an electron from a hydrogen atom at ground state =Eog-El = 0-El = -(-2.18 X 10-18)J =2.18 X 10-8 J

Ionization enthalpy of atomic hydrogen per mole = 2.18 X 10“18 X 6.022 X 1023 = 1312.8 X 103 J.mol-1.

Question 17. How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer: image-

Evident that in both atoms, the valence electrons enter the 3s orbital. However, the nuclear charge of the Mg atom (+12) is greater than that of the Na atom (+11).

Again, the 3s orbital of the Mg atom being filled is more stable than half-filled.

3s -orbital of Na atom. Thus, the first ionization enthalpy of sodium is lower than that of magnesium.

On the other hand, the removal of one electron from the valence shell of the Na atom leads to the formation of the Na+ ion whose electronic configuration is highly stable (similar to inert gas, neon).

So high amount of energy is required to remove the second electron because it disturbs the stable electronic configuration. However, the electronic configuration of Mg+ is not as stable as that of Na+, but electronic configuration of Mg2+ is more stable as it is similar to the electronic configuration of the inert gas, neon.

So, less amount of energy is required to remove an electron from Mg+. Thus, the second ionization enthalpy of sodium is higher than that of magnesium.

Question 18. What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer: The two factors are: atomic size and screening effect.

Question 19.First ionisation enthalpy values (in kjmol-1) of group-13 elements are: B = 801, A1 = 577, Ga = 579, In = 558 and T1 = 589. How would you explain this deviation from the general trend?
Answer: On moving down group-13 from B to Al, ionization enthalpy decreases due to an increase in atomic size and shielding effect which jointly overcome the effect of an increase in nuclear charge. However, ionization enthalpy increases slightly on moving from Al to Ga (2 kj.mol-1).

This is because due to poor shielding of valence electrons by 3d -electrons effective nuclear charge on Ga is slightly more than Al.

On moving from Ga to In, the shielding effect of all the inner electrons overcomes the effect of the increase in nuclear charge. Thus, the ionization enthalpy of In is lower than Ga.

Again, on moving from Into T1, there is a further increase in nuclear charge which overcomes the shielding effect of all electrons present in the inner shells including those of 4f- and 5d -orbitals. So, the ionization enthalpy of T1 is higher than In.

Question 20. Which of the given pairs would have a more negative electron-gain enthalpy: O or F F or Cl?
Answer: 0 and F both belong to the second period. As one moves from O to F, atomic size decreases and nuclear charge increases. Due to these factors, the incoming electron enters the valence shell, and the amount of energy liberated in the case of F is more than that of O.

Again, Fatom (ls22s22p5) accepts one electron to form F- ion (lsz2s22p6) which has a stable configuration similar to neon. However, O-atom when converted to O- does not attain any stable configuration.

Thus energy released is much higher in going from F to F- than in going from 0 to 0-. So, the electron-gain enthalpy of is much more negative than that of O

WBCHSE Class 11 Chemistry Notes For Hydrogen Peroxide

Hydrogen Peroxide And Its Preparation

Hydrogen peroxide (H2O2) is another hydride of oxygen. It was discovered by the French chemist Thenard in 1818. Unlike water, it is very unstable and so, it does not occur as such in nature. Hydrogen peroxide can be prepared by the following methods.

From Sodium Peroxide (Merck’s Process)

An aqueous solution of H2O2 may be prepared by the action of 20% dilute sulphuric acid, cooled to 0°C by freezing mixture, on a calculated amount of Na2O2.

Sodium sulfate produced in the reaction separates out as crystals of Glauber’s salt (Na2SO4.10H2O). The filtrate, when distilled under reduced pressure, produces a 30% H2O2 solution. This solution is known as Merck’s perhydrol.

\(\mathrm{Na}_2 \mathrm{O}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2\)

From Barium Peroxide (Laboratory Preparation)

Principle: The laboratory preparation of hydrogen peroxide involves reacting an ice-cold thin paste of hydrated barium peroxide (Ba02-8H20) with an ice-cold solution of 20% H2SO4.

Hydrogen From barium peroxide

Procedure: In a beaker, a thin paste of hydrated barium peroxide is prepared by adding a small amount of water to it. In another beaker, 20% dilute H2SO4 is taken. Both the beakers are well-cooled by placing them in a freezing mixture.

  • Then the cold paste is slowly added to the cold acid solution, and kept in a freezing mixture, with constant stirring. The addition of paste is stopped when the mixture remains still slightly acidic.
  • As a result of the reaction between barium peroxide and dilute sulphuric acid, H2O2 is produced and white insoluble BaSO4 is precipitated. Solid BaSO4 is then separated by filtration when a dilute aqueous solution (5%) of H2O2 is obtained as the filtrate.

Limitations: Hydrogen peroxide prepared by this method contains appreciable amounts of Ba2+ ions (in the form of dissolved barium persulphate) which may catalyze the decomposition of H2O2. Therefore, H2O2 prepared by this method cannot be preserved for a long time.

  1. The reaction should be carried out at low temperatures because H2O2 is a very unstable compound that dissociates easily into H2O and O2 if the temperature is increased.
  2. Dilute H2SO4 instead of concentrated H2SO4 is to be used because H2O2 may decompose by the heat evolved when cone. H2SO4 reacts with BaO2.
  3. A thin paste of hydrated barium peroxide instead of anhydrous barium peroxide is to be used. This is because the heat generated by the addition of H2SO4 on anhydrous BaO2 decomposes H2O2 to H4O and O2.
    Moreover, if anhydrous Ba02 is used, an insoluble layer of BaSO4 is formed on it which prevents further reaction. However, there is no evolution of heat when a thin paste of Ba02 is used. Also, BaO2-8H2O exists as fine particles and this prevents the formation of insoluble BaSO4, and the reaction proceeds.
  4. The paste of BaO2 is poured into dilute H2SO4. If dilute H2SO4 is added to the paste, the concentration of BaO2 at the beginning of the reaction becomes sufficiently higher as compared to that of the acid. As a consequence, the mixture becomes basic in nature.
    In this basic medium, the decomposition of H2O2 becomes very fast. On the other hand, if the paste is poured into the ice-cold dilute acid, the mixture always contains an excess of the acid and this prevents the decomposition of H2O2.
  5. At the end of the reaction, the mixture should contain a small amount of surplus acid. This is because the excess acid acting as a negative catalyst decelerates the rate of decomposition of H2O2.
  6. In this process, neither HCl nor HNO3 can be used. HCl is not used because it reacts with BaOz to form water-soluble BaCl2 and the separation of H2O2 from the mixture becomes quite difficult HNO3 (an oxidizing agent) is not used as it oxidizes H2O2: H2O2 2HNO3→ 2H2O + 2NO2 + O2. Moreover, due to the formation of water-soluble Ba(NO3)2, the separation of H2O2 becomes difficult.
  7. It is better to use syrupy phosphoric acid instead of dilute sulphuric acid. This is because in the reaction of barium peroxide with dilute H2SO4, water-soluble barium persulphate (BaS208) is produced which decomposes H2O2. However, when syrupy phosphoric acid is used, the phosphate ions \(\left(\mathrm{PO}_4^{3-}\right)\) separate the heavy metal impurities like Pb2+ (which promotes the decomposition of H2O2) present in BaO2 as insoluble phosphates. Therefore, although both H2SO4 and H3PO4 slow down the dissociation of H2O2, it is better to use H3PO4.

The Reaction Of CO2 And Barium Peroxide

When a stream of CO2 is passed through a thin paste of barium peroxide in ice-cold water, H2O2 and BaCO3 are produced. The insoluble BaCO3 is separated by filtration when a dilute solution of H2O2 is obtained.

BaO2 + H2O + CO2→BaCOgl↓ + H2O2

Manufacture Of Hydrogen Peroxide

By the electrolysis of 50% H2SO4: H2O2 is manufactured by the electrolysis of 50% H2S04 solution which is carried out at low temperature (-20°C) using platinum electrodes and a current of high density. Dihydrogen is liberated at the cathode and peroxodisulphuric acid (H2S2O8) or Marshall’s acid is liberated at the anode.

\(2 \mathrm{H}_2 \mathrm{SO}_4 \rightleftharpoons 2 \mathrm{H}^{+}+2 \mathrm{HSO}_4^{-}\)

Cathode: \(2 \mathrm{H}^{+}+2 e \rightarrow 2[\mathrm{H}] \rightarrow \mathrm{H}_2\)

Anode: \( 2 \mathrm{HSO}_4^{-} \longrightarrow \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8+2 e\)
peroxodisulphuric acid

Peroxodisulphuric acid is collected from the anode chamber and then distilled with water under reduced pressure, when it gets hydrolyzed. The low boiling H2O2 distills over along with water leaving behind the high boiling H2SO4 in the flask.

⇒ \(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_5+\mathrm{H}_2 \mathrm{SO}_4\)

⇒ \(\mathrm{H}_2 \mathrm{SO}_5+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2\)

Modification: A recent modification of the above method involves the use of an equimolar mixture of sulphuric acid and ammonium sulfate for electrolysis. Ammonium persulphate [(NH4)2S2O8] formed around the anode is collected and then distilled with water to form H2O2.

⇒ \(\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{NH}_4 \mathrm{HSO}_4\)

⇒ \(\mathrm{NH}_4 \mathrm{HSO}_4 \rightleftharpoons \mathrm{NH}_4 \mathrm{SO}_4^{-}+\mathrm{H}^{+}\)

Cathode: \(2 \mathrm{NH}_4 \mathrm{SO}_4^{-}-2 e \rightarrow\left(\mathrm{NH}_4\right)_2 \mathrm{~S}_2 \mathrm{O}_8\)

Anode: \(\begin{aligned} & 2 \mathrm{NH}_4 \mathrm{SO}_4^{-}-2 e \rightarrow\left(\mathrm{NH}_4\right)_2 \mathrm{~S}_2 \mathrm{O}_8 \\ & \left(\mathrm{NH}_4\right)_2 \mathrm{~S}_2 \mathrm{O}_8+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NH}_4 \mathrm{HSO}_4+\mathrm{H}_2 \mathrm{O}_2 \end{aligned}\)

Deuteroperoxide (D2O2) may be prepared by using K2SO4 instead of (NH4)2SO4 in the above process and distilling the resulting potassium persulphate with D2O.

⇒ \(\mathrm{K}_2 \mathrm{~S}_2 \mathrm{O}_8(s)+2 \mathrm{D}_2 \mathrm{O}(l) \rightarrow 2 \mathrm{KDSO}_4(a q)+\mathrm{D}_2 \mathrm{O}_2(l)\)

By auto-oxidation of 2-ethylanthraquinol: in this process, the air is passed through a 10% solution of 2-ethylanthraquinol in benzene and cyclohexanol when 2-ethylanthraquinol is oxidized by oxygen of air to 2-ethyl anthraquinone and oxygen is reduced to H2O2. The resulting H2O2 is separated from the organic layer by extraction with deionized water and the aqueous solution is distilled under reduced pressure to give a 30% (by weight) H2O2 solution. 2-ethyl-anthraquinone thus obtained is reduced with H2 in the presence of Pd catalyst to give back 2- ethylanthraquinol which is used again. Therefore, the raw materials required in this process are H2 and atmospheric oxygen which are inexpensive. Therefore, this modern industrial method of H2O2 preparation is the most convenient and economical.

Hydrogen By auto-oxidation of 2-ethylanthraquinol

By partial oxidation of 2-propanol:

⇒ \(\underset{\text { 2-propanol }}{\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH}} \underset{\text { [under pressure] }}{\stackrel{\left(\text { little } \mathrm{H}_2 \mathrm{O}_2\right)+\mathrm{O}_2}{\longrightarrow}} \underset{\text { 2-propanone }}{\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O}}+\mathrm{H}_2 \mathrm{O}_2\)

Preparation of H2O2 from Us dilute aqueous solution: Hydrogen peroxide produced by any of the above methods is in the form of a dilute solution. The solution is concentrated simply by heating it because H2O2 readily decomposes below its boiling point (2H2O2—>2H2O + O2). The dilute solution of H2O2 is concentrated carefully to get pure H2O2 by the following steps:

  1. The dilute solution of H2O2 is heated carefully in a water bath. Slow evaporation of water leads to the formation of a 50% H2O2 solution.
  2. The 50% solution thus obtained is placed in a vacuum desiccator over concentrated H2S04 when approximately 90% H2O2 solution is obtained.
  3. The 90% solution is distilled under reduced pressure (10-15 mm) when about 99% pure H2O2 is obtained.
  4. The 99% solution is finally cooled in a freezing mixture of solid CO2 and ether when crystals of H2O2 separate out. These are removed, dried, and remelted to yield very pure H2O2.

WBCHSE Class 11 Chemistry Preparation and Properties of Dihydrogen Notes

Preparation Of Dihydrogen, H2

Preparation Of Dihydrogen In The Laboratory

Principle: In the laboratory, dihydrogen is prepared by the action of dilute sulphuric acid on granulated zinc (commercial variety).

Hydrogen Preparation of dihydrogen

Process: Dilute sulphuric acid is added through a thistle funnel to some pieces of granulated zinc taken in a Wolff’s bottle. The liberated hydrogen gas is collected by the downward displacement of water.

Pure zinc cannot be used in this process because it reacts sluggishly with dilute H2SO4. Impurities like Cu, Cd, etc. present in commercial zinc help to speed up tire reactions by constituting electrochemical couples (local Galvanic cells). If pure zinc is used, a few drops of CuSO4 solution should be added to the reaction mixture to increase the reaction rate.

Concentrated H2SO4 cannot be used in this preparative method because being oxidizing in nature it oxidizes dihydrogen into the water and itself gets reduced to SO2 For the same basic reason, concentrated HNO3 is not used.

⇒ \(\mathrm{Zn}+2 \mathrm{H}_2 \mathrm{SO}_4 \text { (conc.) } \rightarrow \mathrm{ZnSO}_4+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}\)

⇒ \(\left.\mathrm{Zn}+4 \mathrm{HNO}_3 \text { (conc. }\right) \rightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O}\)

Concentrated HCl cannot be used in the preparation of dihydrogen because the produced dihydrogen will contain times of volatile HCl. Moreover, ZnCl2 formed in the reaction is insoluble in HCl and hence the reaction stops after some time

⇒ \(\mathrm{Zn}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_2+\mathrm{H}_2 \uparrow\)

Various Methods Of Preparation Of Dihydrogen

By the reaction of dilute mineral acids with metals:

⇒ \(\mathrm{Zn}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{H}_2 \uparrow ; \mathrm{Mg}+2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_2+\mathrm{H}_2 \uparrow\)

By the reaction of water with metals:

1. By the reaction of certain alkali and alkaline earth metals with water at room temperature:

⇒ \(2 \mathrm{M}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{MOH}+\mathrm{H}_2 \uparrow[\mathrm{M}=\mathrm{Na}, \mathrm{K}]\)

⇒ \(\mathrm{M}^{\prime}+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{M}^{\prime}(\mathrm{OH})_2+\mathrm{H}_2 \uparrow\left[\mathrm{M}^{\prime}=\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\right]\)

Reactions with these very active metals are so vigorous and exothermic that the liberated hydrogen immediately catches fire which may cause accidents. Because of high I.E., Be does not react with water (hot and cold) to produce H2.

2. By the reaction of less active metals like Mg, A1, Zn, etc., with boiling water: Mg, Al, Zn, etc. (less active metals) react with boiling water to produce dihydrogen.

⇒ \(\mathrm{Mg}+\mathrm{H}_2 \mathrm{O} \stackrel{\Delta}{\longrightarrow} \mathrm{Mg}(\mathrm{OH})_2+\mathrm{H}_2 \uparrow\)

⇒ \(2 \mathrm{Al}+6 \mathrm{H}_2 \mathrm{O} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Al}(\mathrm{OH})_3+3 \mathrm{H}_2 \uparrow\)

3. By the reaction of least active metals like iron and zinc with steam: Least active metals like Fe and Zn react with steam to produce dihydrogen.

⇒ \(3 \mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O} \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{H}_2 \uparrow\)

⇒ \(\mathrm{Zn}+\mathrm{H}_2 \mathrm{O} \stackrel{\Delta}{\longrightarrow} \mathrm{ZnO}+\mathrm{H}_2 \uparrow\)

By the reaction of strong alkali with metals and certain non-metals: Metals like Al, Sn, Zn, Si, etc., react with strong alkali to liberate dihydrogen.

⇒ \( 2 \mathrm{Al}+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaAlO}_2+3 \mathrm{H}_2 \uparrow\)
Sodium aluminate

⇒ \(\mathrm{Sn}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{SnO}_2+\mathrm{H}_2 \uparrow\)
Sodium stannate

⇒ \( \mathrm{Zn}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2 \mathrm{ZnO}_2+\mathrm{H}_2 \uparrow\)
Sodium zincate

⇒ \( \mathrm{Si}+2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Na}_2 \mathrm{SiO}_3+2 \mathrm{H}_2 \uparrow\)
Sodium silicate

By the hydrolysis of ionic hydrides:

⇒ \(\mathrm{CaH}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{H}_2 \uparrow\)

⇒ \(\mathrm{LiH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{LiOH}+\mathrm{H}_2 \uparrow\)

Calcium hydride (CaH2) is called hydrolith and the method of preparing dihydrogen from it is known as the hydrolith process.

By the electrolysis of molten ionic hydrides: in a molten state, ionic hydrides except LiH conduct electricity with the liberation of dihydrogen at the anode.

⇒ \(\mathrm{CaH}_2(\text { molten }) \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{H}^{-}\)

Cathode: \(\mathrm{Ca}^{2+}+2 e \rightarrow \mathrm{Ca}\)

Anode: \(2 \mathrm{H}^{-} \rightarrow \mathrm{H}_2 \uparrow+2 e\)

Preparation Of Pure Dihydrogen

Pure dihydrogen (>99.95%) can be prepared by the following methods:

1. By the action of pure dilute H2SO4 on Mg -ribbon:

⇒ \(\mathrm{Mg}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{MgSO}_4+\mathrm{H}_2 \uparrow\)

2. By the hydrolysis of sodium hydride:

⇒ \(\mathrm{NaH}+\mathrm{H}_2 \mathrm{O} \rightarrow\mathrm{NaOH}+\mathrm{H}_2 \uparrow\)

3. By the action of KOH on scrap aluminum (Uyeno’s method):

⇒ \(2 \mathrm{Al}+2 \mathrm{KOH}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KAlO}_2+3 \mathrm{H}_2 \uparrow\)

4. By the electrolysis of a warm dilute solution of barium hydroxide [Ba(OH)2] using orPt-electrode:

Cathode: \(2 \mathrm{H}_2 \mathrm{O}(l)+2 e \rightarrow \mathrm{H}_2(g)+2 \mathrm{OH}^{-}(a q)\)

5. By electrolysis of brine solution:

Cathode: \(2 \mathrm{H}_2 \mathrm{O}(l)+2 e \rightarrow \mathrm{H}_2(g)+2 \mathrm{OH}^{-}(a q)\)

Industrial Preparation Of Dihydrogen

Lane’s process:

This process consists of two stages:

Oxidation stage: Superheated steam is passed over iron filings heated to about 750-800°C. Iron reduces water to dihydrogen and itself gets oxidized to Fe3O4

⇒ \(\begin{gathered} 3 \mathrm{Fe}(s)+\underset{2}{4 \mathrm{H}_2 \mathrm{O}(g) \stackrel{750-800^{\circ} \mathrm{C}}{\longrightarrow}} \mathrm{Fe}_3 \mathrm{O}_4(s)+4 \mathrm{H}_2(g)+\text { heat } \\ \text { steam } \end{gathered}\)

Reduction stage: After the complete oxidation of iron to Fe3O4, the supply of steam is stopped and a steam of water gas (H2 + CO) is passed to reduce Fe3O4 back to iron.

⇒ \(\mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{H}_2 \rightarrow 3 \mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O} ;\)

⇒ \(\mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{CO} \rightarrow 3 \mathrm{Fe}+4 \mathrm{CO}_2\)

In the actual process, dihydrogen gas is manufactured from a small amount of iron by passing steam and water gas over it alternately.

Bosch process:

This process involves the steps as follows:

Preparation of watergate /syngas:

1. Water gas (1:1 mixture of CO and H2 ) is produced by passing superheated steam over red hot coke in the presence of Ni catalyst. This process of producing water gas (also called syngas, a term used to indicate a mixture of CO & H2 in any ratio) from coke or coal is called “coal gasification.”

⇒ \(\mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g) \stackrel{1000^{\circ} \mathrm{C}}{\mathrm{Ni}} \underbrace{\mathrm{CO}(g)+\mathrm{H}_2(g)}_{\text {Water gas }}-121.3 \mathrm{~kJ}\)

2. Syngas may also be produced by heating hydrocarbon with steam at 1000°C in the presence of a nickel catalyst. This is called steam reforming of hydrocarbon.

⇒ \(\mathrm{C}_n \mathrm{H}_{2 n+2}+n \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \stackrel{1000^{\circ} \mathrm{C}}{\mathrm{Ni}} \underbrace{n \mathrm{CO}(g)+(2 n+1) \mathrm{H}_2}_{\text {Syngas }}\)

Example: \(\mathrm{C}_3 \mathrm{H}_8(g)+3 \mathrm{H}_2 \mathrm{O}(g) \stackrel{1000^{\circ} \mathrm{C}}{\longrightarrow} 3 \mathrm{CO}(g)+7 \mathrm{H}_2(g)\)

Separation of dlhydrogen:

When syngas are passed over steam in the presence of iron chromate or a mixture of ferric oxide and chromium oxide heated at 400°C, CO is oxidized and more hydrogen is produced.

Hydrogen Separation ofdlhydrogen

This is called the water gas shift reaction. When the resulting mixture is passed through the water under 30 atm pressure, C2 dissolves leaving behind H2 which is collected.

Electrolysis of water: In this process, electrolysis of a 15-20% NaOH solution is carried out using an iron cathode and nickel-coated iron anode. Pure water is a bad conductor of electricity while water containing a small amount of alkali (or acid) is a good conductor of electricity.

Cathode:\(2 \mathrm{H}_2 \mathrm{O}(l)+2 e \rightarrow 2 \mathrm{H}_2(g)+2 \mathrm{OH}^{-}(a q)\)

Anode:\(2 \mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cl}_2(g)+2 e\)

Overall reaction:

⇒ \(2 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \rightarrow\mathrm{Cl}_2(g)+\mathrm{H}_2(g)+2 \mathrm{OH}^{-}(a q)\)

From methanol: The process involves the catalytic decomposition of methanol. A 1:1 mixture of vaporized methanol and water is passed over a special type of catalyst ‘base metal chromite type’ at 400°C. The mixture of H2 and CO obtained as a result of the decomposition reaction is made to react with steam at the temperature to give CO2 and more H2 gas.

⇒ \(\mathrm{CH}_3 \mathrm{OH} \underset{\text { Catalyst }}{\stackrel{40{ }^{\circ} \mathrm{C}}{\longrightarrow}} \mathrm{CO}+2 \mathrm{H}_2 ; \mathrm{CO}+\mathrm{H}_2 \mathrm{O} \underset{\text { Catalyst }}{\stackrel{400^{\circ} \mathrm{C}}{\longrightarrow}} \mathrm{CO}_2+\mathrm{H}_2\)

CO2 is separated bypassing the gas mixture through cold water under pressure.

At present, about 77% of industrial hydrogen is produced from petrochemicals, 18% from coal, 4% from the electrolysis of aqueous solutions, and 1% from other sources.

Properties Of Dihydrogen

Physical properties

  1. Dihydrogen is a colorless, odorless, tasteless, and highly inflammable gas.
  2. It is the lightest gas known. For example, one liter of H2 gas at STP weighs 0.0899 g. Its density is approximately 1/14 -th of that of air.
  3. It is diatomic gas (r = Cp/Cv = 1.40). The two atoms are joined by a very strong covalent bond (bond dissociation enthalpy = 435.9 kj • mol-1 ).
  4. Since its molecules are non-polar its solubility in water is extremely low.
  5. It can be liquefied under high pressure and at low temperatures.

Hydrogen Some physical constants of atomic and molecular hydrogen

Chemical Properties

  • The chemical properties of dihydrogen depend mainly on its bond dissociation enthalpy. The bond dissociation enthalpy of the H —H bond (435.88kj • mol-1 ) is higher than that of any single bond between two atoms.
  • At 2000K, only 0.081% of it dissociates to form H-atoms. At 5000K, the percentage of dissociation increases to about 95.5%. Hence, the activation energy of the reactions involving H2 is very high and at ordinary temperature is very stable and unreactive.
  • Nearly all of its reactions occur at much higher temperatures or under ultraviolet radiations. Atomic hydrogen having electronic configuration Is1 needs one more electron to complete its orbital. Therefore, atomic hydrogen is very reactive and capable of combining with almost all the elements.

It reacts in three different ways:

  1. By the loss of its single electron to form H+,
  2. By the gain of one electron to form H ion and
  3. By sharing its electrons with other atoms to form single covalent bonds.

Combustion: Dihydrogen is a combustible gas but does not support combustion. It burns with blue flame in air or in oxygen and as a result, water is formed. The reaction is highly exothermic.

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{H}_2 \mathrm{O}(l), \Delta H=-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Reaction with non-metals:

1. Reaction with halogens to form hydrogen halides (HX):

It combines with fluorine in the dark and at ordinary temperatures, with chlorine in the presence of sunlight, with bromine when heated, and with iodine when heated in the presence of Pt -catalyst to form the corresponding hydrogen halides.

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{F}_2(\mathrm{~g}) \stackrel{\text { Dark }}{\longrightarrow} 2 \mathrm{HF}(\mathrm{g}) ; \mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \stackrel{\text { Sunlight }}{\longrightarrow} 2 \mathrm{HCl}(\mathrm{g})\)

⇒ \(\mathrm{H}_2(g)+\mathrm{Br}_2(g) \stackrel{673 \mathrm{~K}}{\longrightarrow} 2 \mathrm{HBr}(g) ; \mathrm{H}_2(g)+\mathrm{I}_2(g) \underset{\mathrm{Pt}}{\longrightarrow} 2 \mathrm{HI}(g)\)

Therefore, the reactivity of halogens towards dihydrogen decreases in the order: \(\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2\)

2. Reaction with nitrogen to form ammonia:

⇒ \(3 \mathrm{H}_2(g)+\mathrm{N}_2(g) \underset{\mathrm{Fe}(\mathrm{Mo})}{\stackrel{673 \mathrm{~K} / 200 \mathrm{~atm}}{\longrightarrow}} 2 \mathrm{NH}_3(g)+\text { heat }\)

3. Reaction with sulfur to form hydrogen sulfide:

⇒ \(\mathrm{H}_2(g)+\mathrm{S}(l) \stackrel{700 \mathrm{~K}}{\longrightarrow} \mathrm{H}_2 \mathrm{~S}(g)\)

4. Reaction with carbon to form methane and acetylene:

⇒ \(2 \mathrm{H}_2(g)+\mathrm{C}(s) \stackrel{1275 \mathrm{~K}}{\longrightarrow} \mathrm{CH}_4(g)\)

⇒ \(\mathrm{H}_2(g)+2 \mathrm{C}(s) \underset{3300 \mathrm{~K}}{\stackrel{\text { Electric spark }}{\longrightarrow}} \mathrm{C}_2 \mathrm{H}_2(g)\)

Reaction with metals: Dihydrogen reacts with strongly electropositive metals like Na, K, Ca, etc., at much higher temperatures to form salt-like (ionic) metal hydrides. In these hydrides, the oxidation state of hydrogen is -1. In these reactions, H2 acts as an oxidizing agent.

⇒ \(\begin{array}{r} \stackrel{0}{\mathrm{H}}_2(g)+2 \mathrm{M}(g) \rightarrow 2 \mathrm{MH}^{-1}(s) ; \stackrel{0}{\mathrm{H}}_2(g)+\mathrm{M}^{\prime}(g) \rightarrow \mathrm{M}^{\prime-1} \mathrm{H}_2(s) \\ {\left[\mathrm{M}=\mathrm{Li}, \mathrm{Na}, \mathrm{K} \text { etc. and } \mathrm{M}^{\prime}=\mathrm{Ca}, \text { Ba etc. }\right]} \end{array}\)

Reaction with metal ions and metal oxides:

Dihydrogen reduces some metal ions (metals lying below hydrogen in the activity series) in aqueous solutions to the corresponding metals.

⇒ \(\mathrm{CuO}(s)+\mathrm{H}_2(g) \stackrel{\text { heat }}{\longrightarrow} \mathrm{Cu}(s)+\mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\mathrm{PbO}(s)+\mathrm{H}_2(g) \stackrel{\text { heat }}{\longrightarrow} \mathrm{Pb}(s)+\mathrm{H}_2 \mathrm{O}(l)\)

Reaction with carbon monoxide to form methanol:

⇒ \(\mathrm{CO}(g)+2 \mathrm{H}_2(g) \underset{\mathrm{ZnO} / \mathrm{Cr}_2 \mathrm{O}_3}{\stackrel{700 \mathrm{~K} / 200 \mathrm{~atm}}{\longrightarrow}} \mathrm{CH}_3 \mathrm{OH}(l)\)

Reaction with organic compounds: Dihydrogen reacts with many organic compounds in the presence of a catalyst to form useful hydrogenated products which are commercially important. For example:

1. When dihydrogen is passed through edible vegetable oils such as soybean oil, groundnut oil, cotton seed oil, etc., at about 200°C under pressure in the presence of finely divided Ni as a catalyst, they undergo hardening and change into edible fats such as vanaspati ghee (Example data) or margarine. In this reaction, the C=C bonds present in glycerides or triglycerides (triesters of glycerol) become saturated.

Hydrogen Reaction with organic compounds

The above process is known as hydrogenation or hardening of oils.

2. Hydroformylation is an important process for the manufacture of aldehydes and alcohols from alkenes. The reaction which involves the addition of a formyl group ( —CHO) and hydrogen atom to a carbon-carbon double bond, is carried out by treating alkenes with CO and H2 at high pressure (10-100 atm) and at temperatures between 40-200°C using transition metals as catalysts. The aldehydes thus obtained further get reduced to alcohols.

Hydrogen Reaction with organic compounds.

Proton (H+) has no chemical existence. Because of higher charge density and higher hydration enthalpy (256 kcal • mol-1), in water, it always remains solvated forming droxoniumion or hydroniumion or oxoniumion \(\left(\mathrm{H}_3 \mathrm{O}^{\oplus}\right)\). This ion is again stabilized by solvation through hydrogen bonding with water molecules. Therefore, in water H+ ions actually exists as [H30(H20)3)+ or [H9O4]+ions.

Allotropy Of Dihydrogen And The Two Active Forms Of Dihydrogen

Ortho- And Para-Hydrogen

  • A dihydrogen molecule contains two H-atoms. The nuclei of both atoms are spinning about their own axis like a top. Depending upon the direction of spin of the nuclei, the molecules of hydrogen are of two types.
  • Dihydrogenin which the spins of both the nuclei are in the same direction (parallel spin) is called ortho-hydrogen and dihydrogen in which the spins of both the nuclei are in the opposite direction (antiparallel spin) is called para-hydrogen.
  • These two types of dihydrogen are called nuclear spin isomers. These are also called allotropes of hydrogen. At ordinary temperatures, dihydrogen contains 75% ortho- and 25% para-hydrogen.
  • As the temperature is lowered the percentage of ortho-hydrogenin tire mixture decreases while that of the para-hydrogen increases. At very low temperatures (20K), it is possible to obtain pure (-100%) para-hydrogen.
  • At high temperatures (>400K), the mixture contains 75% ortho-form. Therefore, the para-form has lower energy as compared to the ortho-form. Both ortho- and para-forms of H2 have the same chemical properties.
  • However, due to different nuclear spins and consequently, different internal energies, they have different physical properties like specific heat, thermal conductivity, boiling point, etc.

Deuterium and helium also exist in ortho and para-allotropic forms.

Hydrogen Spin isomers of hydrogen

Atomic hydrogen

Atomic hydrogen Definition:

Hydrogen in its atomic state, produced by the dissociation of molecular hydrogen at a very high temperature is called atomic hydrogen.

Preparation:

Because of high H—H bond dissociation enthalpy, atomic hydrogen is produced only at a much higher temperature. It is usually obtained by passing dihydrogen at atmospheric pressure through an electric arc struck between two tungsten electrodes. The electric arc produces a temperature around 4000-4500°C.

⇒ \(\mathrm{H}_2(\mathrm{~g}) \stackrel{\text { Electric arc }}{\longrightarrow} 2 \mathrm{H}(\mathrm{g}) ; \Delta H=435.90 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Hydrogen Atomic hydrogen torch

Uses: Atomic hydrogen is very’ reactive (lifetime 0.3 sec). As soon as the hydrogen atoms come in contact with any metallic surface, they immediately get converted into molecular form liberating a large amount of energy (nearly 4000°C), and as a consequence, an extremely hot flame is produced. An atomic hydrogen torch, devised on this principle, is used for welding and cutting purposes

WBCHSE Class 11 Physics For Simple Harmonic Motion: Definition, Examples

Oscillation And Waves

Simple Harmonic Motion Mathematical Analysis Of SHM

Differential Equation: Let x be the displacement from the equilibrium position of a particle executing SHM.

Then the velocity of the particle is, v = \(\frac{d x}{d t}\)

The acceleration of the particle (i.e., the rate of change of velocity), a = \(\frac{d v}{d t}=\frac{d}{d t}\left(\frac{d x}{d t}\right)=\frac{d^2 x}{d t^2}\)

So, the equation of SHM can be written as \(\frac{d^2 x}{d t^2}=-\omega^2 x \text { or, } \frac{d^2 x}{d t^2}+\omega^2 x=0\)…(1)

Equation (1) is called the differential equation of SHM.

Solution Of The Differential Equation: Let x = ept be a solution for equation (1).

∴ \(\frac{d x}{d t}=p e^{p t} \text { and } \frac{d^2 x}{d t^2}=p^2 e^{p t}\)

Putting the values obtained above in equation (1) we have,

⇒ \(p^2 e^{p t}+\omega^2 e^{p t}=0\)

or, \(\left(p^2+\omega^2\right) e^{p t}=0\)

or, \(p^2+\omega^2=0\) (because \(e^{p t} \neq 0\))

or, \(p^2=-\omega^2 \text { or, } p= \pm i \omega \quad[\text { Here, } i=\sqrt{-1}]\)

So, the general solution of equation (1) is x = \(A^{\prime} \sin \omega t+B^{\prime} \cos \omega t\)…..(2)

Where A’ and B’ are Integration constants.

Inserting emx in the equation \(\frac{d^2 y}{d x^2}+c_1 \frac{d y}{d x}+c_2 y=0\), if we get m = a’±ib’, then the general solution of the equation is y = \(e^{a^{\prime} x}\left(A^{\prime \prime} \sin b^{\prime} x+B^{\prime \prime} \cos b^{\prime} x\right) .\)

Let us put A’ = Acosα and B’ = Asinα in equation (2).

Then, A = \(\sqrt{A^{\prime 2}+B^{\prime 2}} \text { and } \alpha=\tan ^{-1} \frac{B^{\prime}}{A^{\prime}}\)

Thus, x = A(sinωtcosα + cosωtsinα) or, x = Asin(ωt+α)…..(3)

This represents the general equation of simple harmonic motion expressing the displacement of the particle.

Differential Equation Special Cases:

1. If A’ = A and B’ = 0, i.e., α= 0, we have from equation (2) or (3), x = Asinωt ……(4)

From this equation, it is seen that, at t = 0, x = 0,

i.e., initially the particle is at its position of equilibrium.

2. If A’ = 0 and B’ = A, i.e., a = \(\frac{\pi}{2}\), we have from equation (2) or (3), x = Asost……..(5)

From this equation, it is seen that when t = 0, x = A, which is the maximum value of displacement of the particle.

So, if a particle executing SHM starts its motion from its equilibrium position, its displacement is expressed as a sine function. If it starts its motion from one end of its path, its displacement is expressed as a cosine function.

If the particle executing SHM starts its motion from another point of its path, then equation (3) is directly used.

Relation Between SHM And Uniform Circular Motion: SHM is the simplest form of linear periodic motion. Again, uniform circular motion is the simplest form of rotational periodic motion.

  • The relation between SHM and uniform circular motion can be shown by a mechanical example.
  • A wheel is rotating with uniform speed about its center. A rod is connected between the circumference of the wheel and the handle of a frictionless piston fitted within a cylinder.
  • With the uniform rotation of the wheel, the piston moves to and fro along the shown path uniformly. As the motion of the wheel is uniform, the motion of the piston will be simple harmonic.

Simple Harmonic Motion Relation Between SHM And Uniform Circular motion

Geometrical Proof: Suppose a panicle is moving with a uniform angular velocity in an anticlockwise direction. This is shown by an arrow, along the circumference A’C’B’D’ of a circle hating its center at O. The foot of the perpendicular, drawn from different positions of the particle on the diameter B’OA’ (or C’OD’), will execute the simple harmonic motion. This is proved below.

Simple Harmonic Motion Geometric Proof

  • When the particle is at P, the foot of the perpendicular drawn from P on the diameter B’OA’ is at N. Now, as the particle moves around the circle, the foot of the perpendicular moves along the diameter A’OB’. Let the particle starting from A’, move around the circle in an anticlockwise direction and come back to A’.
  • Its projection on A’OB’ (N) also moves along A’OB’, reverses its direction, and then comes back along the same path to its starting point simultaneously. This to-and-fro motion of N along A’B’ is a linear periodic motion.

Similarly, if we draw a perpendicular from P on the diameter C’OD’, the foot to the perpendicular will execute a linear periodic motion along the diameter C’OD’. This linear periodic motion will be simple harmonic if it can be shown that the acceleration of N is proportional to its displacement from O and is directed towards the position of equilibrium O.

Let ω be the uniform angular velocity of the particle and r be the radius of the circular path.

∴ The uniform linear speed of the particle is, ν = ωr

The centripetal acceleration of the particle at P along PO is, \(a_r=\frac{v^2}{r}=\omega^2 r\)

Let at any instant, OP make an angle θ with the diameter B’OA’ (taken as x-axis) and the displacement of the foot of the perpendicular (N) be x from the position of equilibrium.

∴ x = ON = OPcosθ = rcosθ

Again, the component of nr along the diameter A’OB’, i.e., die acceleration of N at that instant is a = arcosθ = ω²r cosθ = ω²x

Hence, the acceleration of N is proportional to its displacement from its position of equilibrium.

The component of ar along A’OB’ is related to the motion of N. Obviously, if the particle is at P, this component of acceleration is directed towards O, the position of equilibrium. So the motion of N is simple harmonic.

Therefore, when a particle is in a uniform circular motion, the motion of its projection on any diameter of the circular path is simple harmonic. The circle in the example above is called the circle of reference and the particle is called the reference particle.

Related To SHM Displacement: We know that the general equation for the displacement of a particle executing SHM is, x = Asin(ωt+α)….(1)

Simple Harmonic Motion Quantities Related To SHM

Related To SHM Displacement Special Cases:

  1. If the particle starts its motion from one of the extremities B or C of its path, then the equation becomes x = Acosωt …..(2)
  2. If the particle starts its motion from O, the position of equilibrium, then the equation becomes x = Asinωt……(3)

Related To SHM Velocity: From equation (1) we get,

sin (ωt+a) = \(\frac{x}{A}\)

∴ cos(ωt+ a) = \(\pm \sqrt{1-\frac{x^2}{A^2}}\)

So, the velocity of the particle executing SHM is

∴ v = \(\frac{d x}{d t}=A \omega \cos (\omega t+\alpha)= \pm A \omega \sqrt{1-\frac{x^2}{A^2}}\)

or, v = \(\pm \omega \sqrt{A^2-x^2}\)….(4)

It shows that the velocity of the particle depends on its displacement.

Related To SHM Velocity Special Cases:

  1. When x = 0, i.e., when the particle is at O, the position of equilibrium, v = ± ωA, which is the maximum velocity. ie.., νmax = ± ωA
  2. When x = ±A, i.e., when the particle is at B or C, the two extremities of its path, \(\nu= \pm \omega \sqrt{A^2-A^2}=0\), which is the minimum velocity. i.e., vmin = 0.

So, a particle executing SHM has different velocities at different points on its path. It passes the position of equilibrium with maximum velocity. The magnitude of its velocity gradually decreases as the particle moves towards its extremities from the equilibrium position and it momentarily comes to rest at the extreme points of its path.

Related To SHM Acceleration: From equation (1), we get

velocity, v = \(\frac{d x}{d t}=\omega A \cos (\omega t+\alpha)\)

∴ Acceleration, a = \(\frac{d v}{d t}=-\omega^2 A \sin (\omega t+\alpha)\)

or, a = -ω²x…..(5)

From this equation, we see that the acceleration of the particle depends on its displacement from its mean position. The negative sign indicates that acceleration and displacement are mutually opposite in direction.

Related To SHM Acceleration Special Cases:

  1. When x = 0, i.e., when the particle is at O, the position of equilibrium, a = 0, which is the minimum acceleration.
  2. When x = ±A, i.e., when the particle is at B or C, the extremities of its path, a =\(\mp\)ω²A, which is the maximum acceleration.

A particle undergoing SHM possesses different accelerations at different points on its path. It has zero acceleration at equilibrium and attains the maximum acceleration at the extreme positions of its path.

Instead of equation (1) if we use equation (2) or (3), we get the same equations (4) and (5) for velocity and acceleration.

Related To SHM Time Period And Frequency: Let the equation of a simple harmonic motion: x = Acosωt. In this case, at the beginning of the motion, i.e., at time t = 0, ωt = 0, and x = Acos0 = A. This implies that the particle begins its motion from one extreme end of its path.

Now, as time advances, when t = \(\frac{2 \pi}{\omega}\), again we get, x = Acos2π = A, i.e., the particle returns to the initial point from which it began its motion. Thus, an oscillation is completed.

So, for one complete oscillation, change in ωt = 2π-0 = 2π, i.e., for one complete oscillation, time taken is \(\frac{2 \pi}{\omega}\)(ωt = 2π).

As the total time elapsed for one complete oscillation is called the time period (T) of a SHM, we have, T = \(\frac{2 \pi}{\omega}\)….(6)

Related To SHM Time Period And Frequency Definition: Frequency is defined as the number of complete oscillations per second of a particle executing SHM.

In time T the number of oscillations is 1. Hence in unit time, the number of oscillations is \(\frac{1}{T}\).

∴ Frequency, n = \(\frac{1}{T}=\frac{\omega}{2 \pi}\) …..(7)

or, ω = 2πn

ω is called the angular frequency.

Considering only the magnitude of the acceleration of a particle executing SHM, we get from equation (5), \(\omega^2=\frac{a}{x} \text { or, } \omega=\sqrt{\frac{a}{x}}\)

∴ Time period, T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{x}{a}}\)

= \(2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}}\)…(8)

∴ Frequency, n = \(\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{a}{x}}\)

= \(\frac{1}{2 \pi} \sqrt{\frac{\text { acceleration }}{\text { displacement }}}\)….(9)

Related To SHM Amplitude: The amplitude of simple harmonic motion = OB = OC. Since, -1 ≤ sinθ≤+ 1 and -1≤ cosθ ≤ + 1, we get from equation (1), the maximum value of displacement x = amplitude = |±A| = A

∴ \(\overrightarrow{O B}=\vec{A} \quad \text { and } \overrightarrow{O C}=-\vec{A}\)

From equations (8) and (9) it is evident that the time period and frequency of an SHM do not depend on the amplitude A. So, if the amplitude of oscillation of a simple pendulum diminishes gradually due to air resistance, its time period remains unchanged (law of isochronism). For this, simple harmonic motion is called an isochronous motion.

Related To SHM Phase: The phase of a particle executing SHM at any instant, is defined as its state of motion at that instant. The term ‘state of motion’ indicates displacement, velocity, acceleration, etc., of the particle at any instant.

Suppose in equation (1),  θ= ωt + α……(10)

velocity, v = \(\pm \omega \sqrt{A^2-x^2}= \pm \omega A \cos \theta \text {; }\)

acceleration, a = -ω²x = -ω²Asinθ

Now, ω and A are both constants. So displacement, velocity, and acceleration of the particle at any instant depend entirely on the angle θ. The angle θ, expressed by equation (10), is called the phase angle of the simple harmonic motion. It is seen from equation (10), that, θ depends on time t. Thus, the phase of a particle executing SHM changes continuously with respect to time.

Related To SHM Phase Special Cases:

  1. If θ = 0, then x = Asinθ = 0, i.e., the particle is at O. If the value of θ becomes 90°, then x = Asin90° = A, i.e., the particle is at the end B of its path, i.e., the change of phase = 90° = \(\frac{\pi}{2}\).
  2. If θ = 270°, then x = Asin270° = -A, i.e., the particle is at C. So, when the particle moves from B to C, the change of phase angle = 270° -90° = 180° = π. The positions B and C are then said to be in opposite phases.

If θ = 450° = 360° + 90° , then x = Asin450° = Asin90° = A;

i.e., the particle is at B. So, when the particle starts its motion from B, goes to C, and then returns to B, the change of phase angle = 450° – 90° = 360° = 2π. In this case, the initial and the final positions are in the same phase.

Related To SHM Epoch: It is defined as the initial phase of motion (i.e., at t = 0) of the particle executing SHM.

If the equation of SHM is x = Asin(ωt+ α), then phase angle, θ = ωt+ α.

Putting t = 0 in this equation, we get epoch, θ = α

As a special case, if the particle starts its motion from one extremity of its path we have,

x = \(A \cos \omega t=A \sin \left(\omega t+\frac{\pi}{2}\right)\).

Here phase angle, θ = cot+\(\frac{\pi}{2}\) . If t = 0 , epoch = \(\frac{\pi}{2}\).

Again, if the particle starts its motion from the position of equilibrium, we have, x = Asinωt

Here, phase angle θ= ωt. Putting t = 0, epoch = 0 .

Related To SHM Epoch Phase Difference: In the case of two particles executing SHM, if the phase angle of the first particle is θ1 and that of the second particle is θ2 at an instant, then the phase difference of the SHMs is θ = θ2 – θ1 (or θ1 – θ2). Two simple harmonic motions having the same time period, frequency, and amplitude may have different phases. For example, if at an instant when the first particle reaches C, the second particle is at B, then their phase difference = 180°= π.

If the phase difference between two SHMs remains constant, i.e., it does not change with time, they are said to be coherent; the particles executing these SHMs are said to be in coherent motion.

Relations Among Displacement Velocity And Acceleration: Suppose a particle executing SHM starts its motion from the extremity B of its path. If T is the time period, then the particle crosses the position of equilibrium O in time \(\frac{T}{4}\); reaches C, the other extremity of the path in time \(\frac{T}{2}\) again on its way back crosses O, the position of equilibrium in time \(\frac{3T}{4}\) and returns to B in time T.

Simple Harmonic Motion Displacement Velocity And Acceleration Relations

Using v = \(\pm \omega \sqrt{A^2-x^2} \text { and } a=-\omega^2 x\), we can calculate displacement, velocity, and acceleration of the particle at different times, as shown in the table below.

Simple Harmonic Motion Relation Between Velocity Displacement Acceleration

The graph of displacement and velocity against time is shown, and the graph of displacement and acceleration against time.

Phase Difference Of Velocity And Acceleration With Displacement: Displacement of a particle executing a SHM is, x = Acosωt,

velocity, v = \(-A \omega \sin \omega t=A \omega \cos \left(\omega t+\frac{\pi}{2}\right)\)

and acceleration, a = -Aω² cosωt = Aω²cos(ωt+π)

Simple Harmonic Motion Phase Difference Of Velocity With Displacement

So, the phase difference between velocity and displacement is \(\frac{\pi}{2}\) or 90°, and the difference between acceleration and displacement is π or 180°.

Simple Harmonic Motion Phase Difference Of Acceleration With Displacement

Characteristics Of SHM

  1. Simple harmonic motion is a kind of linear periodic motion, i.e., in this motion, the particle moves to and fro following the same path repeatedly at regular time intervals.
  2. The acceleration of the particle executing SHM is always directed towards the position of equilibrium.
  3. The acceleration of the particle is proportional to its displacement from the position of equilibrium at any instant.
  4. When the particle passes the position of equilibrium, its velocity becomes maximum. The velocity of the particle gradually reduces and momentarily comes to zero at the extremities of its path.
  5. The time period of SHM does not depend on the amplitude. Though the amplitude decreases gradually due to various external resistances, the time period remains unchanged.

It is to be noted that SHM is a special form of periodic motion. If the periodic motion is

  1. Linear and
  2. The acceleration of the particle is proportional to its displacement from the position of equilibrium and is directed towards it, only then the motion of the particle is called simple harmonic.

The motion of the hands of a clock or the motions of the planets and satellites are periodic, but as these motions do not satisfy the above two conditions, they are not considered as simple harmonic. So, it can be said that all simple harmonic motions are periodic, but all periodic motions are not simple harmonic.

 Oscillation And Waves

Simple Harmonic Motion Mathematical Analysis Of SHM Numerical Examples

Example 1. A particle of mass 0.5 g is executing SHM with a time period of 2 s and an amplitude of 5 cm. Calculate its

  1. Maximum velocity,
  2. Maximum acceleration and
  3. Velocity, acceleration, and force acting on the particle when it is at a distance of 4 cm from its position of equilibrium.

Solution:

Amplitude, A = 5 cm; time period, T = 2 s

∴ ω = \(\frac{2 \pi}{T}=\frac{2 \pi}{2}=\pi \mathrm{rad} \cdot \mathrm{s}^{-1}\)

  1. Maximum velocity, \(v_{\max }=\omega A=\pi \cdot 5=5 \pi =5 \times 3.14=15.7 \mathrm{~cm} \cdot \mathrm{s}^{-1}=0.157 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
  2. Maximum acceleration, \(a_{\max }=\omega^2 A=\pi^2 \cdot 5 =5 \times(3.14)^2=49.298 \mathrm{~cm} \cdot \mathrm{s}^{-2} \approx 0.493 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
  3. When x = 4cm,

velocity, v = \(\omega \sqrt{A^2-x^2}=\pi \sqrt{5^2-4^2}\)

= \(3.14 \times 3=9.42 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

= \(0.094 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
\end{aligned}

acceleration, = \(39.438 \mathrm{~cm} \cdot \mathrm{s}^{-2}\)

= \(0.394 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

And force,F = \(m a=0.5 \times 39.438\)

= 19.72 dyn \(\approx 0.197 \times 10^{-3} \mathrm{~N} .\)

Example 2. A particle executing SHM possesses velocities 20cm · s-1 and 15cm · s-1 at distances 6 cm and 8 cm respectively from its mean position. Calculate the amplitude and the time period of the particle.
Solution:

Given:

A particle executing SHM possesses velocities 20cm · s-1 and 15cm · s-1 at distances 6 cm and 8 cm respectively from its mean position.

Velocity of the particle executing SHM, \(\nu=\omega \sqrt{A^2-x^2}\)

In the first case, 20 = \(\omega \sqrt{A^2-6^2}\)….(1)

In the second case, 15 = \(\omega \sqrt{A^2-8^2}\)…(2)

Dividing (1) by (2) we get, \(\frac{20}{15}=\frac{\omega \sqrt{A^2-6^2}}{\omega \sqrt{A^2-8^2}} \text { or, } \frac{4}{3}=\frac{\sqrt{A^2-36}}{\sqrt{A^2-64}} \text { or, } \frac{A^2-36}{A^2-64}=\frac{16}{9}\)

or, \(16 A^2-1024=9 A^2-324\)

or, \(7 A^2=700 \text { or, } A^2=100 \text { or, } A=10 \mathrm{~cm}=0.1 \mathrm{~m}\)

From equation (1) we get,

20 = \(\omega \sqrt{10^2-6^2}=8 \omega \text { or, } \omega=\frac{20}{8}=\frac{5}{2} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

∴ T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{5} \times 2=\frac{4}{5} \times 3.14=2.51 \mathrm{~s} .\)

Example 3. The time period and amplitude of a particle executing SHM are 10 s and 0.12 m respectively. Find its velocity at a distance of 0.04 m from its position of equilibrium.
Solution:

Given:

The time period and amplitude of a particle executing SHM are 10 s and 0.12 m respectively.

Here T = 10 s; A = 0.12 m

∴ \(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{10}=\frac{\pi}{5} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

So, v = \(\omega \sqrt{A^2-x^2}=\frac{\pi}{5} \sqrt{0.12^2-0.04^2}=\frac{\pi}{5} \sqrt{0.0128}\)

= \(0.071 \mathrm{~m} \cdot \mathrm{s}^{-1}\).

Example 4. The frequency of a vibrating wire is 200 Hz. The velocity of a particle on the wire is 4.35m · s-1 when it is at a distance of half its amplitude. Calculate the acceleration of the particle at that instant
Solution:

Given:

The frequency of a vibrating wire is 200 Hz. The velocity of a particle on the wire is 4.35m · s-1 when it is at a distance of half its amplitude.

Here, ω = 2πn = 2π ·200 = 400π rad · s-1 [n = 200 Hz]

Let A be the amplitude of the particle.

If x = \(\frac{A}{2}\), then v = 4.35 m · s-1

∴ v = \(\omega \sqrt{A^2-x^2}\)

or, \(4.35=400 \pi \sqrt{A^2-\frac{A^2}{4}}=400 \pi \cdot \frac{A \sqrt{3}}{2}\)

or, \(A=\frac{4.35 \times 2}{400 \pi \times \sqrt{3}} \mathrm{~m}\).

∴ Required acceleration, a = \(\omega^2 x=\omega^2 \cdot \frac{A}{2}=(400 \pi)^2 \times \frac{1}{2} \times \frac{4.35 \times 2}{400 \pi \sqrt{3}}\)

= \(\frac{400 \pi \times 4.35}{\sqrt{3}}=3154 \mathrm{~m} \cdot \mathrm{s}^{-2} \text { (approx.) }\)

The acceleration of a rapidly vibrating object may reach an exceptionally high value.

Example 5. Two particles executing SUM possess the same frequency. When the first particle just pusses the mean position of its path, the second particle moving lit the same direction is at a distance of 3 cm from its mean position. If the amplitude of vibration of the second particle is 6 cm, what is the phase difference of the two particles?
Solution:

Given:

Two particles executing SUM possess the same frequency. When the first particle just pusses the mean position of its path, the second particle moving lit the same direction is at a distance of 3 cm from its mean position. If the amplitude of vibration of the second particle is 6 cm

If A is the amplitude and θ is the phase angle, then displacement, x = Acosθ

For the first particle, \(x_1=A_1 \cos \theta_1 \text { or, } 0=A_1 \cos \theta_1\)

or, \(\cos \theta_1=0 or, \theta_1= \pm 90^{\circ}\)

For the second particle, \(x_2 =A_2 \cos \theta_2 \text { or, } \cos \theta_2=\frac{x_2}{A_2}=\frac{3}{6}=\frac{1}{2}\)

or, \(\theta_2= \pm 60^{\circ}\)

At the given instant, the two particles are in motion in the same direction.

So,

  1. If θ1 = 90° , then θ2 = 60°; phase difference θ21 = -30°
  2. If θ1 =-90°, then θ2 = -60°; phase difference θ21 = +30°

∴ The required phase difference is ± 30°.

Example 6. A particle executing SHM possesses velocities v1 and v2 when it is at distances x1 and x2 respectively from its mean position. Show that, the time period of oscillation is given by T = \(2 \pi\left(\frac{x_2^2-x_1^2}{v_1^2-v_2^2}\right)^{1 / 2}\)
Solution:

Given:

A particle executing SHM possesses velocities v1 and v2 when it is at distances x1 and x2 respectively from its mean position.

We know, \(v=\omega \sqrt{A^2-x^2}\)

According to the question, \(v_1=\omega \sqrt{A^2-x_1^2} \text { or, } v_1^2=\omega^2\left(A^2-x_1^2\right)\)…..(1)

Also, \(v_2=\omega \sqrt{A^2-x_2^2} or, v_2^2=\omega^2\left(A^2-x_2^2\right)\)

Subtracting (2) from (1) we get, \(v_1^2-v_2^2=\omega^2\left(x_2^2-x_1^2\right) or, \quad \omega^2=\frac{v_1^2-v_2^2}{x_2^2-x_1^2}\)

or, \(\omega=\left(\frac{v_1^2-v_2^2}{x_2^2-x_1^2}\right)^{\frac{1}{2}}\)

We know that, T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{\left(\frac{u_1^2-u_2^2}{x_2^2-x_1^2}\right)^{\frac{1}{2}}}=2 \pi\left(\frac{x_2^2-x_1^2}{u_1^2-u_2^2}\right)^{\frac{1}{2}}\),

Example 7. The equation of a simple harmonic motion is  x = \(10 \sin \left(\frac{\pi}{3} t-\frac{\pi}{12}\right) \mathrm{cm}\). Calculate its

  1. Amplitude,
  2. Time period,
  3. Maximum speed,
  4. Maximum acceleration,
  5. Epoch and
  6. Speed after 1s of initiation of motion

Solution:

x = \(10 \sin \left(\frac{\pi}{3} t-\frac{\pi}{12}\right) \mathrm{cm}\)

Comparing this equation with the equation of SHM, x = A sin(ωt + α) we get,

  1. Amplitude, A = 1.0 cm.
  2. \(\omega=\frac{\pi}{3}\) so time period, T = \(\frac{2 \pi}{\omega}=\frac{2 \pi \times 3}{\pi}=6 \mathrm{~s} \text {. }\)
  3. Maximum speed, \(\omega A=\frac{\pi}{3} \times 10=\frac{10 \pi}{3} \mathrm{~cm} \cdot \mathrm{s}^{-1} \text {. }\)
  4. Maximum Acceleration, \(\omega^2 A=\left(\frac{\pi}{3}\right)^2 \times 10=\frac{10 \pi^2}{9} \mathrm{~cm} \cdot \mathrm{s}^{-2}\)
  5. Epoch = \(-\frac{\pi}{12}=-15^{\circ}\)
  6. Displacement after 1s of initiation of motion, \(x_1=10 \sin \left(\frac{\pi}{3} \cdot 1-\frac{\pi}{12}\right)=10 \sin \left(60^{\circ}-15^{\circ}\right)\)

= \(10 \sin 45^{\circ}=\frac{10}{\sqrt{2}}=5 \sqrt{2} \mathrm{~cm}\)

And speed after 1 s of initiation of motion, \(v_1=\omega \sqrt{A^2-x_1^2}=\frac{\pi}{3} \sqrt{(10)^2-(5 \sqrt{2})^2}\)

= \(\frac{\pi}{3} \sqrt{50}=\frac{5 \sqrt{2}}{3} \pi \mathrm{cm} \cdot \mathrm{s}^{-1}\)

Example 8. Write down the equation of a simple harmonic motion whose amplitude Is 5 cm, epoch Is 0° and the number of vibrations per minute is 150.
Solution:

Given:

Amplitude Is 5 cm, epoch Is 0° and the number of vibrations per minute is 150.

According to the problem, amplitude, A = 0.05 m; epoch, α = 0°; frequency, n = \(\frac{150}{60}=\frac{5}{2} \mathrm{~Hz}\)

∴ \(\omega=2 \pi n=2 \pi \cdot \frac{5}{2}=5 \pi \mathrm{rad} \cdot \mathrm{s}^{-1}\)

So, the equation of the simple harmonic motion is

x = \(A \sin (\omega t+\alpha) \quad \text { or, } x=0.05 \sin 5 \pi t \mathrm{~m}\).

Example 9. The displacement of a vibrating particle at time t is given by x = \(A^{\prime} \sin \left(\frac{\pi}{6} t\right)+B^{\prime} \cos \left(\frac{\pi}{6} t\right)\), where A’ = 0. 03 m, B’ = 0.04m. Calculate the

  1. Amplitude,
  2. Epoch,
  3. Displacement, velocity, and acceleration of the particle after 2 seconds.

Solution:

Given:

The displacement of a vibrating particle at time t is given by x = \(A^{\prime} \sin \left(\frac{\pi}{6} t\right)+B^{\prime} \cos \left(\frac{\pi}{6} t\right)\), where A’ = 0. 03 m, B’ = 0.04m.

x = \(A^{\prime} \sin \left(\frac{\pi}{6} t\right)+B^{\prime} \cos \left(\frac{\pi}{6} t\right)\)….(1)

Let \(A^{\prime}=A \cos \alpha and B^{\prime}=A \sin \alpha\)

∴ \(A^2\left(\cos ^2 \alpha+\sin ^2 \alpha\right)=A^{\prime 2}+B^{\prime 2}\)

or, \(A=\sqrt{A^{\prime 2}+B^{\prime 2}} and \tan \alpha=\frac{B^{\prime}}{A^{\prime}}\)

So, x = \(A \sin \left(\frac{\pi}{6} t\right) \cos \alpha+A \cos \left(\frac{\pi}{6} t\right) \sin \alpha\)

or, x = \(A \sin \left(\frac{\pi}{6} t+\alpha\right)\)….(2)

From this equation we get,

  1. Amplitude, A = \(\sqrt{A^{\prime 2}+B^{\prime 2}}\) = \(\sqrt{(0.03)^2+(0.04)^2}=0.05 \mathrm{~m}\)
  2. Epoch, \(\alpha=\tan ^{-1} \frac{B^{\prime}}{A^{\prime}}=\tan ^{-1} \frac{4}{3}\)
  3. When t = 2 s, we get from equation (1) displacement, x = \(A^{\prime} \sin \left(\frac{\pi}{6} t\right)+B^{\prime} \cos \left(\frac{\pi}{6} t\right)\)

= \(0.03 \sin \frac{\pi}{3}+0.04 \cos \frac{\pi}{3}\)

= \(0.03 \times \frac{\sqrt{3}}{2}+0.04 \times \frac{1}{2}\)

= \(0.04598 \mathrm{~m} \approx 0.046 \mathrm{~m}\)

From equation (2) we get, \(\omega=\frac{\pi}{6} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

So, velocity after \(2 \mathrm{~s}\),

ν = \(\omega \sqrt{A^2-x^2}\)

= \(\frac{\pi}{6} \sqrt{(0.05)^2-(0.046)^2}=1.03 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Again, acceleration after 2s,

a = \(\omega^2 x=\left(\frac{\pi}{6}\right)^2 \times 0.046=1.26 \mathrm{~m} \cdot \mathrm{s}^{-2}\).

Example 10. The equation of motion of a particle executing SHM is expressed by x = \(10 \sin \left(10 t-\frac{\pi}{6}\right)\). Establish an equation to express its velocity and also calculate the magnitude of its maximum acceleration.
Solution:

Given, x = \(10 \sin \left(10 t-\frac{\pi}{6}\right)\)

∴ Velocity, \(\nu=\frac{d x}{d t}=10 \times 10 \cos \left(10 t-\frac{\pi}{6}\right)\)

= \(100 \cos \left(10 t-\frac{\pi}{6}\right) \text { unit. }\)

Again, acceleration, a = \(\frac{d v}{d t}=-10 \times 100 \sin \left(10 t-\frac{\pi}{6}\right)\)

= \(-1000 \sin \left(10 t-\frac{\pi}{6}\right) \text { unit. }\)

Acceleration will be maximum when, \(\sin \left(10 t-\frac{\pi}{6}\right)=1\)

∴ Magnitude of maximum acceleration, \(a_{\max }=1000\) units.

Example 11. Equation of a simple harmonic motion is y = \(2 \sin \left(4 t+\frac{\pi}{6}\right)\).  Find out its time period and initial phase.
Solution:

Given:

Equation of a simple harmonic motion is y = \(2 \sin \left(4 t+\frac{\pi}{6}\right)\).

Comparing the given equation with the general equation of SHM, y = Asin(ωt+θ), we get, ω = 4, i.e., time period, T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{4}=\frac{3.14}{2}=1.57 \mathrm{~s}\)

Phase = \(4 t+\frac{\pi}{6}\); and initial phase, i.e., phase at t = 0

∴ \(\theta=\frac{\pi}{6}=30^{\circ}\)

Example 12. The equations of the two SHMs are x1 = Asin(ωt + δ1) and x2 = Asin(ωt+δ2) respectively. They superimpose on each other. Find the amplitude of the resultant.
Solution:

Given:

The equations of the two SHMs are x1 = Asin(ωt + δ1) and x2 = Asin(ωt+δ2) respectively. They superimpose on each other.

Resultant, x = \(x_1+x_2\)

= \(A\left[\sin \left(\omega t+\delta_1\right)+\sin \left(\omega t+\delta_2\right)\right]\)

= \(2 A \sin \frac{\left(\omega t+\delta_1\right)+\left(\omega t+\delta_2\right)}{2} \cos \frac{\left(\omega t+\delta_1\right)-\left(\omega t+\delta_2\right)}{2}\)

= \(2 A \cos \frac{\delta_1-\delta_2}{2} \sin \left(\omega t+\frac{\delta_1+\delta_2}{2}\right)\)

∴ The amplitude of the resultant SHM is \(2 A \cos \frac{\delta_1-\delta_2}{2}\).

Example 13. The displacements of a particle executing SHM at three consecutive seconds are 6 cm, 10 cm, and 6 cm respectively. Find out the frequency of oscillation of the particle.
Solution:

Given:

The displacements of a particle executing SHM at three consecutive seconds are 6 cm, 10 cm, and 6 cm respectively.

Displacement, x = A cos(ωt+θ)

Let initially at t = 0, displacement = 6 cm.

So, at t = 1 s, displacement = 10 cm

and at t = 2s, displacement is = 6 cm

∴ 6 = A cosθ….(1)

10 = A cos(ω + θ)…(2)

6 = A cos(2ω + θ)…(3)

From (1) and (3) we get,

cos(2ω + θ) = cosθ = cos(-θ)

or, 2ω + θ = -θ or, ω = -θ

∴ From (2), we get,

10 = A cos0 or, A = 10 cm .

From (1), we get,

6 = 10cos(-ω) or, cosω = \(\frac{3}{5}\)

or, ω = 0.927 rad · s-1

∴ Frequency = \(\frac{\omega}{2 \pi}=\frac{0.927}{2 \times 3.14}=0.148 \mathrm{~s}^{-1} \text {. }\)

Example 14. The equation of a SHM is y = 3sin60πt. Calculate its amplitude, time period, and acceleration at its position of maximum displacement.
Solution:

Given:

The equation of a SHM is y = 3sin60πt.

Comparing the given equation with y = Asinωt, we get amplitude, A = 3 units; ω = 60π.

∴ Time period, T = \(\frac{2 \pi}{\omega}=\frac{2 \pi}{60 \pi}=\frac{1}{30} \mathrm{~s}\)

∴ Acceleration at its position of maximum displacement = ω²A = (60π)² x 3 = 1.065 x 105 unit · s-2.

Example 15. Displacement of a particle is given by x = 4(cosπt + sinπt) m. Find its amplitude.
Solution:

Given:

Displacement of a particle is given by x = 4(cosπt + sinπt) m.

x = \(4(\cos \pi t+\sin \pi t)\)

= \(4 \sqrt{2}\left(\cos \pi t \cdot \frac{1}{\sqrt{2}}+\sin \pi t \cdot \frac{1}{\sqrt{2}}\right)\)

= \(4 \sqrt{2}\left(\cos \pi t \cdot \cos \frac{\pi}{4}+\sin \pi t \cdot \sin \frac{\pi}{4}\right)\)

= \(4 \sqrt{2} \cos \left(\pi t-\frac{\pi}{4}\right) \mathrm{m}\)

∴ Amplitude = 4√2 m.

Example 16. Equation of motion of a particle executing SHM is x = \(5 \sin \left(4 t-\frac{\pi}{6}\right) \mathrm{m}\), where x is the displacement. If the displacement is 3 m, find the velocity of the particle.
Solution:

Given:

Equation of motion of a particle executing SHM is x = \(5 \sin \left(4 t-\frac{\pi}{6}\right) \mathrm{m}\), where x is the displacement. If the displacement is 3 m.

Velocity, v = \(\frac{d x}{d t}=5 \times 4 \cos \left(4 t-\frac{\pi}{6}\right)\)

= \(20 \sqrt{1-\sin ^2\left(4 t-\frac{\pi}{6}\right)}=20 \sqrt{1-\left(\frac{x}{5}\right)^2}\)

If x = 3m, then v = \(20 \sqrt{1-\left(\frac{3}{5}\right)^2}=20 \times \frac{4}{5}=16 \mathrm{~m} \cdot \mathrm{s}^{-1} .\)

Example 17. A point mass oscillates along the axis according to the law, x = \(x_0 \cos \left(\omega t-\frac{\pi}{4}\right)\). If the acceleration of the particle is written as a = Acos(ωt+δ), find A and δ.
Solution:

Given:

A point mass oscillates along the axis according to the law, x = \(x_0 \cos \left(\omega t-\frac{\pi}{4}\right)\). If the acceleration of the particle is written as a = Acos(ωt+δ),

x = \(x_0 \cos \left(\omega t-\frac{\pi}{4}\right)\)

∴ Velocity, \(v=\frac{d x}{d t}=-\omega x_0 \sin \left(\omega t-\frac{\pi}{4}\right)\)

and acceleration, \(\alpha=\frac{d v}{d t}=-\omega^2 x_0 \cos \left(\omega t-\frac{\pi}{4}\right)\)

= \(+\omega^2 x_0 \cos \left(\omega t-\frac{\pi}{4}+\pi\right)\)

= \(\omega^2 x_0 \cos \left(\omega t+\frac{3 \pi}{4}\right)\)

Given \(\alpha=A \cos (\omega t+\delta)\)

So comparing the two equations we have, \(A=\omega^2 x_0\) and \(\delta=\frac{3 \pi}{4}\)

Example 18. Two simple harmonic motions are represented by the equations \(y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)\) and y2 = 0.1 cos πt. What is the initial phase difference of the velocity of the first particle with respect to the second?
Solution:

Given:

Two simple harmonic motions are represented by the equations \(y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)\) and y2 = 0.1 cos πt.

∴ \(y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)\);

Velocity, \(v_1=\frac{d y_1}{d t}=100 \pi \times 0.1 \cos \left(100 \pi t+\frac{\pi}{3}\right)\)

= \(10 \pi \cos \left(100 \pi t+\frac{\pi}{3}\right)\)

Phase, ø1 = \(\frac{\pi}{3}\) at time t = 0

y2 = 0.1 cos πt

Velocity, \(\nu_2=\frac{d y_2}{d t}=-0.1 \pi \sin \pi t=0.1 \pi \cos \left(\pi t+\frac{\pi}{2}\right)\)

Phase, \(\phi_2=\frac{\pi}{2}\) at time t = 0

∴ Phase difference = \(\theta_1-\theta_2=\frac{\pi}{3}-\frac{\pi}{2}=-\frac{\pi}{6}\)

Example 19. The displacement of an object attached to a spring and executing SHM is given by x = 2 x 10-2 cosπtm. In what time the object attain maximum speed first?
Solution:

Given:

The displacement of an object attached to a spring and executing SHM is given by x = 2 x 10-2 cosπtm.

x = 2 x 10-2 cosπt

∴ v = \(\frac{d x}{d t}=-\left(2 \times 10^{-2} \pi\right) \sin \pi t\)

v is maximum when sinπt = ±1.

or, \(\pi t=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2} \ldots\)

The speed becomes maximum first when n t = \(\frac{\pi}{2}\)

or, t = \(\frac{1}{2}\) = 0.5

Example 20. If x, v, and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, show that the expressions \(\frac{aT}{x}\) and a²T² + 4π²v² do not change with time.
Solution:

Given:

If x, v, and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T

If A is the amplitude, ω is the angular frequency and x is the displacement of a particle executing SHM then,

velocity v = \(\pm \omega \sqrt{A^2-x^2}\)

acceleration a = \(-\omega^2 x\)

time period T = \(\frac{2 \pi}{\omega}\)

∴ \(\frac{a T}{x}=\frac{-\omega^2 x \cdot \frac{2 \pi}{\omega}}{x}\)

∴ \(a^2 T^2+4 \pi^2 v^2=\left(-\omega^2 x\right)^2\left(\frac{2 \pi}{\omega}\right)^2+4 \pi^2 \omega^2\left(A^2-x^2\right)\)

= \(\omega^4 x^2 \frac{4 \pi^2}{\omega^2}+4 \pi^2 \omega^2 A^2-4 \pi^2 \omega^2 x^2 \)

= \(4 \pi^2 \omega^2 A^2=\text { constant }\)

Example 21. Two simple harmonic motions of angular frequencies 100 and 1000 rad · s-1 have the same displacement amplitude. What is the ratio of their maximum acceleration?
Solution:

Given:

Two simple harmonic motions of angular frequencies 100 and 1000 rad · s-1 have the same displacement amplitude.

Acceleration, f= -ω²x

If the amplitude is a, maximum acceleration =ω²a

∴ Ratio of maximum accelerations = \(\frac{\omega_1^2 a}{\omega_2^2 a}=\left(\frac{\omega_1}{\omega_2}\right)^2=\left(\frac{100}{1000}\right)^2=\frac{1}{100}\)

Example 22. A particle initially at rest at a distance of 5 cm from its mean position performs a SHM completing 60 oscillations in 2 seconds. Find the equation representing the displacement of the particle at any subsequent instant. What will be its equation if initially the particle were at the mean position?
Solution:

Given:

A particle initially at rest at a distance of 5 cm from its mean position performs a SHM completing 60 oscillations in 2 seconds. Find the equation representing the displacement of the particle at any subsequent instant.

Time taken by the particle in SHM to complete 60 oscillations = 2 s.

∴ Time period of the particle, T = \(\frac{2}{60}\) =\(\frac{1}{30}\) s .

Angular velocity of the particle, \(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{\frac{1}{30}}=60 \pi\)

The particle was at rest and about 5 cm away from its equilibrium position.

∴ The amplitude of the particle, a = 5 cm.

Let, at any time t, the particle is x cm away from its equilibrium position.

∴ Displacement of the particle, x = 5cos(60πt+ θ) cm

At t = 0, x = 5 cm

∴ 5 = 5cosθ or, cosθ =1 or, θ = 0

∴ x = 5cos60πt

If the particle starts its oscillation from the equilibrium position, then, x = 5 sin60πt.

Example 23. The equation of two SHMs are \(y_1=10 \sin \left(4 \pi t+\frac{\pi}{4}\right)\), \(y_2=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\), What is thne ratio of therir amplitudes?
Solution:

Given:

The equation of two SHMs are \(y_1=10 \sin \left(4 \pi t+\frac{\pi}{4}\right)\), \(y_2=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\),

The amplitude of the first SHM, A1= 10

Now, \(y_2=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\)

= \(5 \times 2\left(\frac{1}{2} \sin 3 \pi t+\frac{\sqrt{3}}{2} \cos 3 \pi t\right) \)

= \(10\left(\cos \frac{\pi}{3} \sin 3 \pi t+\sin \frac{\pi}{3} \cos 3 \pi t\right) \)

= \(10 \sin \left(3 \pi t+\frac{\pi}{3}\right)\)

∴ The amplitude of the second SHM, \(A_2=10\)

Hence, \(\frac{A_1}{A_2}=\frac{10}{10}=1 or, A_1: A_2=1: 1\).

Graphical Representation Of SHM: If a particle executing simple harmonic motion begins its motion from the position of equilibrium, then the equation of motion of the particle is x = \(A \sin \omega t=A \sin \frac{2 \pi}{T} t\)

During a complete period, i.e., from t = 0 to t = T, the variation of displacement with time is shown in the following table:

Simple Harmonic Motion Graphical Representation Of Simple Harmonic Motion At T Equal To Zero

Now on graph paper, time is placed along the horizontal axis and displacement along the vertical axis. The points O, B, C, D, and E are plotted according to their corresponding coordinates in the above table. The points are connected with a curved line. The graph OBCDE is obtained. This is a sine curve. This curve represents a simple harmonic motion.

Simple Harmonic Motion Graphical Representation Of Simple Harmonic Motion With Sine Curve

If a particle executing simple harmonic motion starts its motion from one end of Its path, then the equation of motion of the particle is x = \(A \cos \omega t=A \cos \frac{2 \pi}{T} t\)

Proceeding similarly, we get the following table and the graph. The graph B’C’D’E’F’ is a cosine curve.

Simple Harmonic Motion Graphical Representation Of Simple Harmonic Motion At T Equal To 1

The sine and the cosine curves are identical except for their initial phase difference. For their special symmetry, sine and cosine functions are called sinusoidal functions. As a simple harmonic motion is represented by any one of these sinusoidal functions, they are also known as harmonic functions.

Simple Harmonic Motion Graphical Representation Of Simple Harmonic Motion

Composition Of Two Colinear SHMs Of The Same Frequency By Graphical And Analytical Methods: Two simple harmonic motions, having the same frequency n (i.e., time period, T = \(\frac{1}{n}\) and angular frequency, ω = 2πn, are being executed along the x-axis, i.e., they are collinear. To obtain the resultant motion due to the superposition of the two SHMs, we have to know their phase difference.

1. Two SHMs Are In Phase: The two SHMs of the same time period are in phase.

Let the equation of the first motion be \(x_1=A \sin \omega t=A \sin \frac{2 \pi}{T} t.\)

and the equation of the second motion be \(x_2=B \sin \omega t=B \sin \frac{2 \pi}{T} t.\)

The resultant displacement at any instant is the vector sum of the displacements of the two individual motions at that instant.

So, the resultant displacement, \(x=x_1+x_2=A \sin \frac{2 \pi}{T} t+B \sin \frac{2 \pi}{T} t=(A+B) \sin \frac{2 \pi}{T} t\)

Therefore, the resultant motion is an SHM of the same frequency with an amplitude equal to (A + B).

The displacements of the two individual motions and their resultant displacement at different times (from t = 0 to t = T) are shown in the table below.

Simple Harmonic Motion Two Simple Harmonic Motion Are In Phase

Plotting the referred values in the table, we get the graph of the resultant displacement x. From the graph, we come to know

Simple Harmonic Motion Graphical Representation Two Simple Harmonic Motion Are In Phase

  1. The resultant motion is also simple harmonic.
  2. The frequency and the time period of the resultant motion are equal to those of the individual motions.
  3. The amplitude of the resultant motion is equal to the sum of the amplitudes of the individual motions.
  4. The resultant motion is also in phase with the individual motions.

2. Two SHMs In Opposite Phase: Two simple harmonic motions have the same frequency and the same time period, but they are in opposite phases.

So, the equation of the first SHM: \(x_1=A \sin \omega t=A \sin \frac{2 \pi}{T} t\).

The equation of the second SHM: \(x_2=B \sin (\omega t+\pi)=-B \sin \omega t=-B \sin \frac{2 \pi}{T} t \text {. }\)

[As the two motions are in opposite phases, the phase difference between them = 180° = π]

∴ Resultant displacement, x = \(x_1+x_2=A \sin \frac{2 \pi}{T} t-B \sin \frac{2 \pi}{T} t=(A-B) \sin \frac{2 \pi}{T} t\)

From t = 0 to t = T, the values of x1, x2, and x are given in the tables below.

Simple Harmonic Motion Two Simple Harmonic Motion Are In Opposite Phase

Plotting the referred values in the table, we get the graph of the resultant displacement x. From the graph, we come to know

Simple Harmonic Motion Graphical Representation Of Two Simple Harmonic Motion Are In Opposite Phase

  1. The resultant motion is also simple harmonic.
  2. The frequency and the time period of the resultant motion are equal to those of the individual motions.
  3. The amplitude of the resultant motion is equal to the difference of the amplitudes of the individual motions.
  4. The resultant motion is in phase with the motion having a larger amplitude but in the opposite phase with the motion having a smaller amplitude.

Due to the superposition of two SHMs of the same amplitude but of opposite phases, the resultant displacement is x = x1 + x2 = 0 for all values of t. It implies that the particle, on which these two SHMs superpose, remains at rest.

Class 11 Chemistry Some Basic Concepts Of Chemistry

Some Basic Concepts Of Chemistry Introduction

Chemistry is an important branch of science. It deals with the source, composition, structure, and properties of matter with special reference to the physical and chemical changes that matter undergoes under different conditions.

With time, the greatest discoveries of chemistry have made human life more comfortable and have facilitated its advancement as well In the last few decades, a tremendous change in the field of chemistry has been observed.

It has become significantly vast and complex. For the convenience of research and a better understanding of the subject, it has been divided into several branches

Class 11 Chemistry Some Basic Concepts Of Chemistry Difference Branches Of Chemistry

Laws Of Chemical Combination, Atomic And Molecular Theory, Equivalent Weight

Importance And Scope Of Chemistry

Chemistry plays amajorrolein science and is often intertwined with different branches of science such as physics, biology, geology, etc. It has made many contributions to human civilization.

Principles of chemistry are found to be very useful in diverse areas such as weather patterns, biochemical processes, functioning of brains, operations of computers, etc.

Chemistry helps to fulfill human needs for food, health care products, and other materials required for improving the quality of life.

Some of the major contributions of chemistry are given below:

  1. Chemistry In Agriculture And Preservation Of Food
  2. Several chemical fertilizers like urea, ammonium sulfate, calcium nitrate, superphosphate of lime, etc. are used for better production of crops.
  3. By hydrogenation of edible oil, artificial fats (such as vanaspati) are prepared.
  4. It helps to protect crops from the harmful effects of insects and bacteria by the use of effective insecticides (such as gammexane, aldrin, parathion, etc.), herbicides, and fungicides.
  5. The use of preservatives {Example sodium benzoate, salicylic acid, potassium nitrate, sodium chloride, etc.) has helped to preserve food materials. like jam, jelly, butter, squashes, fish, offer, etc., for longer periods of time.
  6. Different chemical methods are available to indicate the presence of adulterants in order to ensure the supply of pure foodstuff.

Chemistry In Health Care And Sanitation

  1. Analgesics (For example aspirin, analgin, etc.) are used to give relief from different types of pain.
  2. Antipyretics (For example paracetamol, ibuprofen, etc.) are used to bring down body temperature during fever.
  3. Antiseptics such as Dettol, savlon, tincture of iodine, etc. are used to stop infection of wounds.
  4. Antibiotics (For example penicillin, cephalosporin, tetracycline, streptomycin, chloramphenicol) are used to curb infection and cure diseases like pneumonia, bronchitis, typhoid, tuberculosis, etc.
  5. Tranquilizers such as barbituric acid, veronal, valium, reserpine, etc. are prescribed to patients suffering from mental diseases to reduce their tension or anxiety.
  6. Anesthetics like chloroform, cocaine, novocaine, etc. are applied to patients to make surgical operations painless.
  7. Today dysentery and pneumonia have become curable by the use of penicillin and sulpha drugs.
  8. The widely used drug quinine has now been replaced by some more effective antimalarials like chloroquine, primaquine, etc.
  9. As a preventive measure against various types of diseases, vaccines are found to be used widely Example tetanus toxoid (for tetanus), TABC (for typhoid, paratyphoid A, B, and cholera), oral polio (for polio), etc.
  10. Nowadays life-saving drugs such as taxol and cisplatin are used in cancer therapy; Azidothymidine (AZT) is used for AIDS victims.
  11. Synthetic vitamins and tonics have significant contributions towards the better health of human beings.
  12. Bleaching powder, potassium permanganate, ozone gas, low concentration of chlorine, etc. are used for sterilization of water to make it suitable for drinking.
  13. Disinfectants like phenol and cresols are used to kill the micro-organisms present in drains, toilets, floors, etc.

Chemistry In Comforts, Pleasures And Luxuries

  1. The contribution of chemistry towards the betterment of human society is widely acknowledged by all and it has a profound influence on our daily life.
  2. Synthetic fibers such as terylene, nylon, rayon, dacron, orlon, etc., are used to prepare clothes that are more comfortable, durable, attractive, and easy to wash.
  3. Polythene is used for making toys, bottles, tubes, pipes, kitchen and domestic appliances, sheets for packing materials, and coated wires and cables.
  4. PVC is used for making rain-coats, hose pipes, conveyor belts, radio and TV components, insulating material for wires, cables, and other electrical goods, gramophone records, safety helmets, refrigerator components, bi-cycle, and motor-cycle mudguards, etc.
  5. Phenol-formaldehyde resin and bakelite are used for making combs, fountain pen barrels, electrical goods (switches and plugs), heater handles, telephone parts, cabinets for radio and television, etc. Films used in cameras are made of celluloid coated with suitable chemicals.
  6. Cosmetics such as cream, lipstick, sunscreen lotion, face powder, talcum powder, perfume, toothpaste, nail polish, shampoo, hair dye, etc., are all chemical substances.
  7. Soaps and detergents used for cleaning clothes are chemical substances. Bio-degradable detergents are now in use to avoid environmental pollution.
  8. Paint, varnishes, and lacquer are applied on walls, wooden furniture, and metallic articles to make them more attractive, durable, and resistant to corrosion.
  9. Articles made of iron are electroplated by nickel, chromium, silver, gold, etc., so as to prevent them from rusting and to make them more attractive and durable.
  10. Ammonia, liquid sulfur dioxide CFC, etc. are used as refrigerants in refrigerators and air-conditioners.
  11. Cement, steel, iron, etc., are widely used m the construction of multi-storeyed buildings, dams, and bridges.
  12. LPG and natural gas having high calorific value are used as smokeless fuels for cooking. Compressed natural gas (CNG) is now used as a fuel in public vehicles in metropolitan cities.

Chemistry In Industry

  1. Chemistry plays an important role in the development and growth of a number of industries. Some important examples of manufacturing processes are
  2. Extraction of metals such as iron, aluminum, zinc, copper silver, gold, etc.
  3. Refining of petroleum to produce petroleum ether, gasoline (petrol), kerosene, diesel, paraffin oil, lubricating oil, solvent naphtha, liquid paraffin, petroleum jelly, paraffin wax, etc.
  4. Plastics such as polyethylene, PVC, bakelite, polyurethanes, Teflon, etc.
  5. Synthetic fibers such as nylon, terylene, rayon, etc.
  6. Paints, varnishes, lacquer, and synthetic dyes.
  7. Cement, glass, and ceramic materials.

The dark side of chemistry: Chemistry plays a pivotal role in our daily lives and luxuries. However improper use of chemistry has a negative impact on human society. In modern times, atomic energy is mainly used in chemical warfare.

Different chemical weapons and explosives like RDX are used for terrorist activities. Drugs like cocaine, LSD, and heroin have adverse effects on the youths.

Nature Of Matter

Anything which has mass occupies some space and can be felt by one or more of our senses is called matter.

Everything around us such as aspen, pencil, wood, water, milk, air, etc., and all living beings are composed of matter.

Classification of matter: Matter can be classified in two different ways—Physical classification and Chemical classification.

Physical classification of Matter: At ordinary temperature and pressure, matter can exist in three physical states viz., solid, liquid, and gas. The constituent particles of matter in the three states can be represented as shown in.

Class 11 Chemistry Some Basic Concepts Of Chemistry Arrangement Of Particles In solid, Liquid And Gaseous State

The essential points of differences between the three states of matter are given in the following table:

Class 11 Chemistry Some Basic Concepts Of Chemistry Soild liquid and center

A given substance can be made to exist in the solid, liquid, or gaseous state by changing the conditions of temperature and pressure.

Class 11 Chemistry Some Basic Concepts Of Chemistry Soild liquid or gaseous ,temperature and pressure

Chemical classification of Matter : On the basis of chemical composition, matter can be classified into two major categories such as

  1. Mixtures and
  2. Pure substances.

These can be further subdivided as follows:

Class 11 Chemistry Some Basic Concepts Of Chemistry Matter of Mixture

A mixture is made up of two or more substances (present in any ratio) which are called its components. For example, a sugar solution consists of two components i.e., sugar and water.

In a homogeneous mixture, the components completely mix with each other and its composition remains uniform throughout.

The components of such a mixture cannot be seen even under a microscope. Some examples are air, glucose solution, seawater, petrol, etc.

In contrast to this, in heterogeneous mixtures, the composition is not uniform throughout, and sometimes the different components can be seen even by the naked eye.

For example, the mixtures of sugar and salt, sand, and iron filings are heterogeneous mixtures. The components of such mixtures can be separated by using physical methods such as filtration, crystallization, distillation, chromatography, etc.

Pure substances have characteristics different from that of the mixtures. They have fixed composition throughout the entire mass. Some examples are iron, copper, silver, gold, water, sucrose, etc.

Sucrose contains carbon, hydrogen, and oxygen in a fixed ratio and hence it has a fixed composition.

The constituents of pure substances cannot be separated by simple physical methods. Pure substances are further classified into elements and compounds.

Element: An element consists of only one type of particle. The constituent particles may be atoms or molecules.

Oxygen, nitrogen, sodium, copper, silver, etc., are some examples of elements. They all contain atoms of one type.

The smallest particles (having independent existence) present in metallic elements such as sodium, potassium, etc. are called atoms.

On the other hand, the smallest possible particles (having independent existence) of some other elements (such as oxygen, nitrogen, phosphorus, etc.) are called molecules, which consist of two or more atoms.

Thus, two atoms of oxygen and four atoms of phosphorus combine separately to form molecules of oxygen and phosphorus respectively.

Class 11 Chemistry Some Basic Concepts Of Chemistry Representation of atoms and molecules

Two or more atoms of different elements combine together to form the molecule of a compound.

Examples of some compounds are water, carbon dioxide, ammonia, etc. The molecules of carbon dioxide and water are depicted in.

Class 11 Chemistry Some Basic Concepts Of Chemistry Structure of carbon dioxide and water molecule

A carbon dioxide molecule consists of one carbon atom and two oxygen atoms. Similarly, a molecule of water is composed of two hydrogen atoms and one oxygen atom.

It is thus seen that the atoms of different elements are present in a compound in a fixed ratio and this ratio is the characteristic of a particular compound.

It is needless to mention that the properties of a compound are completely different from those of the constituent elements.

For example, hydrogen and oxygen are gaseous substances while the compound (water) formed by their combination is a liquid at ordinary temperature.

The constituents of a compound cannot be separated by physical methods. They can, however, be separated by chemical methods.

Class 11 Chemistry Some Basic Concepts Of Chemistry Difference between mixture and compound

Physical Quantities

Characteristics of matter that can be examined as a measurable quantity are called physical quantities. Example length, mass, time, temperature, area, volume, velocity, acceleration, force, etc.

Units For Measurement Of Physical Quantities

A unit is defined as the standard of reference chosen for the measurement of any physical quantity.

Example: Suppose the length of a bench is 2 meters. Here length is the physical quantity andmetre is the unit length.

The numerical magnitude ‘2’ implies that the length of the bench is two times that of the value of 1 meter (which is the standard of reference chosen for the measurement of length).

Fundamental Units Used In Different Systems 

Class 11 Chemistry Some Basic Concepts Of Chemistry Fundamental Units used in different systems

 Basic physical quantities and their units in the SI system

Class 11 Chemistry Some Basic Concepts Of Chemistry basic physical quantities and their units in SL System

Some Common Derived Units In CGS And SL System 

Class 11 Chemistry Some Basic Concepts Of Chemistry Some Common Dervied Units in CGS and S1 System

  • The volume of liquids is commonly measured in a liter (L) but this is not a SI unit. 1 L = 1000 mL = 1000 cm3 = 1 dm3
  • Wavelength is expressed in angstrom (A). 1A = 10-10 m

Some Commonly Used Prefixes In CGS And SI Systems 

Class 11 Chemistry Some Basic Concepts Of Chemistry Some Common used prefixed in CGS And SI system

Frequently Used Greek Letters 

Class 11 Chemistry Some Basic Concepts Of Chemistry Frequently Used Greek Letters

Important points regarding the use of SI units:

  1. No dot (•) can be used in between the letters or at the end of the letters used for abbreviations of basic units. Thus, the symbol for centimeter is cm (it is neither c.m. nor cm.)
  2. Abbreviations of units do not have a plural ending. Thus, it is incorrect to write 5 ems or 12 gms. These should be 5 cm and 12 gm respectively.
  3. The abbreviations of units named after scientists start with capital letters and not with small letters. Some examples are Newton (N), Joule (J), Pascal (Pa), Ampere (A), etc. If the names are used in full instead of abbreviations then these start with small letters Example Newton, ampere, Pascal, etc.
  4. Abbreviations of other units such as meter (m), kilogram (kg), second (s), etc., start with small letters.
  5. The temperature in the kelvin scale should not be roprosonled with a degree (°). So, It Is proper to say 290K but not 2H- K.
  6. The derived units such as square meter and cubic centimeter are denoted ns m2 (but not sqm) and cm9 (but not cc) respectively.
  7. To indicate divisions, it is better to use inverse sign. However, the ‘/’ symbol can be used blit once only. One example is Kg-1.K-1 but not J/(Kg . K) or, J/Kg/K.

Conversion of physical quantities in different units involves the following steps:

1. Firstly, we have to determine a unit conversion factor, then

2. The given magnitude of the physical quantity in question, is multiplied by a suitable unit conversion factor such that all units are canceled out leaving behind only the required units.

This is illustrated by the following examples:

To express the length of a wooden pencil (say, 4 inches long) in cm: 

We know, 1 inch = 2.54 cm

∴ \(\frac{1 \mathrm{inch}}{2.54 \mathrm{~cm}}=1=\frac{2.54 \mathrm{~cm}}{1 \mathrm{inch}}\)

Here, both the ratios \(\frac{1 \mathrm{inch}}{2.54 \mathrm{~cm}} \text { and } \frac{2.54 \mathrm{~cm}}{1 \text { inch }}\) and are equal to

‘1’ because the lengths 1 inch and 2.54 cm are exactly equal to each other. Either of these ratios is called unit conversion factor or simply unit factor.

The magnitude of any quantity will remain unchanged when it is multiplied by a suitable unit conversion factor.

Based on these rules, the length of the given wooden pencil can be expressed in cm as follows:

4 inch = 4 inch x 1 (unit factor)

⇒ \(=4 \mathrm{inch} \times \frac{2.54 \mathrm{~cm}}{\text { linch }} 4 \times 2.54 \mathrm{~cm}=10.16 \mathrm{~cm}\)

Here, the quantity inch’ is multiplied by a particular unit conversion factor so that the unit ‘inch’ gets canceled out.

To express the length of an iron rod (say, 30.48 cm long)in inches:

Here, 30.48 cm = 30.48 cm x 1 (unit conversion factor)

⇒ \(=30.48 \mathrm{~cm} \times \frac{1 \mathrm{inch}}{2.54 \mathrm{~cm}}=\frac{30.48}{2.54} \mathrm{inch}=12 \mathrm{inch}\)

In this case, the given length is multiplied by the particular unit conversion factor so that the unit ‘cm’ is canceled out from the numerator and the denominator.

To express a given volume (say, 51) of water in m³

Wo know, 1 L = 1000 cm³

Again, 1m = 100 cm \(\frac{1 \mathrm{~m}}{100 \mathrm{~cm}}=1=\frac{100 \mathrm{~cm}}{1 \mathrm{~m}}\)

⇒ \(\text { So, }\left(\frac{1 \mathrm{~m}}{100 \mathrm{~cm}}\right)^3 \approx 1^3=\left(\frac{100 \mathrm{~cm}}{1 \mathrm{~m}}\right)^3\)

⇒ \(\text { So, }\left(\frac{1 \mathrm{~m}}{100 \mathrm{~cm}}\right)^3=1^3=\left(\frac{100 \mathrm{~cm}}{1 \mathrm{~m}}\right)^3\)

⇒ \(\text { or, } \frac{1 \mathrm{~m}^3}{10^6 \mathrm{~cm}^3}=1=\frac{10^6 \mathrm{~cm}^3}{1 \mathrm{~m}^3}\)

Now, 5L =(5×1000)(5×1000)cm³=5000 cm³

=5000 cm³x1(unit factor)

⇒ \(=5000 \mathrm{~cm}^3 \times \frac{1 \mathrm{~m}^3}{10^6 \mathrm{~cm}^3}\)

⇒ \(=\frac{5000}{10^6} \mathrm{~m}^3=5 \times 10^{-3} \mathrm{~m}^3\)

Accuracy And Precision

The accuracy of a measurement is the agreement of the measured value to the true value. As the difference between the measured value and the true value decreases, the accuracy of the measurement increases.

The degree of accuracy of any measurement depends upon

  1. The accuracy of the measuring device used and
  2. The skill of the operator. The difference between the measured value and the true value is called the absolute error.

Precision refers to the closeness of the results of various measurements for the same quantity.

Good precision does not necessarily mean good accuracy because various measurements may involve the same mistake repeatedly. This can be understood from the given illustration.

Let, the true value for a measurement be 2.00 g. Four different cases may arise when the actual measurements are carried out by different observers A, B, C, and D.

Class 11 Chemistry Some Basic Concepts Of Chemistry Accurancy And Precision

Significant Figures Definition

The total number of digits present in a number (starting from the first non-zero digit) including the last digit whose value is uncertain is called the number of significant figures.

Explanation: Suppose a student is asked to measure the length of a pencil with the aid of a meter scale (in which the closest distance between two successive marks is 0.1 cm). The student reports his experimental result as 15.4 cm.

Here the last digit (i.e., 4) of the reported result is not absolutely correct because there are two possibilities:

  • The length of the pencil may be greater than 15.3 cm but slightly smaller than 15.4 cm or
  • Its length may be much smaller than 15.5 cm but slightly greater than 15.4 cm.

Class 11 Chemistry Some Basic Concepts Of Chemistry Signidficant mirrors

From the above discussion, it can be stated that in the above-reported value (i.e., 15.4cm) there are three significant figures [the first two digits (1 and 5) are certain and the last digit (4) is uncertain].

Example: Suppose the mass of an object measured by an analytical balance is reported to be 12.4567 g. If the accuracy of the balance is 0.0001 g, the actual mass of the object will be (12.4567 ± 0.0001 )g i.e., the value lies between 12.4566g and 12.4568g.

Thus in the reported mass, the first five digits (1,2,4,5 and 6) are certain while the last digit (7) is uncertain. This means that there are significant figures in the reported mass.

Determination of the number of significant figures: The following rules are applied in determining the number of significant figures in a measured quantity.

All non-zero digits are significant.

Examples:

  1. There are two significant figures in the number 57.
  2. In 64.5 cm, there are three significant figures.
  3. In 0.4361g, there are four significant figures.
  4. 2. Zeros between two non-zero digits are significant.

Examples:

  1. There are four significant figures in the number 8005.
  2. 12.032 g has five significant figures.
  3. Zeros to the left of the first non-zero digit are not considered to be significant.

Examples:

  1. 0.53 mL has two significant figures (5 and 3).
  2. 0.0724 kg has three significant figures (7, 2 and 4).
  3. 0.009035 has four significant figures (9, 0, 3 and 5).
  4. If a number ends with one or more zeros and these zeros are to the right of the decimal point then these zeros become significant.

Examples:

  1. 4.0 has two significant (4 and 0).
  2. 2.500 has four significant (2, 5, 0, and 0).
  3. 0.040g has two significant (4 and 0).
  4. 0.4000 km has four significant (4, 0, 0 and 0).

5. If a number ends with one or more zeros but these zeros are not to the right of a decimal point, then these zeros may or may not be significant.

Examples: 10700 g may have three, four, or five significant figures. The ambiguity can be removed by expressing the value in an exponential form of the type N x 10n, where n = an integer and N = a number with a single non-zero digit to the left of the decimal point.

Now, the number, 10700 can be expressed (in scientific notations) in three different exponential forms, thereby indicating the presence of three, four, or five significant figures in the number.

10700 = 1.07 X 104 (Three significant)

= 1.070 X 104 (Four significant)

= 1.0700 X 104 (Five significant)

In these exponential terms, the significant figures of only the first factor {i.e., 1.07 1.070, or 1.0700) are to be counted (remembering that all zeros to the right of a decimal point are significant).

There are three significant in each of the numbers,1.54 x 10-2  and 1.54 x10-6  Similarly, there are four significant in Avogadro’s number 6.022 x 1023.

If a whole number ends with one or more zeros then these zeros are not considered while counting the number of significant. Thus there are only three significant 43700.

If however, the said number expresses the result of any experimental measurement, then such zeros are taken into consideration while counting the number of significant. Thus, if the measured distance between two places is 3200m (taking lm as the least measurable distance) then the number of significant figures in the measured distance is four.

Exactintegralnumbers such as the number of pencils in a dozen the number of grams in a kilogram or the number of centimeters in a meter do not have any uncertainty associated with them and hence these numbers have an infinite number of significant figures.

Examples:

  • A number of pencils in a dozen = 12.0000. has an infinite number of significance.
  • The number of grams in a kilogram = 1000.0000. number of significant.

Rules For Determination Of The Number Of Significant Figures In Final Results Involving Calculations

The observed results of various measurements may have different precisions. Thus, the results obtained at various stages of the calculation are to be rounded off because the final result cannot be more precise than that of the least precise measurement.

Rounding off: The following rules are employed foregrounding offa numbers to the desired number of significant.

1. If the digit, next to the last digit to be retained, is less than 5, the last digit to be retained is left unchanged and all other digits on its right are discarded.

Example: Suppose the result of a measurement is 2.73484. This can be rounded off to give—(a) 2.7348 (for reporting the result upto four decimal places) or, (b) 2.73 (for reporting the result upto two decimal places).

If the digit, next to the last digit to be retained, is greater than 5, the last digit to be retained is increased by 1 and all other digits on its right are discarded.

Example: Suppose the result of a measurement is 2.73687. This can be rounded off to give

2.7369 (for reporting the result upto four decimal places) or, (b) 2.74 (for reporting the result upto two decimal places).

3. If the digit, next to the last digit to be retained, is equal to 5, the last digit is kept unchanged if it is even, and is increased by 1 if it is odd.

Example: Suppose the result of a measurement is 12.63585. This be rounded off to given—(a) 2.6358 (for reporting the result upto four decimal places) or, (b) 2.64 (for reporting the result upto two decimal places).

5. Calculations involving addition and subtraction: The result of an addition or subtraction should be reported to the same number of decimal places as are present in the number having the least number of decimal places. The number of significant figures of different numbers does not play any role.

5. Calculations involving multiplication and division: The result of a multiplication or division should be reported to the same number of significant figures as possessed by the least precise term involved in the calculation.

6. Calculations involving multiple operations: If a calculation involves both multiplication and division, the result should be reported with the same number of significant figures as that of the least precise number involved, other than the integral number.

Leaving the integral number 4, the least precise number 0.62 has only two significant figures. So, the final result should be reported as 0.74 (two significant figures).’

Numerical Examples

Question 1. The density of a metallic substance is 7.2 g- cm-3. Find its density in the SI unit.
Answer:

Given:

The density of a metallic substance is 7.2 g- cm-3.

⇒ \(d=\frac{7.2 \mathrm{~g}}{1 \mathrm{~cm}^3}=\frac{\frac{7.2}{1000} \mathrm{~kg}}{\left(10^{-2}\right)^3 \mathrm{~m}^3}=\frac{7.2}{1000 \times 10^{-6}}\)

= 7.2×103=7200 kg.m-3

Question 2. The wavelength of radiation is 643.5 nm. Find its wavelength in the SI unit.
Answer:

Given:

The wavelength of radiation is 643.5 nm.

We know, and = 10-9m

⇒ \(\text { Unit conversion factor }=\frac{10^{-9} \mathrm{~m}}{1 \mathrm{~nm}}\)

So, the wavelength of the radiation = 643.5 nm

⇒ \(=643.5 \mathrm{~nm} \times \frac{10^{-9} \mathrm{~m}}{1 \mathrm{~nm}}=6.435 \times 10^{-7} \mathrm{~m}\)

[Here, we choose the conversion factor that has nm in the denominator]

Question 3. In diamond, the average distance between two carbon atoms is 1.54A. Express the distance between two C-atoms in the SI unit.
Answer:

Given:

In diamond, the average distance between two carbon atoms is 1.54A.

1A = 10-7m

⇒ \(\text { Unit conversion factor }=\frac{10^{-10} \mathrm{~m}}{1}\) [Here, we choose the conversion factor that has Ain the denominator]

So, the average distance between two carbon atoms

⇒ \(=1.54=1.54\times \frac{10^{-10} \mathrm{~m}}{1}=1.54 \times 10^{-10} \mathrm{~m}\)

Question 4. The atomic mass of nitrogen is 14.00674u. Find out the mass of one nitrogen atom (up to 3 significant figures).
Answer:

Given:

The atomic mass of nitrogen is 14.00674u.

Mass of 6.022 x 1023 no. of-atoms = 14.00674g

⇒ \(\text { Mass of } 1 \mathrm{~N} \text {-atom }=\frac{14.00674}{6.022 \times 10^{23}} \mathrm{~g}=2.3259 \times 10^{-23} \mathrm{~g}\)

=2.32 x 10-23g

Question 5. If the density of water is Ig/mL, then find the number of H-atoms in 45 mL water (up to 3 significant figures).
Answer:

Given:

The density of water is Ig/mL

Mass of 45 mL water = 45 g [ density of water= Ig/mL]

Number of water molecules in 18 g of water = 6.022 x 1023

Number of-atoms in 18 g ofwater= 2 x 6.022 x 1023

[ 1 molecule of H2O contains 2 H-atoms]

Number of atoms in 45g of water

⇒ \(=\frac{2 \times 6.022 \times 10^{23} \times 45}{18}=3.011 \times 10^{24}=3.01 \times 10^{24}\)

[After after rounding up to 3 significant figures]

Question 6. The dimension of an iron block is 4.6in X 3.0in X 1.9in and the density of iron is 7.87g/cm3. Find out the mass of the iron block. [Given 1 in = 2.54 cm]
Answer:

Given:

The dimension of an iron block is 4.6in X 3.0in X 1.9in and the density of iron is 7.87g/cm3.

Mass of the iron block

⇒ \(\begin{aligned} =\left(4.6 \mathrm{in} \times \frac{2.54 \mathrm{~cm}}{1 \mathrm{in}}\right) \times(3.0 \mathrm{in} & \left.\times \frac{2.54 \mathrm{~cm}}{1 \mathrm{in}}\right) \\ & \times\left(1.9 \mathrm{in} \frac{2.54 \mathrm{~cm}}{1 \mathrm{in}}\right) \times\left(\frac{7.87}{1 \mathrm{~cm}^3}\right) \end{aligned}\)

= 3.381 x 103g = 3.4 X 103g [Afterrounding off]

Question 8. Express 2.64 km distance in inches. [Given 1km = 1000m, lm – 1.094 yd, 1 yd = 36 in]
Answer:

\(264 \mathrm{~km}=2.64 \mathrm{~km} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{1.094 \mathrm{yd}}{1 \mathrm{~m}} \times \frac{36 \mathrm{in}}{1 \mathrm{yd}}\)

= 1.0397 x 105in = 1.04 X 105in

[After rounding off up to 3 significant figures]

Laws Of Chemical Combination

Two or more substances react to form new substances. Such chemical reactions take place according to certain laws called the laws of chemical combination. These are

  1. Law of conservation of mass
  2. Law of constant proportions
  3. Law of multiple proportions
  4. Law of reciprocal proportions

Gay Lussac’s law of gaseous volumes. The first four laws deal with mass relationships while the fifth deals with the volumes of the reacting gases and products involved in the reaction.

Law Of Conservation Of Mass

Postulated by: French chemist, A. Lavoisier 1774.

Law Of Conservation Of Mass Statement

In any physical or chemical change, the total mass of the reactants is equal to that of the products,

Law Of Conservation Of Mass Explanation: Suppose two substances A and B react together to form two new substances C and D. According to the law, the sum of the masses of A and B will be equal to the sum of the masses of C and D.

Thus there will be no increase or decrease in the total mass of matter during a chemical reaction or a physical change.

So, the law can alternatively be stated as—Matter can neither be created nor be destroyed. Hence, the law is also known as the law of indestructibility matter.

Law Of Conservation Of Mass Example: Aqueous solutions of sodium chloride and silver nitrate are taken in two separate conical flasks and the flasks are weighed together in a balance.

Then, the contents of the flasks are mixed together. Consequently, a curdy white precipitate is found to be formed due to the following chemical reaction.

⇒ \(\mathrm{NaCl}+\mathrm{AgNO}_3 \rightarrow \mathrm{AgCl} \downarrow+\mathrm{NaNO}_3\)

The flasks along with the contents are again weighed together and it is noticed that there is no change in mass. This justifies the law of conservation of mass.

Law Of Conservation Of Mass Limitations:

Modification of the law of conservation of mass: According to Einstein’s theory of relativity, mass and energy are interconvertible. Mass (m) gets converted into energy (E) according to Einstein’s equation, E = me2 (where c = velocity of light).

In ordinary chemical reactions, the amount of energy released is very small and hence the law of conservation of mass holds good. In nuclear reactions, however, the change in mass is quite significant because a tremendous amount of energy is released during these reactions. So, the law of conservation of mass does not hold good.

In such cases, the total sum of mass and energy remains constant. Thus, the law of conservation of mass has been modified and the modified law is known as the law of conservation of mass energy.

The law states that mass and energy are interconvertible but the total sum of mass and energy of a system before and after any physical or chemical change remains constant.

Law Of Constant Proportions Or Definite Proportions

Postulated: French chemist, Louis Proustin 1799.

Statement A pure chemical compound always consists of the same elements (irrespective of their sources & method of preparations) combined together in the same definite proportions by mass.

Law Of Constant Proportions Explanation: Suppose compound AB is prepared by two different methods. In one method, x gram of A combines a co with y gram of B while in the other method m gram of A combines with n gram of B to form the compound AB. According to the law of constant proportions,

⇒ \(x: y=m: n, \text { or, } \frac{x}{y}=\frac{m}{n}\)

Law Of Constant Proportions Example: Pure water obtained from any natural source (For example well, river, lake, etc.) or prepared artificially (For example bypassing H2 gas over heated CuO) is always found to be made up of only two elements i.e., hydrogen and oxygen combined together in the same definite ratio of 1: 8 by mass.

The converse of the law of constant proportion is not always true: The converse of the law of constant proportions can be stated as—”When the same elements combine in a constant proportion by mass, the same compound will always be formed.”

This statement is not always correct, especially for isomeric compounds. Although the isomeric compounds have the same molecular formula, their properties are not similar. For example,

Combination of carbon, hydrogen, and oxygen in the ratio of 12: 3: 8 by mass may produce either ethyl alcohol (C2H5OH) or dimethyl ether (CH3OCH3) under different experimental conditions. But their properties are different.

A combination of carbon, hydrogen, nitrogen, and oxygen in the ratio of 12: 4: 28: 16 by mass may produce two different compounds urea (NH2CONH2) and ammonium cyanate (NH4CNO) under different experimental conditions. These two compounds have different properties.

Again, the converse of the law of constant proportion is not true for monomers and polymers. Thus, acetylene (C2H2) on polymerization gives benzene (C6H6). These two compounds contain carbon and hydrogen in the same ratio by mass (12: 1) but have different properties.

Imitations of the law of constant proportions:

If two or more isotopes of an element take part separately in the formation of a particular compound, then the same compound will contain different proportions by mass of Ihe elements depending upon its isotopic mass.

For example, In 12CO2, the ratio of the masses is C:0 = 12:32 whereas in 14C02, the ratio of the masses is C.0 = 1 4: 32.

This shows different sources of carbon dioxide may contain carbon and oxygen present in different mass ratios.

There are some compounds that have variable molecular compositions. The law of constant proportions is not applicable to such compounds.

For example, cuprous sulfide may have a molecular composition from and titanium oxide may have a composition. Such compounds are called non-stoichiometric compounds.

Numerical Examples

Question 1. Analysis of 30g of compound D was found to contain 10g of element A and 20g of element IS. Again analysis of 45 g of another compound E was found to contain 15g of element B and 30g of element C. Calculate the amounts of D and E formed if 15g of A, 60g of B, and 15g of C are mixed together and allowed to react with each other. Also, calculate the total mass of the mixture after the completion of the reaction. Assume that no other reaction is possible except the reaction of B with A and separately. State which laws of the chemical combination can be utilized in the calculation.
Answer:

In compound D, mass-ratio of A to B = 10: 20 =1: 2

In compound E, the mass ratio of C = 15: 30 =1:2

Now applying the law of constant proportions we have, 15g of A combined with 2 x 15 = 30 g of B to form (15 + 30) = 45 g compound D.

Similarly, 15g of C combines with x 15 = 7.5 g of B to form (15 + 7.5) = 22.5 g compound E

∴ Amount of B remaining unreacted in the mixture

= [60 -(30+7.5)] = 22.5 g

The total mass of the mixture after completion of the reaction = Mass of D + Mass of + Mass of remaining unreacted

= 45 + 22.5 + 22.5 = 90 g.

Concepts of the law of mass action and the law of constant proportions are utilized in the calculations.

Question 2. 5g of pure MgO (obtained by reaction of metallic magnesium with oxygen) contains 3g of Mg. Again 8.5 g of pure MgO (obtained by heating MgCO3) contains 5.1g of Mg. Show that these results are in accordance with the law of constant proportions.
Answer: In the first variety of MgO, the ratio of masses of Mg to O = 3 : (5-3) = 3: 2

In the second variety of MgO, the ratio of the masses of Mg to O

= 5.1: (8.5 -5.1) =5.1 : 3.4 = 3: 2

So independent of its source, MgO always contains Mg and O in the mass ratio of 3: 2 and this is in accordance with the law of constant proportions.

Law Of Multiple Proportions

Postulated by: John Dalton in 1803.

Law Of Multiple Proportions Statement

when two elements combine with each other to form two or more compounds, then the different masses of one of the elements which combine with a fixed mass of the other, bear a simple whole number ratio (Example1: 2,1 : 2: 3,1: 3: 4 etc.).

Law Of Multiple Proportions Explanation:

Let a fixed mass of the element X combine separately with a, b, and c parts by masses of another element Y to form three different compounds A, B, and C.

So according to the law of multiple proportions, the ratio a: b: c will be a simple whole number ratio.

Law Of Multiple Proportions Example: Carbon and hydrogen combine with each other to form ethane (C2Hg), ethene (C2H4), and ethyne (C2H2).

In the formation of these three compounds, 24 parts by mass of carbon combine separately with 6 parts, 4 parts, and 2 parts by masses of hydrogen respectively.

Thus, the ratio of the masses of hydrogen which combine separately with the fixed mass of carbon (24 parts) in these compounds is =6:4:2 =3:2:1 which is a simple whole number ratio.

Exception of the law of multiple proportions: in case of simple hydrocarbons—methane (CH4), ethane (C2H6), propane (C3H8), butane (C4H10), etc., the ratio of different masses of hydrogen which combines separately with 12 parts by masses of carbon is 4: 3: 2.67: 2.5. This is not a simple whole number ratio.

Numerical Examples

Question 1. Two compounds A and B consist of tin and oxygen. Compound A contains 78.77% of tin and 21.23% of oxygen while compound B contains 88.12% of tin and 1 1.88% of oxygen. Show that these data illustrate the law of multiple proportions.
Answer: In the formation of compound B, 11.88 parts by mass of oxygen combine with 88.12 parts by mass of tin.

21.33 parts by mass of oxygen combined with \(\frac{88.12 \times 21.23}{11.88}=157.47\) parts by mass of tin.

Thus, the ratio of the masses of tin which combine separately with fixed mass (21.33 parts) of oxygen to form the compounds A and B is given by = 78.77: 157.47 = 1:2 (approx) which is a simple whole number ratio. So, the given data illustrates the law of multiple proportions.

Question 2. Two oxides of a metal, M were heated separately in hydrogen. The water obtained in each case was carefully collected and weighed. It was observed that— O 0.725 g of the first oxide gives 0.18 g of water and 2.86 g of the second oxide gives 0.36 g of water. Show that these results are in accordance with the law of multiple proportions.
Answer: The amount of water obtained by the reduction of the first oxide =0.18 g.

Now, 18 g water contains = 16 g of oxygen

∴ 0.18g of water contains = 0.16 g of oxygen.

∴ 0.725g of the first oxide contains = 0.16 g of oxygen

So, the mass of metal in the first oxide = (0.725- 0.16)

= 0.565 g

∴ The mass of oxygen which combines with 0.565g of metal, M =0.16g.

Again, 0.36 g of water is obtained by a reduction of 2.86 g of the second oxide.

Now, 0.36 g ofwater contains \(=\frac{16 \times 0.36}{18}=0.32 \mathrm{~g}\) oxygen.

So, the amount of metal in the second oxide

=(2.86-0.32) =2.54 g

The mass of the oxygen winch combines with 2.54g of metal, M = 0.32 g.

The mass of oxygen which combines with 0.565 g of metal, \(\mathrm{M}=\frac{0.32 \times 0.565}{2.54}=0.071 \mathrm{~g}\)

Thus, the ratio of the masses of oxygen which combine separately with a fixed mass (0.565 g) of the given metal to form two different oxides is given by =0.16: 0.071 = 2:1 (approx), which is a simple whole number ratio. So, the given data are in accordance with the law of multiple proportions.

Law of reciprocal proportions

Postulated by: Richter 1792

Law of reciprocal proportions Statement

When two different elements combine separately with a fixed mass of a third element, then the ratio of their masses is either the same or some simple whole number multiple of the ratio in which they combine directly with each other.

Law of reciprocal proportions Explanation:

Suppose parts by mass of the element A and b parts by mass of the element B combine separately with a fixed mass of the element C to form the compounds X and Y respectively.

Thus, the ratio of the masses of the elements A and B which combine separately with the fixed mass of C is given by a: b. Now according to the law of reciprocal proportions, if the elements A and B combine with each other, the ratio of their masses in the resulting compound will be either a: b or xa:yb (where x andy are simple whole numbers).

Law of reciprocal proportions Example: The elements, carbon and oxygen combine separately with the third element, hydrogen to form methane (CH4) and water (H2O) respectively. Analysis shows that

1. in methene (CH4), 4 parts by mass of hydrogen combines with 12 parts by mass of carbon i.e., 1 part by mass of hydrogen combines with 3 parts by mass of carbon;

2. in water (H2O), 2 parts by mass of hydrogen combined with 16 parts by mass of oxygen i.e.,1 part by mass of hydrogen combined with 8 parts by mass of oxygen.

Class 11 Chemistry Some Basic Concepts Of Chemistry law of reciprocal proportions

Thus, the masses of carbon and oxygen, which combine separately with a fixed mass (1 part) of hydrogen are in the ratio of 3:8.

So, according to the law of reciprocal proportions, if carbon and oxygen combine directly with each other, the ratio of their masses in the resulting compound will be 3: 8 or some simple whole number multiple of it.

Again, carbon and oxygen directly combine to form carbon dioxide (CO2). Analysis shows that in this compound, the ratio of the masses of C to 0 is = 12: 32 = 3: 8.

This ratio is found to be the same as that has been predicted earlier. Thus, the law of reciprocal proportions is illustrated.

Again, C and 0 directly combine with each other to form another compound called carbon monoxide (CO).

Analysis shows that the ratio of the masses to O in this compound is 12: 16 = 3: 4. This is a simple whole number multiple of the ratio 3: 8 as predicted earlier, 3:4 = (3×2:8xl)

Numerical Examples

Question 1. Show that the following experimental data are in agreement with the law of reciprocal proportions :

  1. 0.46 g of Mg on burning in air forms 0.76 g of MgO.
  2. 0.41 g of Mg in reaction with excess acid produces 380 cm³ of H2 at STP.
  3. 0.16 g of H2 reacts with excess oxygen to produce 1.45g of water.

Answer:

1. Mass of oxygen 0.76 g of MgO = (0.76- 0.46) = 030g

∴ Mass of oxygen combining within of Mg \(=\frac{0.30}{0.46}\) =0.652g

2. Mass of 22400 cm³ of H2 gas (atSTP)=2g

∴ Mass of 380 cm³ of H2 gas (at STP) =\(\frac{2 \times 380}{22400}=0.034 \mathrm{~g}\)

So, the amount of H2 gas produced by 0.41g of Mg=0.034 g

Amount of H2 gas produced by lg of Mg \(=\frac{0.034}{0.41}=0.083 \mathrm{~g}\)

From (1) and (2), it is found that the masses of hydrogen and oxygen combined with or replaced by a fixed mass are in the ratio, 0.083: 0.652 1: 8 Now, according to the law of reciprocal proportions, if the elements H and O combine, the ratio of their masses in the resulting compound will be either 1: 8 or any simple multiple of it.

0.16 g of hydrogen reacts with excess oxygen to form 1.45 g of water.

∴ Mass of oxygen combined with hydrogen

= 1.45- 0.16 = 1.29 g

So, the mass ratio of H to O in water

= 0.16: 1.29=1: 8

This is the same ratio as predicted earlier.

Thus, the given data are in agreement with the law of reciprocal proportions.

Question 2. Ammonia contains 17.65% of hydrogen, water contains 11.11% of hydrogen and nitrous oxide contains 36.36% of oxygen. Show that these data illustrate the law of reciprocal proportions.
Answer: In ammonia, the amount of 11 = 17.65 %.

Amount of = (100- 17.65) = 82.35 %.

So, the mass of N that combines with 1 part by the mass of H

⇒ \(=\frac{82.35}{17.65}=4.66 \text { parts. }\)

In water, the amount of present =11.11 %,

Amount of C-present = (100- 11.11) = 88.89 %

So, the mass of O that combines with part by the mass of H

⇒ \(=\frac{88.89}{11.11}=8 \text { parts. }\)

So, according to the law of reciprocal proportions, if the elements, N and O combine together, the ratio of their masses in the die compound so formed will be either 4.66: 8 (=0.5825: 1) or any simple multiple of it.

Now in nitrous oxide, the amount of present = 36.36 %,

amount of present = (100- 36.36) = 63.64 %

So, die mass-ratio of N to O in nitrous oxide

= 63.64: 36.36 = 1.75: 1 = 3 X 0.5825: 1

It is a simple multiple of the ratio, 0.5825:1 as predicted earlier. So, the given data illustrate the law of reciprocal proportions.

Gay-Lussac’s law of gaseous volumes

Postulated by: French chemist, Gay-Lussacin 1808.

Gay-Lussac’s law of gaseous volumes Definition

When gases react with each other, they always do so in volumes that bear a simple whole number ratio to one another and to the volumes of the products, if these are also gaseous, provided all volumes are measured under similar conditions of temperature and pressure.

The volumes of gaseous reactants and products are considered to be at constant temperature and pressure because the volumes of Gay-Lussac’s gases are dependent on both temperature and pressure.

So, GayLussac’s law will not be valid if volumes are not measured under the same conditions of temperature and pressure.

Gay-Lussac’s law of gaseous volumes may be considered the law of definite proportions in terms of volume.

The law of definite proportions, discussed earlier, was with respect to mass. Gay-Lussac’s law was justified theoretically by Avogadro in 1911.

Gay-Lussac’s law of gaseous volumes Example: 1 has been experimentally found that under the same conditions of temperature and pressure, 2 volumes of hydrogen reacts with 1 volume of oxygen to form 2volume of steam. So, the volumes of the reactants (H2 and O2) and the product (steam i.e., H2O) measured under identical conditions of temperature and pressure are in the proportion of 2: 1: 2 which is a simple ratio.

It has been experimentally observed that 1 volume of hydrogen gas reacts with 1 volume of chlorine gas to form 2volume of hydrogen chloride.

So, under identical conditions of temperature and pressure, the ratio of the volumes of gaseous hydrogen, chlorine, and hydrogen chloride is 1: 1: 2, which is a simple ratio.

Characteristics of Lussac’s law of gaseous volumes:

The other laws of chemical combinations interpret the chemical combinations in terms of masses of reactants and products.

But, Gay-Lussac’s law of gaseous volumes establishes the relation between the gaseous reactant(s) and product(s) in terms of their volumes.

This law cannot be explained with the help of Dalton’s atomic theory whereas the other laws of chemical combination can be successfully explained by this theory.

Numerical Examples

Question 1. 2 volumes of 03 produce 3 volumes of O2 on complete decomposition. 40 mL of a mixture of O3 and O2 is heated at first and then brought back to the previous temperature and pressure.

The volume of the gaseous mixture is now found to be 42 mL. Find the percentage composition of Og in the gas mixture by volume.

The volume of all gases is measured under the same conditions of temperature and pressure.

Answer: Let, the volume of O3 present in the mentioned gas mixture is x mL. Volume of O2 in the gas mixture = (40- x) mL. Now, the decomposition of O3 can be represented as

⇒ \(2 \mathrm{O}_3 \longrightarrow 3 \mathrm{O}_2\)

⇒ \(\begin{array}{ll}
2 \text { volume } & 3 \text { volume } \\
2 \mathrm{~mL} & 3 \mathrm{~mL} \\
1 \mathrm{~mL} & 3 / 2 \mathrm{~mL}
\end{array}\)

Hence, x mL of 03 yields (3/2)xml of O2.

So,the total volume of O2 = \((40-x)+\frac{3}{2} x=\left(40+\frac{x}{2}\right) \mathrm{mL}\)

According to the question, 03 is completely decomposed.

So, \(40+\frac{x}{2}=42\) Or,x=4

Amount of O3 in the initial mixture =4×100/40=10%

Question 2. What is the minimum volume of oxygen that must be mixed with 100mL of carbon monoxide to convert it completely into carbon dioxide in an explosion? Find the volume of carbon dioxide formed at the same temperature and pressure. The volume of all gases is measured under the same conditions of temperature and pressure.
Answer: The formation of CO2 by the explosion of a mixture of CO and O2 follows the equation given below:

⇒ \(2 \mathrm{CO}+\mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2\)

2 volume 1 volume 2 volume

(under identical conditions of pressure and temperature)

The above equation shows that under similar conditions of pressure and temperature, 2 volumes of CO react with 1 volume of02 to produce 2 volumes of C02

According to Gay-Lussac’s law of gaseous volumes, the ratio of the volumes of CO, 02, and C02 is:

2: 1: 2 = 2 x 50: 1 X 50: 2 X 50 = 100: 50: 100

Hence, 50 mL of 02 must be muted with 100 mL of CO so that CO2 formed as a result of the reaction will be 100 mL.

Question 3. Under the same pressure and temperature, a mixture of 100 mL of water gas and 100 mL of 02 is subjected to explosion. Find the composition of the gas mixture formed by an explosion under the same conditions as AllS. pressure water gas and is temperature.
Answer: Water is gas and is a temperature. m DALTON’S ATOMIC THEORY (mixture of the same volume of CO and H2 ).

So, 100 mL of water gas contains 50 mL each of CO and H2. The reactions caused by the explosion are:

⇒ \(2 \mathrm{CO}+\mathrm{O}_2 \longrightarrow 2 \mathrm{CO}_2 \cdots[1] \quad 2 \mathrm{H}_2+\mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}\)

From equation

1. it is seen that under the same conditions of temperature and pressure, 1 Volume of 02 reacts with 2 vols. of CO to form 2 vols. of CO2.

So, under the same conditions of temperature and pressure, the volume of 02 required for 50 mL of CO = 25 in and the volume of CO2 formed = 50 mL.

According to equation [2], under the same conditions of temperature and pressure, the volume of 02 required to react with 50 mL of H2 = 25 mL and the volume of water vapor, produced by the reaction = 50 mL.

Total volume of 02 used up in the two reactions = (25 + 25) = 50 mL.

Hence after explosion, both the volumes of CO2 and H20 produced is 50 mL while the volume of unreacted 02 =(100-50) = 50 ml

Dalton’s Atomic Theory

Atoms are the building blocks of matter. John Dalton developed the concept of the atom and put forward a scientific theory regarding the constitution of matter called Dalton’s atomic theory.

The theory is based on some postulates. [A postulate means a statement accepted without proof.]

Postulates Of Dalton’s Atomic Theory

  1. Matter is composed of very tiny particles called atoms.
  2. The smallest particles of an element were termed ‘simple atoms’ and that of a chemical compound were called ‘compound atoms’.
  3. Atoms are indivisible and cannot be divided by any physical or chemical means.
  4. Atoms can neither be created nor destroyed.
  5. Atoms of the same element are identical in all respects i.e., in mass, size, and other properties.
  6. Atoms of different elements are different in all respects.

Atoms take part in chemical reactions. During chemical reactions, they combine with one another in simple whole-number ratios such as 1: 1, 1: 2, 2 : 3, etc to form compound atoms (nowadays called molecules). [When we say ‘atom,’ we imply a simple atom.]

Postulates Of Dalton’s Atomic Theory Definition

The smallest indivisible, indestructible, and discrete particle of an element that retains all the physical and chemical properties of that element and takes part in chemical reactions is called an atom.

Importance of Dalton’s atomic theory

  1. According to this theory, an atom is the ultimate building block of matter. It is the first scientific approach toward the constitutional aspects of matter.
  2. The idea that all atoms of the same element are identical in mass helped to determine the atomic masses of elements.
  3. It successfully explains the laws of chemical combinations involving masses.
  4. The idea that atoms combine with one another in a simple ratio helped to determine the formulae of chemical compounds and to express chemical reactions in the form of balanced chemical equations.
  5. The idea of the indivisibility of atoms has made chemical calculations easier. Still now, in all chemical calculations, the atom is considered an indivisible unit.
  6. This theory helped Avogadro to formulate the concept of molecules and to propose molecular theory.

Limitations Of Dalton’s Atomic Theory

  1. Dalton’s atomic theory did not make any distinction between the smallest particle of the elements and that of the chemical compounds having free existence.
  2. So, this theory created confusion regarding the nature of the ultimate particles of matter.
  3. Later this confusion was dispelled by Avogadro when he first introduced the concept of molecules in the substances.
  4. According to Dalton’s atomic theory, the atom is indivisible. However, after some fundamental scientific discoveries, it has been found that an atom is composed of sub-atomic particles like electrons, protons, neutrons, etc., i.e., the atom cannot be regarded as Indivisible.
  5. In the opinion of Dalton, the atom can neither be created nor be destroyed. But this proposition is not correct as an atom of an element can be transformed artificially into an atom of another element by nuclear reactions.
  6. According to Dalton, atoms of the same element are identical in all respects while atoms of different elements are different. This postulate was proved wrong after the discoveries of isotopes and isobars.
  7. Isotopes are the atoms of the same element having different atomic masses and physical properties while isobars are the atoms of different elements having the same atomic mass.
  8. During the formation of a chemical compound, the atoms unite together in simple whole-number ratios.
  9. This statement is not valid in all cases. In the case of compounds like protein, starch, cellulose, etc.,
  10. The atoms combine In the ratio of whole numbers but the ratios are not simple. Besides, in Berthollide compounds, the atoms do not combine in the die ratio of whole numbers.
  11. Gay-Lussac’s law of gaseous volumes cannot be explained due to the absence of molecular concepts in Dalton’s atomic theory.
  12. Compounds in which die atoms of the constituent elements are present in a simple ratio of their numbers are called Daltonide compounds. For example, CO2, H2O, FeO etc.
  13. There are certain compounds in which atoms of the constituent elements do not exist in a simple ratio of their numbers. These are known as Berthollide compounds such as Cu, 7S, TI07SO, etc.

Modified Form Of Dalton’s Atomic Theory

Atom is no longer considered to be indivisible. With the discovery of radioactivity, cathode rays, etc., it has been well established that atoms are composed of minute sub-atomic particles like electrons, protons, neutrons, etc.

Atoms of an element with similar chemical properties may possess different physical properties and masses ( For Example  Isotopes).

Atoms of different elements with dissimilar properties may have Identical masses (For example Isobars such as 40Ca and 40Ar).

During the formation of a chemical compound, atoms of different elements may not combine in the ratio of simple whole numbers (e.g., sucrose: C2H22O).

Atom is no longer Indestructible. With the discovery of artificial radioactivity has been possible to convert atoms of one element into atoms of another element. For example

\({ }_7^{14} \mathrm{~N}+{ }_2^4 \mathrm{He} \rightarrow{ }_8^{16} \mathrm{O}+{ }_1^2 \mathrm{H}\)

This is called a nuclear reaction. However, the chemical reactions fail to effect any such change.

According to Dalton’s atomic theory, atoms take part in chemical reactions—which is true even today.

But, now it has been slightly modified and it is established that the electrons in the outermost shell of an atom take part in chemical reactions.

Concept Of Molecules And Avogadro’s Hypothesis

In order to correlate Dalton’s atomic theory and Gay-Lussac’s law of gaseous volumes, Berzelius, a Swedish chemist made a generalization known as Berzelius’ hypothesis.

It states that, under the same conditions of temperature and pressure equal volumes of all gases contain the same number of atoms. Application of this hypothesis to some gaseous reactions leads to the conclusion that atoms are divisible.

This is In direct conflict with DaJton’s atomic theory, which states that atoms are the smallest particles of elements and are indivisible.

Hence, scientists discarded Berzelius’ hypothesis. While Investigating the cause of the failure of Berzelius’ hypothesis an Italian scientist Amadeo Avogadro (1811) announced that it would be possible to correlate Gay-Lussac’s law of gaseous volumes with DaJton’s atomic theory if the existence of another type of minute particle, besides atom, was conjectured.

He named this minute particle a molecule. By applying this concept of molecule, he introduced the molecular theory wherein the distinction between ‘atom’ and ‘molecule’ was mentioned explicitly. In Avogadro’s opinion—

The building blocks of matter are of two kinds—one is the atom as mentioned by Dalton while the other ultimate particle molecule as mentioned by Avogadro.

A molecule refers to the ultimate particle of a substance (element or compound) that has free existence and possesses all the characteristic properties of that substance.

An atom is the ultimate particle of an element, which takes part in a chemical reaction and may or may not exist in a free state.

Molecules may be of two types viz., elementarymolecule and compoundmolecule. Elementarymolecule is formed by atoms of the same element On the other hand, atoms of different elements form a compound molecule.

Unlike an atom, a molecule may be divided into its constituent atoms which take part in any chemical reaction.

Avogadros hypothesis

Avogadros hypothesis Statement: The same conditions of temperature and pressure, equal volumes of all gases (element or compound) contain the same number of molecules.

Avogadro’s hypothesis Explanation: If ‘n’ is the number of molecules presenting 1L of hydrogen at pressure P and temperature T, then at the temperature and pressure, 1L of carbon dioxide or 1L of any other gas will also contain the ‘ri number of molecules.

Postulates Of Dalton’s Atomic Theory

The converse statement of Avogadro’s hypothesis: All gases containing the same number of molecules will occupy the same volume under the same temperature and pressure.

Hence, if the ‘n ‘ number of hydrogen molecules occupy V volume under certain conditions of temperature and pressure, then the ‘n ‘ number of molecules of nitrogen, carbon dioxide, or ammonia will also occupy the same volume ( V), provided the temperature and pressure remain the same.

Elementary Molecule And Compound Molecule

Elementary molecule: Molecules composed of atoms of the same element are known as elementary molecules or homoatomic molecules. For example, hydrogen (H2) oxygen (O2), chlorine (Cl2), etc.

There are some solid non-metals whose molecules are composed of a single atom; example carbon (C), silicon (Si), etc., and thus monoatomic. Gases like O2, Cl2, H2, etc., are diatomic.

Again, molecules of some non-metals contain more than two atoms viz., phosphorus (P4), sulfur (S0), etc., which are polyatomic.

Compound molecule: Molecules that are composed of atoms of two or more different elements are called compound molecules or heteroatomic molecules.

For example, a water (H2O) molecule consists of 2 atoms of hydrogen and 1 atom of oxygen. Again, the sulphuric acid (H2SO4) molecule is composed of 2 atoms of hydrogen, 1 atom of sulfur, and 4 atoms of oxygen.

The number of atoms present in an elementary molecule is called the atomicity of the molecule. Thus the atomicities of argon, nitrogen, and phosphorus are 1, 2, and 4 respectively.

Correlation between Dalton’s atomic theory and Gay-Lussac’s law of gaseous volumes

Avogadro’s hypothesis helps to correlate Dalton’s atomic theory with Gay-Lussac’s law of gaseous volumes.

Formation of hydrogen chloride from hydrogen and chlorine gases: From actual experiments, it is known that under the same conditions of temperature and pressure, 1 volume of hydrogen (H2 ) combines with the volume of chlorine (Cl2) to produce 2 volumes of hydrogen chloride (HC1) gas; i.e., 1 volume of H2 +1 volume of Cl2 = 2 volume of HCl

Class 11 Chemistry Some Basic Concepts Of Chemistry At the same T and P, 1 volume of hydrogen reacts with 1 volume of cholorine to form 2 volume of hydrogen choloride

If under the experimental conditions of temperature and pressure, 1 volume of H2 contains the ‘n’ number of molecules, then according to Avogadro’s hypothesis, at the same temperature and pressure, 1 volume of Cl2 and 2 vols. of HC1 will contain ‘n ’ number of chlorine molecules and 2n number of hydrogen chloride molecules respectively

So, ‘n’ molecules of H2 +’n’ molecules of Cl2 = 2n molecules of HC1

or, 1 molecule of H2 +1 molecule of Cl2 = 2 molecules of HC1

i.e.,1/2 molecule of H2 molecule of Cl2 =1 molecule of HC1.

It has been later proved by Avogadro’s hypothesis and other experiments that elementary gases such as hydrogen, chlorine, nitrogen, oxygen, etc. are diatomic, i.e., each molecule of these gases contains two atoms only.

Therefore, 1/2 molecule of hydrogen =1 atom of hydrogen and, 1/2 molecule of chlorine = 1 atom ofchlorine Thus, the combination of one atom of hydrogen with an atom of chlorine yields one molecule of hydrogen chloride.

This deduction does not. contradict Dalton’s atomic theory because an atom is indivisible but a molecule may be divisible.

Indeed, during the chemical reaction, the molecules of hydrogen and chlorine split into their respective atoms and these atoms combine in a simple ratio to form a hydrogen chloride molecule.

Again, when Gay-Lussac’s law is applied to this gaseous reaction, the ratio of the volume of the reactants and the product becomes, H2: Cl2: HC1 =1:1: 2—it is a simple whole number ratio.

Thus in the case of the above chemical reaction, it is seen that Avogadro’s hypothesis successfully correlates Gay-Lussac’s law of gaseous volumes with Dalton’s atomic theory.

Deduction of Gay-Lussac’s law of gaseous volumes with the help of Avogadro’s hypothesis

Gay-Lussac’s law of gaseous volumes can be deduced with the help of Avogadro’s hypothesis in the following way.

Let, a molecules of a gas A, react with b molecules of another gas B, to form c molecules of another gas C at a particular temperature and pressure, where a, b, and c are small whole numbers.

Now, let’s assume that under the experimental conditions of temperature and pressure, the unit volume of gas A contains n number of molecules. So, according to Avogadro’s hypothesis, at the same temperature and pressure, the unit volume of each of the gases B and C will also contain n number of molecules.

Let n number of molecules of gas A occupy1 volume.

A number ofmolecules of gas A occupy \(\frac{a}{n}\) volume.

Similarly, b number of molecules of gas B occupy \(\frac{b}{n}\) volume.

and c number ofmolecules ofgas C occupy \(\frac{c}{n}\) volume.

Thus, the ratio of the volumes of the reacting gases, A and B to that of the gaseous product, \(C=\frac{a}{n}: \frac{b}{n}: \frac{c}{n}=a: b: c\) which is a simple ratio because a, b and c are small whole numbers.

Thus, it is observed that under the same conditions of temperature and pressure, the reacting gases combine in a simple proportion by volume and the volumes of the ga product(s) also maintain a simple ratio with the volumes of the gaseous reactants.

Therefore, the law of gaseous volumes as expressed by Gay-Lussac is established.

Modified Form of Dalto’s atomic theory based on Avogadro’s hypothesis: Molecular concept of matter.

With the introduction of Avogadro’s concept, modification of the atomic concept of matter became unavoidable.

The modified concept came to be known as the molecular concept or atomic-molecular concept of matter.

The newly embodied concept regarding the constitution of matter and its related properties is summarised below.

  1. The smallest particles of an element that take part in chemical reactions are known as atoms. Atoms may or may not have independent existence.
  2. The ultimate particles of a substance, element, or compound that can exist in the free state and possess all the physical and chemical properties of that substance are called molecules.
  3. Generally, molecules are composed of two or more atoms. Atoms of the same element form elementary’ molecules (For example H2 Cl2, N2, etc.) while atoms of different elements constitute a compound molecule (for example; H2O HNO3, etc.).
  4. Molecules are divisible.
  5. Molecules ofthe same substance are identical in mass and properties but the molecules of different substances differ in mass and properties.
  6. During chemical reactions, the participating molecules react in a simple ratio of their numbers to form molecules of new substances. However, the molecules do not react directly. At first, the reacting molecules split into their respective atoms which in turn combine mutually in a simple ratio to form molecules of a new substance.
  7. The formation of HCI gas by the combination of HCl2 is shown below pictorially.

Class 11 Chemistry Some Basic Concepts Of Chemistry At the same T and P, 1 molecucule of H2 combines with 1 molecule of CL2 to form 2 Molecules of HCL

Definition of molecule on the basis of Avogadro’s hypothesis

The ultimate particle of a substance (element or compound) that can exist in the free state and possesses all the properties of that substance is called the molecule of that substance.

Can Avooadro’s hypothesis be considered as a law?

Avogadro’s hypothesis originated from mere imagination. Even till now, it has not been verified by any direct experiment But the validity of this hypothesis has been well established indirectly with the help of various experiments.

The conclusions resulting from the application of this hypothesis have always been proved errorless.

No experimental results ever challenged the validity of this hypothesis. That is why, Avogadro’s hypothesis is, nowadays, called Avogadro’s law.

Atomic Mass Or Atomic Weight

The absolute mass of an atom of any element is so small that it cannot be weighed directly with the help of a chemical balance.

Moreover, it is inconvenient to express such a small mass. Thus, the mass of an atom of an element is expressed in terms of relative mass and this relative mass is called atomic mass.

Different Scales Of Atomic Mass

  • In order to determine the relative mass of an atom of any element, it is necessary to take an element as a standard of reference.
  • For this purpose, elements like hydrogen, and carbon are considered as the standard elements.

Hydrogen scale: At first, hydrogen was regarded as the standard element for determining the atomic masses of elements.

Hydrogen scale Definition The Atomic mass of an element may be defined as a relative number that shows how many times an atom of the element is heavier than one atom of hydrogen, taking the mass of hydrogen as unity.

⇒ \(\text { Atomic mass of an element }=\frac{\text { mass of } 1 \text { atom of the element }}{\text { mass of } 1 \text { atom of hydrogen }}\)

Example: The statement ‘atomic mass of sodium is 23 signifies that one atom of sodium is 23 times heavier than one atom of hydrogen.

Oxygen scale: Later on, instead of hydrogen, oxygen was considered the standard. In the oxygen scale, the atomic mass of an element is defined as follows.

Oxygen scale Definition The Atomic mass of an element is a relative number that denotes how many times an atom of the element is heavier than the l/16th part of the mass of an oxygen atom.

\(\begin{aligned}
\text { Atomic mass of an element } & =\frac{\text { mass of } 1 \text { atom of the element }}{\frac{1}{16} \times \text { mass of } 1 \text { atom of oxygen }} \\
& =\frac{\text { mass of } 1 \text { atom of the element }}{\text { mass of } 1 \text { atom of oxygen }} \times 16
\end{aligned}\)

Example: The statement ‘the atomic mass of nitrogen is 14’ implies that an atom of nitrogen is 14 times heavier than 1/16 the part of the mass of an oxygen atom.

Reasons for taking oxygen as standard Instead of hydrogen:

  1. Most of the elements, metals, in particular, react with oxygen compared to hydrogen to form stable compounds.
  2. As hydrogen is the lightest of all elements, slight experimental errors in the determination of atomic masses in the H-scale become erroneous.
  3. Atomic masses of elements determined in the O-scale are mostly whole numbers compared to the fractional values as obtained from H -the scale.

Carbon (12C) scale: At present, carbon has been accepted as the standard element. This scale has been approved by the international organization 1UPAC.

On the die scale, the mass of one 12C atom is taken as 12. [On this basis the relative mass of hydrogen comes out to be 1,008 and that ofoxygen is 15.994 (or roughly 16).

Carbon (12C) scale Definition: The atom’s mass of an element is a relative number which denotes how many times an atom of that particular element is heavier than l/12th part of the mass of one 12 C atom.

That is the atomic mass of an element

⇒ \(\begin{aligned}
& =\frac{\text { mass of } 1 \text { atom of the element }}{\frac{1}{12} \times \text { mass of } 1 \text { carbon }\left({ }^{12} \mathrm{C}\right) \text { atom }} \\
& =\frac{\text { mass of } 1 \text { atom of the element }}{\text { mass of } 1 \text { carbon }\left({ }^{12} \mathrm{C}\right) \text { atom }} \times 12
\end{aligned}\)

Since the atomic mass of an element is the ratio of two Atomic masses of some elements in different scales masses, it is in fact relative atomic mass, it has no unit and is expressed as a pure number.

Physical and chemical scales of atomic mass: Natural oxygen consists of 3 isotopes:leO (99.759%),170 (0.037%) and 180 (0.204%). So, the true atomic mass i.e., the average atomic mass of natural oxygen

= 16×0.99759 +17×0.00037 +18 X 0.00204 = 16.00204

But the atomic mass ofthe most abundant isotope (160) of natural oxygen = 16. Chemists take the average atomic mass of natural oxygen as the standard of reference to prepare the chemical scale of atomic mass and physicists take the atomic mass of the most abundant isotope of natural oxygen as the standard of reference to prepare the physical scale of atomic mass.

Almost all elements have isotopes, therefore, in place of ‘mass of atom; ‘average mass of an atom’ is to be used.

The chemical scale of atomic mass: The scale of atomic mass, which is obtained by taking the average mass of an atom of natural oxygen as 16.0000, is called the chemical scale of atomic mass and the atomic mass, as obtained by this scale, is known as the chemical atomic mass of that element.

Physical scale of atomic mass: The scale of atomic mass, which is obtained by taking the mass of a lsO isotope in natural oxygen as 16.0000, is called the physical scale of atomic mass and the atomic mass, as obtained by this scale, is known as physical atomic mass.

According to the physical scale, the average atomic mass of natural oxygen = 16.00447. However according to the chemical scale, the average atomic mass of natural oxygen = 16.0000. So, 16.0000 units in the chemical scale = 16.00447 units in the physical scale.

1 unit in chemical scale =16.00447/16.0000=1.0002794 in physcial Scale.

Thus, the magnitude of the average atomic mass of an element on the chemical scale is slightly less than that of the mass of an atom ofthe element’s physical scale of atomic mass.

1.0002794 is the conversion factor that is used to convert the chemical atomic mass of an element to its physical atomic mass and vice-versa.

Atomic mass on a physical scale = 1.0002794 x Atomic mass on a chemical scale.

Class 11 Chemistry Some Basic Concepts Of Chemistry Atomic Masses Of Some elements In Different Scales

Atomic mass unit (AMU)

The atomic mass of an element is a relative number and it has no unit. So, the atomic mass of an element does not stand for the absolute or actual mass of an atom of that element order to express the actual mass of an atom, scientists introduced another unit. This unit is termed an atomic mass unit (AMU).

Atomic mass unit Definition: The unit with respect to which the actual atom of any element is expressed and whose value is equal to the mass of l/12th part of the mass of one 12C atom is called the atomic mass unit.

Atomic mass unit= \(\frac{1}{12}\) x actual mass of one C atom.

Mathematical expression: Actual mass of 6.022 x 1023 atoms of 12C isotope = 12 g.

⇒ \(\begin{aligned}\quad \text { Actual mass of one }{ }^{12} \mathrm{C} \text { atom } & =\frac{12}{6.022 \times 10^{23}} \mathrm{~g} \\
\text { So, atomic mass unit }(1 \mathrm{amu}) & =\frac{1}{12} \times \frac{12}{6.022 \times 10^{23}} \mathrm{~g} \\
& =1.6605 \times 10^{-24} \mathrm{~g}
\end{aligned}\)

1 amu = 1.6605×10-24g=1.6605×10-27kg

The actual mass of an atom: Atomic mass of an element

⇒ \(\begin{aligned}
& =\frac{\text { mass of } 1 \text { atom of the element }}{\text { mass of } 1 \text { carbon }\left({ }^{12} \mathrm{C}\right) \text { atom } \times \frac{1}{12}} \\
& =\frac{\text { mass of } 1 \text { atom of the element }}{1 \mathrm{amu}}
\end{aligned}\)

The actual mass of an atom of an element
= atomic mass ofthe element x 1 amu
= atomic mass ofthe element x 1.6605 x 10
100-24g

Thus, the atomic mass of an element, multiplied by 1 amu gives the actual mass of an atom of that element.

Examples: 1 Actual mass of atom of hydrogen

= 1.008 amu = 1.008 x 1.6605 x 10-24 g

2. Actual mass of atom of nitrogen

= 14 amu = 14 x 1.6605 x 10-24 g

The actual mass of an atom of oxygen

= 16 amu = 16 x 1.6605 x 10-24 g

In recent times, a new symbol ‘u’ (which signifies unified mass) is used in place of amu (Le., atomic mass unit)

Therefore Mass of1 H-atom = 1.008u (le., 1.008 amu)

The atomic mass of an element and the actual mass of an atom of an element are completely different: The atomic mass of an element indicates how many times an atom of that element is heavier than 1/12 part of an atom of a C isotope. It is a relative number and it has no unit.

The actual mass of an atom of an element indicates the exact mass of an atom of that element and has definite units (for example 8kg).

Example: Atomic mass of oxygen is 16 but the actual mass of an atom ofoxygen is 16 x 1.6605 x 10-24g = 2.656 X 10-23g

Average Atomic Mass

Atomic masses of most of the elements are fractional numbers because they actually represent their average atomic masses. In nature, most of the elements exist as a mixture of two or more isotopes.

The relative abundance of the isotopes of particular natural elements is more or less fixed. The atomic mass of any element is determined by taking the average of the atomic masses obtained on the basis of the abundance of various isotopes of the element in nature.

Thus, the estimated atomic mass ofthe element is a fractional value although the atomic masses of different isotopes are whole numbers.

Average atomic mass

⇒ \(=\frac{\Sigma(\text { natural abundance of isotope }(\%) \times \text { its atomic mass })}{100}\)

Let the natural abundance of the three isotopes of an element be x %, y%, and z % and their atomic masses be a, b, and c respectively.

The atomic mass (average atomic mass) of the element

⇒ \(=\frac{x \times a+y \times b+z \times c}{x+y+z}=\frac{x \times a+y \times b+z \times c}{100}\)

Example: Natural chlorine contains 2 isotopes: 35C1 and 37C1 as a mixture of 75% and 25% respectively. The atomic masses of these isotopes are 35 and 37 respectively, both of which are whole numbers. But the atomic mass ofchlorine.

⇒ \(=\frac{(35 \times 75)+(37 \times 25)}{(75+25)}=35.5, \text { which is a fraction. }\)

Gram-atomic mass and gram-atom

Gram-atomic Mass Definition The Gram-atomic mass of an element is defined as the atomic mass expressed in grams.

Atomic mass has no unit while the unit of gram-atomic mass is gram. For example, the atomic masses of nitrogen and oxygen are 14 and 16 respectively but the gram-atomic masses of these elements are 14 grams and 16 grams respectively.

The gram atomic mass is best defined as the mass in grams of an element that contains the same number of atoms as 12 presenting 12 grams of C atom.

Gram-atom: One gram-atom of an element is defined as the quantity in gram which is numerically equal to its atomic mass. For example, one gram-atom of nitrogen means 14 g of nitrogen and one gram-atom of oxygen signifies 16 g of oxygen.

One gram-atom of an element also referred to as the mass in gram of the element contains 6.022 x 1023 number (Avogadro number) of atoms.

A number of gram-atom: The given mass of an element expressed in gram, when divided by its gram-atomic mass, gives the number of gram-atoms present in that quantity ofthe element.

Therefore, the number of gram-atom of the element

⇒ \(=\frac{\text { mass of the element }(\mathrm{g})}{\text { gram-atomic mass of that element }}\)

Examples: Number of gram-atom in 42 g of N2 \(=\frac{42 \mathrm{~g}}{14 \mathrm{~g}}=3\)

Number of gram-atom in 64 g of O2 \(=\frac{64 \mathrm{~g}}{16 \mathrm{~g}}=4\)

The discussion done till now about atomic mass indicates the relative atomic mass of an element.

However, according to IUPAC, the atomic mass of any element is the mass of one atom of that element expressed in the atomic mass unit (u).

The mass of 1 atom of any element with respect to the mass of 1 atom of 12C isotope as 12u is considered as the atomic of the corresponding element.

The atomic mass of different elements can be precisely determined by using a mass spectrograph.

According to the 12C scale, atomic masses of different elements are tabulated below—

Class 11 Chemistry Some Basic Concepts Of Chemistry Gram atomic mass and gram atom

Numerical Examples

Question 1. The atomic weight of ordinary hydrogen is 1.008. Ordinary hydrogen contains two isotopes JH and 11H. What is the weight percentage of 21H in ordinary hydrogen?
Answer: Let in ordinary hydrogen, 11H = X%

Percentage of 21H = 100- x

Atomic mass of ordinary hydrogen \(=\frac{x \times 1+(100-x) \times 2}{100}\)

As per given data, \(\frac{x+(100-x) \times 2}{100}=1.008\)

or, 200-x = 1.008 x =99.2

∴ Ordinary hydrogen contains 99.2% of 11H and (100-99.2) = 0.8 % of 21H.

Question 2. Chlorine occurs in nature in the form of two isotopes with atomic masses 34.97 and 36.97 respectively. The relative abundance of the isotopes are 0.755 and 0.245 respectively Find the atomic mass of chlorine.
Answer: Atomic mass of chlorine

⇒ \(=\frac{34.97 \times 0.755+36.97 \times 0.245}{0.755+0.245}=35.46\)

Question 3. Determine the mass of1 F-atom in gram (F = 19)
Answer: Gram-atomic mass of fluorine = 19 g. The mass of1 gram-atom of fluorine= 19 g Number of atoms in 1 gram-atom fluorine = 6.022 x 1023 Mass of 6.022 x 1023 atoms of fluorine = 19g

Hence, the mass of1 atom of fluorine = \(\frac{19}{6.022 \times 10^{23}} \mathrm{~g}\)

=3.1550×10-23g

Question 4. Calculate the atomic volume of sodium (atomic weight = 23 ). Density of sodium=0.972 g rnL-1.
Answer: In the case of monatomic elements (like Na), the volume of 1 gram-atom is called its atomic volume.

1 gram-atomNa =23gofNa(v atomic mass of Na = 23)

Atomic volume of sodium = \(\begin{aligned}
& =\frac{\text { gram-atomic mass }}{\text { density }} \\
& =\frac{23 \mathrm{~g}}{0.972 \mathrm{~g} \cdot \mathrm{mL}^{-1}}=23.66 \mathrm{~mL}
\end{aligned}\)

Question 5. Find out the highest and lowest masses from the following:

  1. 25.6g oxygen (atomic mass = 16)
  2. 2.86 gram-atom of sodium (atomic mass = 23)
  3. 0.254 gram-atom of iodine (atomic mass = 127

Answer: Mass of 2.86 gram-atom of sodium = 2.86 x 23 = 65.78g

Mass of 0.254 gram-atom ofiodine = 0.254 x 127 = 32.258g

∴ Oxygen has the lowest and sodium has the highest mass.

Question 6. A compound contains 28% of nitrogen and 72% of metal by weight. In the compound, 3 atoms of metal remain combined with 2 atoms of nitrogen. What is the atomic mass of the metal?
Answer: Each molecule of the compound contains 3 atoms of metal and 2 atoms of nitrogen. If the symbol ofthe metal is M, then the molecular formula ofthe compound will be M3N2.

Molecular mass ofthe compound = 3a + 2 X 14 = 3a + 28

[where a is the atomic mass ofthe metal]

∴ Quantity of nitrogen in the compound \(=\frac{28}{(3 a+28)} \times 100 \%\)

Now, according to the problem, \(\frac{28}{3 a+28} \times 100=28\)

or, 3a +28= 100 or, a= 24

Molecular Mass Or Molecular Weight

A molecule is the smallest particle ofthe substance (element or compound) which has independent existence. Molecules are formed by the combination of atoms of the same or different elements.

So, like the atomic mass, some standard should be taken to express the molecular mass. At present, the 12C isotope is taken as the standard to express both the atomic mass and molecular mass.

Molecular mass with respect to 12 C-atoms

The molecular mass of a substance (element or compound) is a relative number that denotes how many times a molecule of the substance is heavier than the l/12th part of the mass of a 12C -atom.

∴ Molecular mass

⇒ \(\begin{aligned}
& =\frac{\text { mass of } 1 \text { molecule of the element or compound }}{\frac{1}{12} \times \text { mass of } 1 \text { carbon }\left({ }^{12} \mathrm{C}\right) \text { atom }} \\
& =\frac{\text { mass of } 1 \text { molecule of the substance }}{\text { mass of } 1 \text { carbon }\left({ }^{12} \mathrm{C}\right) \text { atom }} \times 12
\end{aligned}\)

On the carbon scale, molecular masses of nitrogen and oxygen are 28.013 and 31.998 respectively. It means that one molecule of each of nitrogen and oxygen is respectively 28.0 til and 31.990 times heavier than 1/12 the part of the mass of one,2C -atom.

Determination of the molecular mass of an element and a compound: The molecular mass of a substance (element or compound) can be determined by adding the atomic of all the atoms present in a molecule of the substance (element or compound).

The molecular mass of an element: Let us consider, the molecular formula of an element to be A,(, where n = Atomicity (i.e., no. of atoms present in the molecule of the element). Therefore, in the case of an element molecular mass = atomic, mass ofthe element x its atomicity.

  1. In the case of monoatomic elements (n = 1), molecular mass and atomic mass will be the same. Most of the metal elements and noble gas elements belong to this group.
  2. In the case of diatomic elements (n =2), the molecular mass is twice its atomic mass. Most ofthe gaseous elements (H2, N2, O2, Cl2, etc.) belong to this group.
  3. In the case of triatomic elements (n =3), the molecular mass is thrice its atomic mass. For example, ozone (O3 ).
  4. In the case of tetra-atomic elements(n = 4), the molecular mass is ( four times its atom mass. For example, phosphorus (P4).

The molecular mass of the compound: Let, the molecular formula of a compound be A2 C2, where the number of atoms of A, B, and C in a molecule of the compound are x, y,z respectively.

If the atomic masses are a, b, and c respectively, then the molecular mass ofthe compound =axx+bxy+cxz.

Class 11 Chemistry Some Basic Concepts Of Chemistry Molecular Mass Of Some Compounds

Molecular Mass Of Some Elements And Compounds In Unified Scale 

Class 11 Chemistry Some Basic Concepts Of Chemistry Molecular Mass Of Some Element And Compounds In Unified Scale

Formula Mass Or Formula Weight

Formula mass represents the sum of atomic masses of the Number of gram-molecule or gram-mole: The given present in the formula with which a substance is expressed. For example, the formula mass of sodium chloride, gram-molecular mass, gives the number of gram-mole or NaCl= (23 + 35.5) = 58.5.

Molecular mass and formula mass are not always synonymous: Molecular mass can be determined from the formula of an element or compound.

So sometimes, the formula mass is called the molecular mass. But these two terms are not always synonymous.

When a substance contains discrete molecules, then only these two terms can be used in the same sense.

Again, there are certain substances which do not exist as molecules. For example, the compound sodium chloride is represented by the formula NaCl but the existence of discrete molecules of this compound is conspicuously absent from the crystal of sodium chloride,

Na+ and Cl- ions exist in a state of aggregation where one Na+ ion is surrounded by six Cl- ions and one Cl- ion is similar and pressure, molar volumes of all gases are the same and it does surround by six Na+ ions to give rise to an octahedral not depend on the nature of the molecular mass ofthe gas.

As a result, there is no existence of Gram-molecular volume or molar Thus the statement—”the molecular mass of sodium chloride is 58.5″—has no logical basis because sodium chloride never forms a molecule.

The correct statement should be—”The formula mass of sodium chloride is 58.5″ Usually the term formula mass’ is applied to ionic compounds that do not exist as discrete molecules even in the solid state while the term ‘molecular mass’ is applied to the case of covalent compounds which remain in the molecular state even in aqueous solutions.

Gram-molecular mass

Gram-molecular mass: Molecular mass. of an element or compound expressed in gram is called gram-molecular N2 28 g 56 g —5628 = 2 2 x 22.4 = 44.8 L mass or gram-mole.

Unlike molecular mass molecular mass has a unit. For example, the gram-molecular masses of nitrogen and carbon dioxide are 28 g and 44 g respectively.

Gram-molecule or gram-mole: The quantity of a substance (element or compound), expressed in gram, which is numerically equal to its molecular mass, represents one gram-molecule or one gram-mole of that substance.

For example, 28 g of nitrogen, 32 g of oxygen, and 18 g of water represent gram-molecule or 1 gram-mole of each of the respective compounds.

Several gram-molecule or gram-mole: The given present in the formula with which a substance is expressed in gram when divided by its Therefore, number of gram-mole of substance.

⇒ \(=\frac{\text { mass of the substance }(\mathrm{g})}{\text { gram-molecular mass of that substance }}\)

Examples: 1. Number of gram-mole in 88 g of C02 \(=\frac{88 \mathrm{~g}}{44 \mathrm{~g}}=2\)

Number of gram-molein 45 g of H2O \(=\frac{45 \mathrm{~g}}{18 \mathrm{~g}}=2.5\)

Gram-molecular volume or molar volume at STP: The volume occupied by the gram-mole of all gases at STP (standard temperature and pressure) is 22.4L.

Example: Each of 2g hydrogen, 28 g nitrogen, 44 g carbon dioxide, and 18 g water vapor occupy a volume of 22.4L at STP. Conversely, it can be stated that 22.4 L of any gas at STP contains 1 gram-mole of that gas.

Volumes of different amounts of gases at STP

Class 11 Chemistry Some Basic Concepts Of Chemistry Volumes of different amount of gases at STP

Numerical Examples

Question 1. Calculate the number of gram-molecules present in 14.7gH2SO4.
Answer: Number of gram-molecules of a substance

⇒ \(=\frac{\text { given mass }(\text { in } \text { g) }}{\text { gram-molecular mass }}=\frac{14.7}{98}=0.15\)

Question 2. Calculate the mass of 1.5 gram-molecule of glucose.
Answer: Gram-molecular mass of glucose(C6H12O6)

= 6X12 + 12X1 + 6X16 = 180g

Mass of 1.5 gram-molecule of glucose = 1.5 x 180 = 270g

Question 3. If the number of gram-molecules present in 4.8g of oxygen and rg of nitrogen are equal, calculate the value of x.
Answer: Gram-molecular mass of oxygen (O2 ) and nitrogen (N2 ) are 32 g and 28g respectively.

Number of gram-molecule of oxygen in 4.8g = 4.8/32 and number of gram-molecules of nitrogen in xg = \(\frac{x}{28}\)

⇒ \(\text { Given, } \frac{x}{28}=\frac{4.8}{32} \text { or, } x=4.2\)

Question 4. Calculate the volume of 3.6 g of water vapor at 273 K temperature and 1 atm pressure.
Answer: Molecular mass of water vapor (H2O) =18 No. of gram-molecule in 3.6g water vapor =3.6/18 At STP (273K, late), the volume of gram-molecule of water vapour= 22.4L

At STP, the volume of 0.2 gram-molecule of water vapor = 0.2X22.4 = 4.48L

Question 5. If the density of water at 273K is i.0g-cm-3, calculate its molar volume at that temperature.
Answer: Density of water at 273K = 1.0 g-cm Gram-molecular mass of water = 18g Hence, at 273K the molar volume of water.

\(=\frac{\text { gram-molecular mass }}{\text { density }}=\frac{18 \mathrm{~g}}{1.0 \mathrm{~g} \cdot \mathrm{cm}^{-3}}=18 \mathrm{~cm}^3\)

Question 6. How many moles of water molecules are present in 1.8 ml of water?
Answer:

At ordinary temperature, the density of water =1g mL 1.

Mass of 1.8 mL ofwater = 1.8 X 1 = 1.8 g.

Now, 1.8 g of H2O =\(\frac{1.8}{18}\)=0.1 mol H2O

Question 7. Calculate the number of moles and the volume at STP of 0.53g of acetylene.
Answer: Gram-molecular mass of acetylene is 26g.

Number of gram-mole present in 0.52g of acetylene

= 0.52/26g = 0.02

At STP, 26 g of acetylene occupies a volume of 22.4 L.

At STP, 0.52g of acetylene occupies a volume of

⇒ \(=\frac{22.4 \times 0.52}{26} \mathrm{~L}=0.448 \mathrm{~L}\)

Question 8. At STP, the volume of lg of a gaseous substance is 280 mL. Find its relative molar mass.
Answer: At STP, the mass of 22400 mL of the gas \(=\frac{1}{280} \times 22400=80 \mathrm{~g}\) Now, on the basis of Avogadro.s hypothesis, 22400mL (at STP) of any gas contains1 gram-mole ofthe substance.

The molar mass of the gas = 80

Question 9. The volume of one atom of a metal M is 1.66x 10-23 cm3. Find the atomic mass of M (Given: density of M = 2.7 g.cm-3).
Answer: Volume of 6.022 x 1023 atoms of the metal, M

= 6.022 X 1023 X 1.66 X 10-23 =9.99652 cm3

Mass of 6.022 x 1023 atoms of the metal M

= density x volume = 2.7 x 9.99652 = 26.990 g

= gram-atomic mass ofthe metal M

Hence, the atomic mass ofthe metal = 2.990.

Question 10. Haemoglobin was found to contain 0.335% iron (Atomic weight of Fe = 56). The molecular weight of haemoglobin is 1.67 X 104. Find the number of iron atoms in hemoglobin.
Answer: Molecular mass of hemoglobin = 1.67 x 104

∴ 1 gram-mole of hemoglobin = 1.67 x 104 g.

Here, Fe -content haemoglobin = 0.335%

∴ Fe-content 1.67 x 104 g of haemoglobin

⇒ \(=\frac{0.335 \times 1.67 \times 10^4}{100}=55.945 \mathrm{~g} .\)

Hence, the number of gram-atom of Fe present in haemogobin= \(=\frac{55.945}{56} \approx 1\)

The atomic weight of =

∴ In 1 gram-mole of hemoglobin, the number of gram atoms of Fe =1

So, 1 mol of hemoglobin contains only one atom of Fe.

Question 11. The mass of 0.1 mol of X2Y is 4.4 g and the mass of 0.05 of XY2 is 2.3g. Find the Atomic mass of X and Y.
Answer:

Mass of 0.1 mol of X2Y = 4.4 g.

∴ Mass of 1 mol of X2Y \(=\frac{4.4}{0.1}=44 \mathrm{~g}\)

Similarly, mass of 1 mol of XY2 =2.3/0.05 =46g

So, the molecular mass of X2Y and XY2 are 44 and 46 respectively. Let, the atomic masses of X and Y be a and b respectively. So, the molecular mass of X2Y = (2a + b) and the molecular mass of XY2 =(a + 2b).

According to the question, 2a + b =44 and a + 2b = 46.

By solving the equations, a = 14 and b = 16.

∴ The atomic masses of X and Y are 14 and 16 respectively.

Question 12. A plant virus is found to consist of uniform cylindrical particles whose diameter is 150A and length is 5000A. The specific volume of the virus is 0.75cm³/g. If the virus is considered to be a single particle, then find its molecular mass.
Answer: Volume ofthe virus =nr2 xl

⇒ \(=3.14 \times\left(\frac{150}{2} \times 10^{-8}\right)^2 \times 5000 \times 10^{-8}=0.884 \times 10^{-16} \mathrm{~cm}^3\)

⇒ \(\begin{aligned}
\text { Mass of single virus } & =\frac{\text { volume }}{\text { specific volume }} \\
& =\frac{0.884 \times 10^{-16}}{0.75}=1.178 \times 10^{-16} \mathrm{~g}
\end{aligned}\)

The molar mass of virus = Mass of single virus x NA

= 1.78 x 10-16x 6.023 x 1023 =7.095 x 10 7

Avogadro’s number

Avogadro’s number Definition: Avogadro’s number may be defined as the number of molecules present in one gram-mole of any substance, or element Similarly, mass of  1 mol of XY2 =2.3/0.05 =46g or compound (solid, liquid, or gas).

Avogadro’s number is usually denoted by the letter ‘N’ or ‘Na’- and its value is 6.022 x 1023. R. A. Millikan determined its value by the oil drop experiment’ in 1913.

Example: 28 g of N2, 32 g of O2, 18 g of H2O, or 100 g of CaCO3 —each indicates 1 gram-mole of substance containing 6.022 x 1023 number of molecules.

The value of Avogadro’s number does not depend on temperature and pressure because the mass and the number of molecules do not change with the variation in temperature and pressure.

Alternative definition of Avogadro’s number: The number of atoms present in the gram-atom of an elementary substance is called Avogardro’s number.

Example: 16 g O2 contains 6.022 x 1023 number atoms and in 12g carbon, the number of constituent atoms is 6.022 x 1023.

As 1 gram-mole of any gaseous substance (element or compound) occupies a volume of 22.4 L at STP, Avogadro’s number can further be expressed as—the number of molecules present in 22.4 I. of any gaseous substance (element or compound) at STP is called Avogadro’s number.

Modern Definition Of Avogadros Number

The Number of Od atoms Present In exactly 12 g carbon (12C) is designated as Avogadro’s number.

Class 11 Chemistry Some Basic Concepts Of Chemistry Modern Definition of Avogadros number

Airogadro’s constant: ‘Avogadro’s number/mole’ is called Avogadro’s constant i.e., ‘6.022 X 1023 ‘.

Avogadro’s number has no unit. But Avogadro’s constant has the unit ‘per mole’ or mol-1.Itis auniversal constant.

Definition of Gram-mole, Gram-atom, Molecular mass, Atomic mass in terms of Avogadro’s number.

Gram-mole: The quantity of a substance (element or compound) expressed in gram which contains 6.022 x 1023 molecules is defined as 1 gram of that substance.

Gram-atom: The quantity of an element expressed in gram which contains 6.022 x 1023 atoms is called 1 gram-atom of that element.

Molecular mass: The molecular mass of a substance (element or compound) may be defined as the number which when expressed in grams contains 6.022 x 1023 number of molecules that substance.

Atomic mass: The atomic mass of an element indicates the number which when expressed in grams contains 6.022 x 1023 atoms of that element.

Applications of avocados number

Calculation of actual mass of a molecule: if the molecular mass of a substance is known, the actual mass of one molecule of that substance can be calculated in the following way: Let, the molecular mass of a substance be M.

∴ 1 gram-mole ofthe substance =M g of that substance. Now, we know that 1 gram-mole of a substance contains Avogadro’s number of molecules.

∴ Mass of 6.022 x 1023 molecule M g.

∴ Mass of 1 molecule = \(\frac{M}{5.022 \times 10^{23}} \mathrm{~g}\)

∴ The actual mass of a molecule of any substance

⇒ \(=\frac{\mathrm{gram}-\text { molecular mass of the substance }}{\text { Avogadro’s number }}\)

Examples: Actual muss of a molecule of oxygen: The molecular mass of oxygen = 32.

∴ Actual mass of I oxygen molecule

⇒ \(\begin{aligned}
=\frac{\text { Gram-molecular mass of oxygen }}{\text { Avogadro’s number }} & =\frac{32 \mathrm{~g}}{6.022 \times 10^{23}} \\
& =5.313 \times 10^{-23} \mathrm{~g} .
\end{aligned}\)

Actual mass of 1 molecule of water:

The molecular mass of water = 18.

∴ Actual mass of I molecule of water

⇒ \(=\frac{18 \mathrm{~g}}{6.022 \times 10^{23}}=2.989 \times 10^{-23} \mathrm{~g}\)

Calculation of actual mass of an atom: From the known value of the atomic mass of an element, the actual mass of an atom of that element can be determined in the following way: Let, the atomic mass of an element = A.

1 gram-atom of that element = A g.

We know, 1 gram-atom of an element contains Avogadro’s number of atoms, i.e., 6.022 x 1023 atoms.

Mass of 6.022 x 1023 atoms =A g.

∴ Actual mass of 1 atom of an element

= \(=\frac{\text { gram-atomic mass of the element }}{\text { Avogadro’s number }}\)

Example 1. The actual mass of 1 atom of nitrogen:

Atomic mass ofnitrogen = 14.

∴ The actual mass of 1 atom of nitrogen:

⇒ \(\begin{aligned}
& =\frac{\text { gram-atomic mass of nitrogen }}{\text { Avogadro’s number }} \\
& =\frac{14 \mathrm{~g}}{6.022 \times 10^{23}}=2.324 \times 10^{-23} \mathrm{~g}
\end{aligned}\)

An alternative method for determination of the actual mm of an atom: Let the molecular mass of a substance= M 1 gram-molecule ofthe substance = Mg.

Now, 1 gram-molecule of any substance contains

Avogadro’s number of molecules.

Hence, mass of 6.022 x 1023 molecules =M g

Mass of 1 molecule \(=\frac{M}{6.022 \times 10^{23}} \mathrm{~g} .\) If the atomicity of the element is n, then one molecule ofthe element consists of n number of atoms.

Hence, the mass of n atoms of the element \(=\frac{M}{6.022 \times 10^{23}} \mathrm{~g}\) and the mass of I atom of the element \(\frac{M}{n \times 6.022 \times 10^{23}} g\)

Hence, the actual mass of an atom of the element

⇒ \(=\frac{\text { gram-molecular mass of the element }}{\text { atomicity of a molecule of the element } \times \text { Avogadro’s number }}\)

Example: Determination of the actual mass of a nitrogen atom—Gram-molecular mass of nitrogen = 28 and atomicity of a nitrogen molecule =2. Hence, the actual mass of a nitrogen atom.

⇒ \(=\frac{\text { gram-molecular mass of nitrogen }}{\text { atomicity of a nitrogen molecule } \times \text { Avogadro’s number }}\)

⇒ \(=\frac{28 \mathrm{~g}}{2 \times 6.022 \times 10^{23}}=2.324 \times 10^{-23} \mathrm{~g}\)

Number of molecules in a definite mass of a substance: Let, the mass of a certain amount of an element or a compound = W gram and its gram-molecular mass M gram. Now, the number of molecules presenting M gram of the substance = 6.022 X 1023.

W gram ofthe substance will contain \(=\frac{W}{M} \times 6.022 \times 10^{23}\) number of molecules.

Some important relations: If Avogadro’s number =NA and gram-molecular mass of an element or a compound =M, then—

Mass of 1 atom of an element \(=\frac{\text { gram-atomic mass }}{N_A}\)

Mass of 1 molecule of the substance \(=\frac{M}{N_A}\)

No. of molecules in W gram ofthe substance \(=\frac{W \times N_A}{M}\)

Number of molecules present in V L of gas at STP \(=\frac{V \times \mathrm{N}_{\mathrm{A}}}{22.4}\)

Class 11 Chemistry Some Basic Concepts Of Chemistry Number of molecules

No. of molecules in n mole of any substance = n x NA

lu (or, 1 amu)=\(\frac{1}{N_A}\)– g. = 1.6605 x lO-24 g (This amount is called 1 program or 1 dalton)

1 gram-mole of any substance contains the same number of molecules:

1. Let, the actual mass of atom hydrogen =x g

∴ Actual mass of a molecule hydrogen -2xg hydrogen is diatomic]

∴ Now,1 gram-mole ofhydrogen = 2 g ofhydrogen

∴ Number of molecules present in 1 gram-mole (or 2g) ofhydrogen= 2/2x = l/x

The molecular mass of oxygen = 32

2. A molecule of oxygen is 32 times heavier than 1 atom of hydrogen.

If the actual mass of an atom of hydrogen is xg, then the actual mass of 1 molecule ofoxygen will be 32x g.

Now,1 gram-mole ofoxygen =32 g of oxygen.

∴ Number of molecules present in 1 gram-mole (or 32g) of oxygen of oxygen

⇒ \(=\frac{32}{32 x}=\frac{1}{x}\)

Let, the molecular mass of a substance be M. Therefore, a molecule of that substance will be M times heavier than an atom of hydrogen.

If the actual mass of 1 atom of hydrogen is x g, then the actual mass of 1 molecule of that substance will Now, be 1 gram-mole of that substance =M g Number of molecules constituting 1 gram-mole of the substance = \(=\frac{M}{M \times x}=\frac{1}{x}\)

Therefore, it can be concluded that 1 gram-mole of any substance contains the same number of molecules.

The ratio of the number of molecules present In equal masses of two different substances (solid, liquid, or gas) with different molecular masses: Let the molecular masses of two substances A and B be MA and MB respectively and the mass of each of them be = W g.

1 gram-mole of any substance contains = 6.022 x 1023 number of molecules.

∴ Number of molecules present in MA g of A = 6.022 x 1023 Number of molecules presentin’ W g of \(A=\frac{6.022 \times 10^{23} \times W}{M_A}\)

Similarly, the number of molecules present in Wg of B \(=\frac{6.022 \times 10^{23} \times W}{M_B}\)

The ratio of the number of molecules present in equal masses of A and B

⇒ \(=\frac{6.022 \times 10^{23} \times W}{M_A}: \frac{6.022 \times 10^{23} \times W}{M_B}=M_B: M_A\)

Thus, the ratio of the number of molecules in equal masses of two different substances Is equal to the inverse ofthe ratio of their molecular masses.

Mole Concept

Mole is a Latin word meaning quantity, heap, or collection. In the SI system, mole (symbol: mol) was introduced as the seventh base quantity for the amount of a substance.

Mole Concept Definition: A mole is defined as the number of particles (like atoms, molecules, ions, or radicals) that is exactly equal to the number of 12C atoms present in 0.012kg of carbon.

Mole Concept Explanation: 0.012kg of 12C atoms contains Avogadro’s number of carbon atoms. So, 1 mol of any substance i.e., molecule, atom, ion, or radical contains Avogadro’s number of the specified species.

Irrespective of the ultimate particles, one mole always contains the same number of constituent particles i.e., 6.022 x 1023.

Number of molecules in 1 mol of molecule 1

= Number of atoms in 1 mol of the atom
= Number of mol of ion
= Number ofradicalsin1 mol of radical
= 6.022 X 1023

Mole Concept Discussion: While using the word ‘mole’, the name and nature (i.e., molecule, atom, ion, or radical) should be mentioned.

For example, the number of molecules in 1 mol of an oxygen molecule is equal to the number of atoms present in 1 mol of oxygen atom and in both cases, this number is 6.022 x 1023.

But oxygen is diatomic and hence the number of oxygen atoms in a 1 mol oxygen molecule is double the number of oxygen atoms in a 1 mol atom of oxygen.

So, the statement “1 mol oxygen” may lead to unnecessary confusion because it indicates both 1 mol of oxygen molecule and 1 mol of oxygen atom. But quantitatively they are altogether different although the number of ultimate particles in both cases is the same.

Often, quantity of substance are expressed in decimole, centimole or millimole where, 1 decimol = 10-1 mol, 1 centimol = 10-2 mol,1 millimol = 10~3 mol.

‘ Mole’ is also used in the case of electrons e.g., 1 mol electron = 6.022 X 1023 electrons and 1 millimol electron =6.022 X 1020 electrons.

Useful relations used in the mole-related calculation

Relation between ‘1 mol’ molecule and gram-mole: Mol molecule of any substance (element or compound) denotes 6.022 x 1023 molecules of that substance.

Again, the mass of 6.022 x 1023 molecules = 1 gram molecular mass or 1 gram-mole.

Hence, 1 mol molecule of an element or compound — 1 gram-mole of that substance.

Mole Concept Examples:

l mol nitrogen molecule = 6.022 x 102J nitrogen molecules= 28 g nitrogen (=1 gram-mole of nitrogen)

1 mol carbon dioxide molecule = 6.022 x 1 023 carbon dioxide molecules= 44 g carbon dioxide (= 1 gram-mole carbon dioxide)

1 million oxygen molecule = 10~3 mol oxygen molecule = 10-3 X 6.022 x 1023 oxygen molecules = 10~3 X 32 g oxygen = 3.2 x 10-2 g of oxygen

1 mol sodium chloride = 6.022 X 1023 formula units of sodium chloride

Relation between ‘1 mol’ atom And gram-atomic mass or gram-atom: 1 mol of an element denotes = 6.022 x 1023 atoms.

Again, a mass of 6.022 x 1023 atoms =1 gram-atomic mass (or, 1 gram-atom) So,1 mol atom of an element = 1 gram-atom of that element.

Mole Concept Examples:

  • l mol oxygen atom = 6.022 x 1023 oxygen atoms =16 g oxygen (=1 gram-atom ofoxygen)
  • 1 mol sodium atom = 6.022 x 1023 sodium atoms =23 g sodium(=1 gram-atom of sodium)
  • Relation between ‘1 mol’ ion and gram-ion: 1 mol ion indicates 6.022 x 1023 ions.
  • Again, number ofionsin1 gram-ion = 6.022 x 1023. 1 million =1 gram-ion (gram-formula mass of)

Examples:

l mol Cl- ion = 6.022 x 1023 Cl- ions

= 35.5 g Cl- ions (=1 gram-ion of Cl-) (v mass of an electron is negligible)

1 mol \(\mathrm{SO}_4^{2-}\) ion = 6.022 x 1023 \(\mathrm{SO}_4^{2-}\) ions = 96 g

\(\mathrm{SO}_4^{2-}\)  ions (=1 gram-ion of \(\mathrm{SO}_4^{2-}\) )

1 mol Al3+ ion = 6.022 x 1023 Al3+ ions =27 g Al3+ ions (=1 gram-ion of Al3+)

The volume occupied by the ‘1 mol’ molecule of a gas at STP: For any substance, 1 mol molecule stands for 6.022 x 1023 number of molecules. Moreover, 22.4 L of any gas at STP contains 6.022 x 1023 molecules.

Hence, the volume occupied by 1 mol molecule of any gaseous substance at STP = 22.4 L.

Examples: 1 mol oxygen molecule = 6.022 x 1023 oxygen molecules =22.4L of oxygen at STP

1 mol carbon dioxide molecule= 6.022 x 1023 carbon dioxide molecules =22.4 L ofcarbon dioxide at STP.

Various relationships regarding the mole concept

Class 11 Chemistry Some Basic Concepts Of Chemistry Various relationships regrading mole concepts

Number of electrons in terms of mol: 6.022 x 1023 number of electrons are present in 1 mol of the electron. The charge carried by 1 mol electron is 96500 coulomb or faraday.

Calculation of mole number

In the case of monatomic elements: 23 g sodium (1 gram atom sodium) = 1 mol sodium

W g sodium = mol\(=\frac{W}{23}\) sodium

Mole number monatomic element

⇒ \(=\frac{\text { mass of the element }(\mathrm{g})}{\text { gram-atomic mass of the element }}\)

In the case of ions: 35.5 g (1 gram-ion) of Cl =1 mol Cl.

Wg Cl- ions\(=\frac{W}{35.5}\) mol CP ions.

Similarly, 96 g (1 gram-ion) of \(\mathrm{SO}_4^{2-}\) ions = 1 mol \(\mathrm{SO}_4^{2-}\)

W g \(\mathrm{SO}_4^{2-}\) ions \(=\frac{W}{96}\) mole \(\mathrm{SO}_4^{2-}\) Ions

Mole number of ion = \(=\frac{\text { mass of ion (g) }}{\text { mass of } 1 \text { gram-ion }}\)

2. In case of any gas: 22.4 L of any gas at STP =1 mol of gaseous molecules.

VL of any gas at STP =\(=\frac{V}{22.4}\) mol of gaseous molecules.

In case of any gas: 22.4 L of any gas at STP =1 mol of
gaseous molecules.

VL of any gas at STP = mol of gaseous molecules.

Mole number of gaseous molecule

⇒ \(=\frac{\text { volume of the gas at STP (L) }}{22.4 \mathrm{~L}}\)

Calculation of mass of substance from mole number

Mass of monoatomic element = mole number of molecule or atom of the element X gram-atomic mass of the element.

Mole number of gaseous molecules

⇒ \(=\frac{\text { volume of the gas at STP }(\mathrm{L})}{22.4 \mathrm{~L}}\)

Calculation of mass of substance from mole number

1. Mass of monoatomic element = mole number of molecule or atom of the element X gram-atomic mass of the element.

For example, 0.5 mol of Na = ( 0.5 x 23) = 11.5 g Na

2. Mass of polyatomic substance (element or compound) = mole number of the substance X gram-molecular mass of that substance

For example, 0.5 mol H2SO4 =(0.5 x 98) = 49g H2SO4

Mass of ions = mole number of Ions x mass of 1 gram-ion or gram-ionic mass

For example, 0.5 mol SO2-4 =(0.5 x 96) =48 gSO2- ion

Determination of the volume of a gas from mole number:

The volume of any gas at STP (L) = a number of moles of gaseous molecules x 22.4 L.

Determination of number of particles (molecules, atoms, or ions): Number of particles (molecule, atom, or ion) = mole number x 6.022 x 1023.

Advantages of the mole concept

Chemical calculations can be worked out in a much simpler way by using mole numbers instead of the masses or volumes of the reactants and products. To be more specific, (2 x 2.016) g of 112 react with 32 g of O2 to form (2 x 18.016) g of water (H2O). If the reaction is expressed in terms of mole, then we will say that 2 mol of hydrogen molecules combine with 1 mol of oxygen molecules to produce 2 mol of water molecules.

⇒ \(2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}\)

Valuable Information like the number of molecules, atoms, or ions present In a certain amount of any substance is represented by the number of moles.

The word ‘mole’ finds extensive use in expressing the quantity of molecules, ions, or other tiny particles present in the solution.

If the mole number of molecules of a gaseous substance is known, then its volume at STP can be determined.

Numerical Examples

Question 1. If we spend 10 lakh rupees per second then how much time will be required to spend an amount of money which is equal to Avogardo’s number?
Answer: 10000000 rupee will be spent in 1 sec

6.022x 1023 rupees will be spend in ,\(\frac{1 \times 6.022 \times 10^{23}}{10^6} \text { sec. }\)

⇒ \(=\frac{6.022 \times 10^{17}}{60 \times 60 \times 24 \times 365} \text { years }=1.91 \times 10^{10} \text { years }\) years = 1.91×10 10 years

Question 2. Find the weight of 12.046 X 1025 number of ammonia molecules.
Answer: 1 gram-mole ammonia = 17 g of ammonia.

∴ Number of molecules contained in 1 gram-mole of ammonia = 6.022 X 1023.

Hence, 6.022 x 1023 molecules weigh 17 g

∴ 12.046 X 1025 molecules weight

⇒ \(=\frac{17 \times 12.046 \times 10^{25}}{6.022 \times 10^{23}}=3400 \mathrm{~g}=3.4 \mathrm{~kg}\)

Question 3. What is the quantity of charge carried by 1 mol of electron?
Answer: 1 mol electron =6.022×1023 number of electrons

Charge carried by 1 electron =1.6x 10-19 coulomb

The total charge carried by 1-mole electron

=(6.022x 1023×1.6×10-19)= 96352 coulomb.

Question 4. Calculate the number of molecules left when 1021 molecules of CO2 are removed from 200 mg of CO2.
Answer: 200mg of CO2 =0.2 g of (Gram – molecular mass of CO2=44g). 44g CO2 Contains 6.022×1023 molecules.

∴ 0.2g CO2 Contains \(\begin{aligned}
& =\frac{6.022 \times 10^{23} \times 0.2}{44} \text { molecules } \\
& =2.7372 \times 10^{21} \text { molecules }
\end{aligned}\)

On removing 1021 molecules, number of C02 molecules remaining = 2.7372 X 1021 – 1 x 1021 = 1.7372 X 1021

Question 5. Find the number of atoms of hydrogen and oxygen present in one spherical drop of water with radius I mm at 4°C.
Answer:

The volume of one spherical drop of water

⇒ \(=\frac{4}{3} \times \pi \times(0.1)^3=\frac{4}{3} \times \frac{22}{7} \times 10^{-3}=4.19 \times 10^{-3} \mathrm{~cm}^3\)

Further, the density of water at 4°C = lcm-3

Mass of 4.19 x 10-3cm3 water at 4°C =4.19 x 10~3 g.

Now, the mass of 1 gram-molecule of water = 18 g.

∴ Number of molecules in 18g water =6.022 x 1023

∴ Number of molecules present in 4.19 x 10-3 g of water

⇒ \(=\frac{6.022 \times 10^{23} \times 4.19 \times 10^{-3}}{18}=1.4017 \times 10^{20}\)

1 molecule of water contains 2 hydrogen and 1 oxygen atom.

∴ The number of hydrogen atoms in 1.4017 x 1020

∴ Molecules of water = 2 x 1.4017 x 1020 = 2.803 x 1020

So, the number of oxygen atoms present in 1.4017 x 1020

molecules of water = 1.4017 x 1020

Question 6. Find the number of electrons in a drop of sulphuric acid weighing 4.9 x 10-3 mg [assume it to be cent percent pure].
Answer: 4.9 X 10-3 mg = 4.9 X 10-6 g Molecular mass of H2S04 = 98 Its gram-molecular mass = 98 g. Number of molecules in 98 g of H2S04 = 6.022 x 1023 So, number of molecules in 4.9 x 10-6 g of H2S04

⇒ \(=\frac{6.022 \times 10^{23} \times 4.9 \times 10^{-6}}{98}=3.011 \times 10^{16}\)

Atomic numbers of H, S, and O are 1, 16, and 8 respectively. So, the number of electrons in them is 1, 16, and 8.

Total number of electrons present in 1 molecule of H2SO4 =(2×1=1×16=4×8)=50

3.011 x 1016 number of H2S04 molecules contain = 3.011 X 1016 x 50 = 1.5055 X 1018 electrons.

Thus, 4.9 x 10-3 mg of H2S04 has a 1.5055 X 1018 number of electrons.

Question 7. Find the number of 0 atoms and 0 molecules present in 1 g of oxygen.
Answer: 1 g oxygen= gram-mole ofoxygen [v = 32 ]

Number of moleculesin1 g ofoxygen = 6.022 x 1023 x — = 1.882 x 1022

As oxygen molecule is diatomic, the number of atoms present

in 1g of oxygen = 2 X 1.882 X 1022 = 3.764 X 1022

Question 8. Calculate the number of O-atoms present in 112 L of COz gas at STP.
Answer: At STP, the number ofmoleculespresentin 22.4 L of C02 gas
= 6.022 x l023

Number of molecules present in 112L of C02 gas =112. x 6.022 X 1023 = 5 X 6.022 X 1023 22.4

Now, each molecule of C02 contains 2 atoms of oxygen.

Number ofO-atoms presenting 112L of C02 gas (at STP)

= 2 X 5 X 6.022 X 1023 = 6.022 X 1024

Question 9. Find the number of neutrons present in 5 x 104 mol of 14 C Isotope.
Answer: Atomic number of carbon = 6

Number of neutrons in one C atom =14-6 = 8.

Now, in 1 mol of 14C, the number of atoms = 6.022 x 1023

In 5 x 10 4 mol of C, the number of atoms

= (6.022 X 1023 x 5 X 10-4) = 3.011 X 1020 14

Again, one C atom contains 8 neutrons.

3.011 x 1020 number of atoms of 14C contain

= (8 x 3.011 x 1020) =2.4088 x 1021 neutrons.

Number of neutrons in 5 x 10-4 mol of 14C =2.4088 x 1021.

Question 10. Find the number of hydrogen and oxygen atoms present in 0.09 g of water.
Answer: -09 g water =\(=\frac{0.09}{18}\) =5×10 3 gram-mole ofwater [ Molecular mass ofwater = 18

Number of molecules in 1 gram-mole water = 6.022 x 1023

Number of molecules in 5 x 10-3 gram-mole water

⇒ \(=6.022 \times 10^{23} \times 5 \times 10^{-3}=3.011 \times 10^{21}\)

The number of hydrogen and oxygen atoms in 1 molecule of water (H2O) are 2 and 1 respectively.

The number of hydrogen atoms in 3.011 x 1021 number

of water molecules = 2 x 3.011 x 1021 =6.022 x 1021

The number ofoxygen atoms =1 x 3.011 x 1021

= 3.011 X 1021

Question 11. What is the mass of 1 millimol of ammonia? Also, find the number of ammonia molecules presenting it.
Answer: 1 millimol = 10-3 mol.

Mass of1 mol of ammonia = 17 g

Mass of 10-3 mol of ammonia = 17 x 10-3g

Again, the number of ammonia molecules present in 1 mol of ammonia = 6.022 x 1023

The number of ammonia molecules in 10-3 mol of ammonia = 6.022 x TO23 x 10-3 = 6.022 x 1020

Hence, the number of ammonia molecules present in millimolar of ammonia =6.022 X 1020

Question 12. What will be the number of

  1. Moles of ethylene,
  2. Molecules of ethylene,
  3. Atoms of carbon and

Atoms of hydrogen in 0.28 g of ethylene contained in a cylinder?

Answer: 0.28 g of ethylene \(=\frac{0.28}{28}\) = 011 gram-mole ethylene [since molecular mass of ethlene= 28]

Quantity of ethylene in the cylinder= 0.01 mol.

1 gram-mole contains 6.022 x 1023 ethylene molecules.

1 gram-mole contains 6.022 x 1023 ethylene molecules.

1 gram-mole contains 6.022 x 1023 ethylene molecules.

1 gram-mole contains 6.022 x 1023 ethylene molecules.

2 x 6 022 X 1°21 carbon atoms

So, the number of carbon atoms present in the cylinder

⇒ \(=2 \times 6.022 \times 10^{21}=1.2044 \times 10^{22}\)

1 molecule of ethylene contains 4 atoms of hydrogen.

Number of hydrogen atoms in 6.022 x 1021 ethylene molecules = 4 x 6.022 x 1021 =2.4088 X 1022 Thus, the number of hydrogen atoms present in the cylinder = 2.4088 x 1022.

Question 13. Find the number of moles and quantifying gram, contained in 100 m of ammonia at STP.
Answer: At STP, the volume of gram-mole of ammonia = 22.4L

∴ Number of the in 100 of ammonia \(=\frac{1 \times 100}{22.4}=4.464\)

Now 1 gram-mole of ammonia = 17 g of ammonia [since the molecular mass of ammonia = 17 ]

∴ 4.464 gram moles of ammonia = (17 x 4.464) =75.888g of ammonia

Thus, 100 l of ammonia at STP contained = 75.888g

Question 14. Suppose the human population of the world is 3 X 1010. If 100 molecules of sugar are distributed per head, what is the total quantity of sugar required for distribution?
Answer: Gram-molecular mass of sugar

=(12×12+22×1+16×11)=(144+22+176)= 342g

Number of molecules in 342 g of sugar =6.022×1023

∴ Mass of 100 molecules of sugar \(\begin{aligned}
& =\frac{342 \times 100}{6.022 \times 10^{23}} \\
& =5.679 \times 10^{-20} \mathrm{~g}
\end{aligned}\)

So, 5.679 x 10-20 g of sugar will be required per head.

For 3 x 1010 number of people, the quantity of sugar

required =5.679 x 10-20 X 3 x 1010 = 1.7037 X 10-9 g.

Question 15. Find the number of atoms of nitrogen in 1 grantmole of NO and 0.5 gram-mole of N02. Which one will be heavier—1 gram-mole of NO or 0.5 gram-mole of N02?
Answer: Number of molecules present 1 gram-mole of NO = 6.022 X 1023.

Each molecule of NO contains a nitrogen atom.

therefore 6.022×1023 No molecules will contain 6.022×1023 atoms of nitrogen.

Again, the number of N02 molecules present in 0.5 gram-mole of N02 =\(=\frac{1}{2} \times 6.022 \times 10^{23}\) x 6.022 x 1023 =3.011 X 1023

Again, 1 gram-mole NO2= 30g of NO [∴ MNO = 30]

∴ 0.5 gram-mole of NO2 = 0.5×46 =23g NO2 [since M NO2=46]

So, 1 gram- mole NO2 is heavier than 0.5 Gram mole NO2.

Question 16. A mixture contains 02 and N2 in the proportion of 1: 4 by weight. What will be the ratio of the number of molecules of 02 and N2 in the mixture?
Answer: Let, the mass ofthe mixture be W g

In the mixture, mass of 02 =\(\frac{W}{5} \mathrm{~g}\) and mass of N2 \(=\frac{4 W}{5} \mathrm{~g}\)

Number of gram-mole \(\mathrm{N}_2=\frac{4 W / 5}{28}=\frac{4 W}{140}\)

So, number of gram-mole of \(=\frac{4 W / 5}{28}=\frac{4 W}{140}\)

So, number of 002 molecules = \(=\frac{W}{160} \times 6.022 \times 10^{23}\)

number of N2 molecules = x 6.022 x 1023

The ratio of the number of molecules of 02 to the number of molecules of N2 in the mixture

⇒ \(=\frac{W \times 6.022 \times 10^{23}}{160}: \frac{4 W \times 6.022 \times 10^{23}}{140}\)

⇒ \(=\frac{1}{160}: \frac{4}{140}=14: 64=7: 32 \text {. }\)

Question 17. A young man has given ills bride a tin engagement ring containing a 0.50-carat diamond. How many atoms of carbon are present in that ring? |1 carat = 200 mg).
Answer: 1 carat = 200 mg = 0.2 g. So,0.5 carat = 0.5 x 0.2 = 0.1 g Diamond is composed of only carbon atoms

i.e., 0.1 g diamond = 0.1 g carbon

Number of carbon atoms in 12 g of carbon = 3.022 x 1023

[ since Atomic mass of carbon = 12 ]

Number of carbon atoms in 0.1 g of carbon

⇒ \(=\frac{6.022 \times 10^{23}}{12} \times 0.1=5.018 \times 10^{21}\)

Therefore, the young man gave his bride-to-be 5.0018×1021 atoms of carbon

Question 18. What is the number of O-atoms present in 44.8 L of Ozone gas at STP?
Answer:

At STP, the number of molecules present in 22.4 L of ozone gas = 6.022 X 1023

Number of molecules present in 44.8 L of ozone gas at

⇒ \(\mathrm{STP}=\frac{6.022 \times 10^{23} \times 44.8}{22.4}=2 \times 6.022 \times 10^{23}\)

∴ Number of oxygen atoms present in the given volume of ozone gas = 3 X 2 X 6.022 X 1023 = 3.6132 X 1024

[Since each molecule of ozone contains 3 O-atoms]

Applications Of Avogadro’s Hypothesis

The following important corollaries which are of great significance in chemistry have been established by the application of Avogadro’s hypothesis:

Molecules complementary gases (e.g., hydrogen, oxygen, etc.) except inert gases are diatomic.

The molecular mass of a gaseous substance (element or compound) is twice its vapor density.

The gram-molecular volume of any gaseous substance (element or compound) at STP is 22.4L.

The molecular formula of any gaseous compound can be determined from its volumetric composition.

The atomic masses of elements can be determined from the value of their vapor density.

All elementary gases are diatomic except inert gases

1. Hydrogen and chlorine molecules are diatomic: From experiments, it is observed that under the same conditions of temperature and pressure, 1 volume of hydrogen combines chemically with 1 volume of chlorine to form 2 volume of hydrogen chloride gas.

∴ 1 volume of hydrogen + 1 volume ofchlorine = 2 volumes of hydrogen chloride Let, ‘n’ be the number of hydrogen molecules in 1 volume of hydrogen gas under the experimental condition.

According to Avogadro’s hypothesis, at the same temperature and pressure, 1 volume of chlorine and 2 volumes of hydrogen chloride will contain ‘n’ molecules of chlorine and ‘2n’ molecules of hydrogen chloride respectively.

Therefore, ‘n ‘ molecules of hydrogen +’n’ molecules ofchlorine = 2n molecules of hydrogen chloride. or, 1 molecule of hydrogen +1 molecule ofchlorine = 2 molecules of hydrogen chloride.

molecule ofhydrogen +i molecule of chlorine =1 molecule ofhydrogen chloride.

Hence, 1/2 molecule of hydrogen and 1/2 molecule of chlorine are present molecules of hydrogen chloride.

We know, a molecule of hydrogen chloride is composed of hydrogen and chlorine atoms only.

Therefore, 1 molecule of hydrogen chloride must contain at least 1 atom of hydrogen and 1 atom of chlorine because the atom is indivisible.

Thus 1 atom of hydrogen and 1 atom of chlorine must have come from 1/2 molecule of hydrogen and 1/2 molecule of chlorine respectively i.e., 1/2 molecule of hydrogen and 1/2 molecule of chlorine contain one hydrogen atom and one chlorine atom respectively.

So, two atoms of hydrogen and two atoms of chlorine are present in hydrogen and chlorine molecules respectively i.e., hydrogen and chlorine molecules are diatomic.

2. Nitrogen molecule is diatomic: Actual experiments show that under the same conditions of temperature and pressure, 3 volumes of hydrogen and 1 volume of nitrogen combine chemically to produce 2 volumes of ammonia.

1 vol. of nitrogen + 3 vol. hydrogen = 2 vol. of ammonia If under the experimental conditions of temperature and pressure 1 volume of nitrogen contains ‘n’ nitrogen molecules, then according to Avogadro’s hypothesis, at the same temperature and pressure 3 volumes of hydrogen and 2 volume of ammonia will also contain ‘3n’ and ‘2n’ molecules of hydrogen and ammonia respectively.

‘n’ molecules ofnitrogen +’ 3n’ molecules ofhydrogen = ‘2n’ molecules ofammonia

i.e.,1 molecule ofnitrogen + 3 molecules ofhydrogen = 2 molecules of ammonia

or, 1/2 molecule ofnitrogen + 3/2 molecules ofhydrogen =1 molecule ofammonia.

Now, at least 1 atom of nitrogen must be present in 1 molecule of ammonia and this atom of nitrogen must come from 1/2 molecule of nitrogen.

So, in a molecule of nitrogen, the number of nitrogen atoms must be at least 2 i.e., the nitrogen molecule is diatomic.

3. Oxygen molecule is diatomic: From actual experiments, it has been found that under identical conditions of temperature and pressure, 2 volumes of hydrogen react with 1 volume ofoxygen to give 2 volumes of steam.

2 vol. of hydrogen +1 vol. ofoxygen = 2 vol. of steam If at the experimental conditions of temperature and pressure, 1 volume of oxygen gas contain ‘n ‘ molecules, then according to Avogadro’s hypothesis, at that temperature and pressure 2 volume of each of hydrogen and steam must contain’ 2n’ molecules each. Therefore, ‘ 2n’ molecules of hydrogen +’n’ molecules of oxygen

= ‘2n’ molecules of steam or, 2 molecules of hydrogen +1 molecule of oxygen

= 2 molecules of steam.

According to Dalton’s atomic theory, the atom is indivisible. Therefore, at least one atom of oxygen must be present in one molecule of steam and this atom of oxygen must come from 1/2 molecule of oxygen,

So, the number of oxygen atoms present in one molecule of oxygen is two. Hence, oxygen molecules are diatomic.

The molecular Mass Of A Gas Is Twice Its Vapour Density

Absolute density of a gas: The mass (in grams) of L of a gas at a certain temperature and pressure is called its absolute density at that temperature and pressure. Absolute density changes with temperature and pressure.

The normal density of a gas: The normal density of a gas may be defined as the mass (in grams) of 1 L of the gas at STP; e.g., the normal density of hydrogen is 0.089 g- L-1.

Relative density or vapor density of a gas: The vapor density of a gas is a relative value that shows how many times it is heavier than the equal volume ofthe lightest gas i.e., hydrogen under similar conditions of temperature and pressure. Its value is independent of temperature and pressure.

Relative density or vapor density of a gas Definition: Vapour density or relative density of a gaseous substance is defined as the ratio of the mass of a certain volume of the gas to the mass of the same volume of hydrogen, measured under the same conditions of temperature and pressure.

Relation between normal density & relative density of a gas: For any gas, relative density of a gas: for any gas, relative density

⇒ \(\begin{aligned}
& =\frac{\text { mass of } V \text { volume of the gas at STP }}{\text { mass of } V \text { volume of } \mathrm{H}_2 \text { gas at STP }} \\
& =\frac{\text { mass of } 1 \mathrm{~L} \text { of the gas at STP }}{\text { mass of } 1 \mathrm{~L} \text { of } \mathrm{H}_2 \text { gas at STP }} \\
& =\frac{\text { normal density of the gas }}{\text { normal density of } \mathrm{H}_2 \text { gas }}=\frac{\text { normal density of the gas }}{0.089} \\
& \quad\left[\text { Normal density of } \mathrm{H}_2 \text { gas }=0.089 \mathrm{~g} \cdot \mathrm{L}^{-1}\right]
\end{aligned}\)

Thus for any gas, normal density = 0.089 x relative density

Relation between molecular mass and vapor density (D):

Vapour density of a gas \(=\frac{\text { mass of } V \text { volume of a gas }}{\text { mass of } V \text { of } \mathrm{H}_2 \text { gas }}\)

[At same temperature and pressure]

According to Avogadro’s hypothesis, under the conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

So, If the V volume of a gas under certain conditions of temperature and pressure contains ‘n ‘ molecules, then at the same temperature and pressure, the V volume of hydrogen gas will also contain ‘n’ number of molecules.

Vapor density of a gas(D)

⇒ \(\begin{aligned}
& =\frac{\text { mass of ‘ } n \text { ‘ molecules of a gas }}{\text { mass of ‘ } n \text { ‘ molecules of hydrogen }} \\
& =\frac{\text { mass of } 1 \text { molecule of a gas }}{\text { mass of } 1 \text { molecule of hydrogen }}
\end{aligned}\)

⇒ \(\begin{aligned}
& =\frac{\text { mass of } 1 \text { molecule of a gas }}{\text { mass of } 2 \text { atoms of hydrogen }}[\text { hydrogen is diatomic }] \\
& =\frac{1}{2} \times \frac{\text { mass of } 1 \text { molecule of a gas }}{\text { mass of } 1 \text { atom of hydrogen }}
\end{aligned}\)

Now by definition, the molecular mass (M) of a substance

⇒ \(=\frac{\text { mass of } 1 \text { molecule of the substance }}{\text { mass of } 1 \text { atom of hydrogen }}\)

Therefore, vapour density of a gas (D) = \(=\frac{1}{2} x\)

its molecular mass (M) i.e., D= M/2 Or, M=2D

So, the molecular mass of any gas is twice its vapor density. [If the atomic mass of hydrogen is taken as 1.008, molecular mass (M)’= 2.016 x vapor density (D) ]

Effect of temperature and pressure on the vapor density of a gas: The absolute density of a gas depends on both temperature and pressure.

This is because even though the mass of gas is kept fixed, the volume changes with variations in temperature and pressure.

But the vapor density of a gas is independent of temperature and pressure because it is a ratio ofmass of a certain volume of gas to the mass of the same volume of hydrogen at a fixed temperature and pressure.

Thus, vapor density is a mere number. Temperature or pressure has no effect on it.

Vapor density of a gas with respect to another gas: Let MA and MB be the molecular masses of the gases A and B respectively. So, the vapor density of gas A with respect to gas B, t(DA)B] is given by

⇒ \(\left(D_A\right)_B=\frac{M_A}{M_B}\)

Examples: Vapour density of 02 with respect to N2

⇒ \(=\frac{\text { molecular mass of } \mathrm{O}_2}{\text { molecular mass of } \mathrm{N}_2}=\frac{32}{28}=1.14\)

Vapor density of NH3 with respect to air

⇒ \(=\frac{\text { molecular mass of } \mathrm{NH}_3}{\text { average molecular mass of air }}=\frac{17}{29}=0.59\)

The gram-molecular volume of any (JASOOIIS substance (element or compound) at STP Is 22.4 L

With the help of Avogadro’s hypothesis, It can be deduced that the molar volume of any gaseous substance at STP Is 22.4 L

Vapour density (D) of any gas \(=\frac{\text { mass of } V \text { volume of the gas }}{\text { mass of } V \text { volume of } \mathrm{H}_2 \text { gas }}\)

If the gas is kept at STP, then the vapor density (D) of the gas

⇒ \(\begin{aligned}
& =\frac{\text { mass of } V \text { volume of the gas at STP }}{\text { mass of } V \text { volume of } \mathrm{H}_2 \text { gas at STP }} \\
& =\frac{\text { mass of } 1 \mathrm{~L} \text { gas at STP }}{\text { mass of } 1 \mathrm{~L} \mathrm{H}_2 \text { gas at STP }}=\frac{\text { mass of } 1 \mathrm{~L} \text { gas at STP }}{0.089 \mathrm{~g}} \\
& \qquad\text { Mass of } 1 \mathrm{~L} \text { of } \mathrm{H}_2 \text { gas at STP }=0.089 \mathrm{gl}
\end{aligned}\)

Mass of L of the gas at STP =D x 0.009 g

According to Avogadro’s hypothesis, D= M/2

[where, M = molecular mass of the gas]

Mass of1 L ofthe gas at STP = \(=\frac{M \times 0.089}{2} \mathrm{~g}\)

Or, Volume occupied by \(\frac{M \times 0.089}{2}\)g of the gas at STP =1l

Volume occupied by M gram gas at STP = \(=\frac{2}{0.089}\)= 22.4 L

The exact atomic mass of hydrogen on a 12 C -scale is 1.008.

Thus, the true molecular mass of hydrogen is 2.016.

For an accurate result, the correct value of the normal density of hydrogen i.e., 0.09 g-L-1 is to be considered.

Therefore, the volume occupied by M gram of the gas at
g of the gas at STP \(=\frac{2.016}{0.09} \mathrm{~L}=22.4 \mathrm{~L}.\)

So,1 gram-mole ofthe gas occupies 22.4L at STP.

Thus, it is proved that the molar volume of any gaseous substance at stp =22.4 l.

The molar volume of all gases Is same under the identical Conditions of temperature and pressure: On the basis of

Avogadro’s hypothesis, this important corollary can established in the following way:

Let, the actual mass of 1 atom of hydrogen = x g.

Mass of1 hydrogen molecule = 2x g [hydrogen is diatomic] Now, 1 gram-mole hydrogen =2 g of hydrogen

Number of hydrogen molecules in 1 gram-mole (or 2g) of hydrogen \(=\frac{2}{2 x}=\frac{1}{x}\)

The molecular mass of oxygen = 32

So, the actual mass of ] molecule of oxygen = 32 x actual

mass of1 atom ofhydrogen = (32 X x)g

Now, I gram-mole oxygen 32 % oxygen

Number of oxygen molecule* in J gram-mole for 32 g oxygen = \(=\frac{32}{32 x}=\frac{1}{x}\)

let, the molecular: m of any gaseous substance = M

Therefore, the I molecule of the above gas is A/ times heavier than the I atom of hydrogen.

So, the actual mass of the molecule of the gas

= M x actual mass of hydrogen atom =(M x x)g

Now, I gram-mole of the gas =M g

Number of molecules in 1 gram- mole of the gaseous substance = \(=\frac{M}{M x}=\frac{1}{x}\)

Thus, it shows that 1 gram-mole of any gaseous substance contains the same number of molecules.

Again according to Avogadro’s law, at the same temperature and pressure, the volume occupied by this same number of molecules Is the same.

Hence, under identical conditions of temperature and pressure, the volume occupied by 1 gram-mole of all gaseous substances. i.e., the molar volume of all gases is the same.

Volume ovmple At STP,l gram-mole of any gas occupies a volume of 22.4L Again, 1 gram-mole of any gas contains 6.022 x 1023 molecules.

1 molecule occupies the voulume of \(\frac{22.4}{5.022 \times 10^{23}} \mathrm{~L}\)

1 molecule occupies the volume of \(\frac{22.4}{6.022 \times 10^{23}} \mathrm{~L}\) = 3.719 x 10-23 l at STP.

Determination of the molecular formula of any gaseous compound from its volumetric composition

If the volumetric compositions ofthe constituent elements and the vaPour density of the gaseous compound are known, the the molecular formula of the compound can easily be determined by the application of Avogadro’s hypothesis.

Determination of molecular formula of hydrogen chloride: From experiments, it is found that at a certain temperature and pressure, 1 volume of hydrogen combines with 1 volume of chlorine to form 2 volumes of hydrogen chloride.

If under the experimental conditions of temperature and pressure, ‘n ‘ molecules of hydrogen are present in its I volume, then according to Avogadro’s law, under the same conditions of temperature and pressure, ‘n ‘ molecules ofchlorine and ‘2n’ molecules of hydrogen chloride will be present in 1 volume of chlorine and 2 volumes of hydrogen chloride respectively.

‘ n ‘ molecules ofhydrogen+’n’ molecules ofchlorine

= ‘2n’ molecules of hydrogen chloride

or, 1 molecule of hydrogen + 1 molecule ofchlorine

= 2 molecules of hydrogen chloride

or, I molecule of hydrogen + 1/2 molecule ofchlorine

= 1 molecule of hydrogen chloride

According to Avogadro’s law, both hydrogen and chlorine are diatomic.

atom of hydrogen +1 atom of chlorine

=1 molecule of hydrogen chloride

The molecular formula of hydrogen chloride =HC1.

The molecular mass, determined from the deduced formula =1 + 35.5 = 36.5

[ V Atomic masses of hydrogen and chlorine are 1 and 35.5 respectively]

Again, the vapor density of hydrogen chloride = 18.25

Thus, the molecular mass of hydrogen chloride = 2X18.25 =36.5

So, the molecular mass, determined from the deduced formula of hydrogen chloride is exactly the same as calculated from the measured vapor density of hydrogen chloride.

Therefore, it is proved that the molecular formula of hydrogen chloride is HCl.

2. Determination of molecular formula of carbon dioxide: Carbon dioxide is made up of carbon and oxygen.

From experiments, it is found that under similar conditions of temperature and pressure, a certain volume of carbon dioxide dissociates to give an equal volume of oxygen.

So, the volume of carbon dioxide contains the volume of oxygen.

If at a certain temperature and pressure, 1 volume of carbon dioxide contains an ‘n’ number of molecules, then according to Avogadro’s law, at that temperature and pressure, 1 volume of oxygen also contains an ‘n’ number of molecules.

So, 1 molecule of carbon dioxide contains 1 molecule of oxygen or 2 oxygen atoms (as oxygen is diatomic).

Let, the molecular formula of carbon dioxide be Cx02 where x stands for the number of carbon atom(s) present in 1 molecule of carbon dioxide.

Now, Molecular mass of CxO2 =12X x+16X2 =(12x+32)

[since atomic masses of and O are 12 and 16 respectively]

Further, the vapor density of earphone dioxide is 22, and hence its molecular mass = 2 x 22 = 44 Therefore, 12x + 32 = 44 or, x =

The molecular formula of carbon dioxide is CO2

3. Determination of molecular formula of ammonia:

Experimental observations show that under similar conditions of temperature and pressure, 1 volume of nitrogen and 3 volumes of hydrogen combine chemically to produce 2 volumes of measure, the number of molecules present in 3 volummonia.

Let, under experimental conditions of temperature and pressure, ‘n’ be the number of molecules present In I volume of nitrogen.

Then according to Avogadro’s law, under the same conditions of temperature and pressure of hydrogen, 2 volumes of ammonia will be 3n and 2n respectively.

‘n’ molecules of nitrogen + ‘3n ‘ molecules of hydrogen.

= ‘2n’ molecules of ammonia

i.e.,1/2 molecule of nitrogen +3/2 molecules of hydrogen

=1 molecule of ammonia

or, 1 atom of nitrogen + 3 atoms of hydrogen = 1 molecule of ammonia [v hydrogen and nitrogen are diatomic So, 1 molecule of ammonia is composed of 1 atom of nitrogen and 3 atoms of hydrogen.

The molecular formula of ammonia =NH3

The molecular mass calculated from the deduced formula = 14 +(1×3) = 17

[ since Atomic masses of and H are 14 and 1 respectively]

Again, the vapor density of ammonia = 8.5

therefore Molecular mass of ammonia = 8.5 x 2 = 17

So, the molecular mass determined from the deduced formula of ammonia is exactly the same as the molecular mass calculated from the value of the vapor density of ammonia.

Hence, it is proved that the molecular formula of ammonia is NH3

Determination of the atomic mass of a gaseous element from its vapor density

If the vapor density of a gaseous element is known then its atomic mass can be estimated with the help of Avogadro’s law.

The molecular mass of a gaseous element = atomic mass of the element x atomicity of its molecule.

The atomic mass of the gaseous element.

⇒ \(\begin{aligned}
& =\frac{\text { molecular mass of gaseous element }(M)}{\text { atomicity }} \\
& =2 \times \frac{\text { vapour density of gaseous element }(D)}{\text { atomicity }}[\quad M=2 D]
\end{aligned}\)

Example: Vapour density and atomicity of oxygen are 16 and 2 respectively.

Therefore Atomic mass of oxygen \(=2 \times \frac{16}{2}=16\)

Importance of Avogadro’s hypothesis

The contributions of Avogadro’s hypothesis towards the development of chemistry are mentioned below:

This hypothesis made a clear distinction between the ultimate particle of matter (atom) and the smallest particle having independent existence (molecule).

This hypothesis modified the drawbacks of Dalton’s atomic theory and proposed the molecular theory.

This theory successfully explained Gay-Lussac’s law of gaseous volumes.

The different corollaries derived from this hypothesis served as important tools for the purpose of chemical calculations.

Determination of the atomic mass of an element, the molecular formula of a gaseous compound, and the expression of a chemical reaction by an equation was made possible with the help of this hypothesis.

Numerical Examples

Question 1. The vapor density of a gaseous element is 5 times that of oxygen. If the element is triatomic, find its atomic mass.
Answer: Vapour density of oxygen \((D)=\frac{M}{2}=\frac{32}{2}=16\)

Therefore Vapour density of the gaseous element = 5×16 =80

Therefore Molecular mass of the gaseous element = 80 x 2 =16

So, atomic mass of the element \(\frac{\text { molecular mass }}{\text { atomicity }}\)

=160/3=53.33

Question 2. 100 mL of ay weighs 0.144 g at STP. What is the vapor density of the gas?
Answer: Mass of 1 00mL of the gas at STP = 0.144 g

therefore 22400mL ofthe gas at STP weighs \(=\frac{0.144 \times 22400}{100}\)

= 32.256 g

Hence, the molecular mass of the gas = 32.256 g

Vapour density of the gas = \(\frac{\text { molecular mass }}{2}\)

⇒ \(=\frac{32.256}{2}=16.128\)

Question 3. The vapor density of sulfur relative to nitrogen gas at STP is 9.143. Determine the molecular formula of sulfur vapor
Answer: If the molecular masses of two gases are M1 and M2 then
the vapor density of the first gas, relative to the second gas [(D1)2]=M1/M2

So, vapor density of sulfur vapor relative to nitrogen gas \(=\frac{\text { molecular mass of sulfur vapor}}{\text { molecular mass of nitrogen gas }}\)

∴ Molecular mass of sulphur vapour = 9.143×28 =256.004

Now, the atomic mass of sulfur = 32

∴ Atomicity of sulphur vapour= \(=\frac{256.004}{32} \approx 8\)

Hence, the molecular formula of sulfur vapor =S8

Question 4. At STP, 250 cm3 of a gas weighs 0.7317 g. If the density of H2 gas at STP is 0.08987 g L”1 then what will be the vapor density of the gas? Determine the molecular mass of the gas.
Answer: Mass of250 cm3 of H2 gas at STP
⇒ \(=\frac{0.08987 \times 250}{1000} \mathrm{~g}=0.0224 \mathrm{~g}\)

[since At STP, the density of H2 gas = 0.08987 g.L-1 1

∴ The vapor density of the gas

⇒ \(=\frac{\text { mass of } 250 \mathrm{~cm}^3 \text { gas at STP }}{\text { mass of } 250 \mathrm{~cm}^3 \mathrm{H}_2 \text { gas at STP }}=\frac{0.7317}{0.0224}=32.66\)

Thus, its molecular mass = 2×32.66 =65.32.

Question 5. Volumes of N2 and 02 in any gas mixture are 80% and 20% respectively. Determine the average vapor density of the gas mixture.
Answer: Vapour density of N2 \(=\frac{M_{\mathrm{N}_2}}{2}=\frac{28}{2}=14 \text { and }\)

Vapour density of 02 \(=\frac{M_{\mathrm{O}_2}}{2}=\frac{32}{2}=16\)

Since the Average vapor density of the gas mixture

⇒ \(\begin{aligned}
& =\frac{80 \times \text { vapour density of } \mathrm{N}_2+20 \times \text { vapour density of } \mathrm{O}_2}{100} \\
& =\frac{80 \times 14+20 \times 16}{100}=14.4
\end{aligned}\)

Question 6. At 26.7°C, the vapor density of a gaseous mixture containing N2O4 and N2O4 is 38.31. Calculate the number of moles of N2O4 in 100g of that mixture.
Answer: Let the amount of N02 in 100 g of the mixture x g

Therefore Amount of N204 = (100 -x) g

Therefore Number of moles ofN02 in mixture \(=\frac{x}{46}\)

And number of moles of N90i( in mixture \(=\frac{(100-x)}{92}\)

since MNO2 = 46 MN2O4=92]

Total number ofmoles ofN02 and N204 in the mixture \(=\frac{x}{46}+\frac{100-x}{92}=\frac{100+x}{92} \cdots(1)\)

Now,molecular mass ofthe mixture = (2 x 38.3) = 76.6

Total number of moles of the mixture \(=\frac{100}{76.6}\)

Now from 1 and 2 \(\frac{100+x}{92}=\frac{100}{76.6} \quad \text { or, } x=20.1\)

So, number of moles of N02 in mixture \(=\frac{20.1}{46}=0.4369\)

Question 7. The vapor density of a gas, relative to air is 1.528. What is the mass of 2L of the gas at 27°C temperature and 750 mm Hg pressure? [Vapour density of air, relative to hydrogen = 14.4.]
Answer: Vapour density of a gas \(=\frac{\text { mass of certain volume of a gas }}{\text { mass of same volume of } \mathrm{H}_2 \text { gas }}\)
Therefore Vapour density of the gas

⇒ \(=\frac{\text { mass of } V \mathrm{~cm}^3 \text { of gas }}{\text { mass of } V \mathrm{~cm}^3 \text { of air }} \times \frac{\text { mass of } V \mathrm{~cm}^3 \text { of air }}{\text { mass of } V \mathrm{~cm}^3 \text { of } \mathrm{H}_2 \text { gas }}\)

[at same temperature and pressure]

=1.528×14.4 =22

Therefore Molecular mass ofthe gas = 2 x vapour density = 44
Again, volume of1 gram-mole of any gas at STP = 22.4 L
So, mass of22.4L of given gas at STP = 44 g
Now, let the volume ofthe gas be V L at STP. i.e.,

P1 = 750 mm; P2 = 760 mm; Vx = 2L; V2 = VL;

Tx =27 °C =(27 + 273) =300 K; T2 = 273 K

⇒ \(\quad \frac{750 \times 2}{300}=\frac{760 \times V}{273} \quad \text { or, } V=\frac{750 \times 2 \times 273}{300 \times 760}=1.796 \mathrm{~L}\)

Now,mass of 22.4 L ofthe gas at STP = 44 g

Therefore Mass of 1.796l of the gas at stp \(=\frac{44 \times 1.796}{22.4}=3.528 \mathrm{~g}\)

Hence, the mass of the gas= 3.258g.

Question 8. Under the same conditions of temperature and pressure, complete combustion of the volume of a gaseous hydrocarbon produces 3 volumes of carbon dioxide and 4 volumes of steam. What is the die formula of die hydrocarbon?
Answer: Volume ofhydrocarbon+ 02 =3 volumes of C02 +4 volumes of H.,0 (steam) [under same conditions of temperature and pressure]

If 1 volume of hydrocarbon contains ‘n’ molecules, then under identical conditions of temperature and pressure 3 volumes of C02 will contain ‘3n’ molecules and 4 volumes of steam will contain ‘4n’ molecules.

Therefore, ‘n’ molecules hydrocarbon = ‘3n’ molecules C02 + ‘4n’ molecules H2O Now, 3 molecules of C02 = 3 atoms of C

[ ∴ 1 atom of C is present in 1 molecule of C02 and 4 molecules of H2O = 8 atoms of H [v 2 atoms of I-I are presentin1 molecule of H2O

So, 1 molecule of hydrocarbon contains 3 atoms of C and 8 atoms of H.

Molecular formula of the hydrogen = C3H8

Question 9. Under the same conditions of temperature and pressure, a gaseous hydrocarbon contains hydrogen which is twice its volume. If the vapor density of that hydrocarbon is 14, then what will be its molecular formula?
Answer: According to a problem, at the same conditions of temperature and pressure, 1 volume of gaseous hydrocarbon contains 2 volumes of hydrogen.

Since n molecules of gaseous hydrocarbon contain 2n molecules of hydrogen or, 1 molecule of gaseous hydrocarbon contains 2 molecules of hydrogen =4 atoms of hydrogen.

(According to Avogadro’s hypothesis)

Therefore Molecular formula hydrocarbon is CXH4

[Where x stands for the number of c-atoms]

The molecular formula of hydrocarbon is CXH4 = 12xx+4

[since the atomic mass of C=12, H=1]

Again, the vapor density of the hydrocarbon = 14

So, molecular formula of the hydrocarbon = 2 x 14 = 28

CxH4 = 14 x 2

therfore, 12x+4=28 or, x=2

The molecular formula hydrocarbon is C2H4.

Question 10. The weight of 350mL of a diatomic gas at 0°C temperature and 2 atm pressure is lg. Calculate the weight of its own atom.
Answer: According to the given condition, the volume of diatomic gas at 0°C and 2 atm pressure = 350 mL.

Let, the volume ofthe gas at 0°C and atm pressure be V mL

Now, according to Boyle’s law, P1 V1 =P2 V2

or, 2 X 350 =1 X V V = 700 mL

i.e, mass of 700mL of the gas at 0°C and 1 atm pressure =1 g

∴ Mass of 22400mL ofthe gas at 0°C and 1 atm pressure

⇒ \(=\frac{22400}{700} \times 1=32 \mathrm{~g}\)

Molar volume of all gases at STP = 22400 mL.

Hence, the gram-molecular mass ofthe given gas = 32 g.

So, the mass of 6.022 x 1023 molecules ofthe given gas= 32 g.

Therefore, the mass of 2 x 6.022 x 1023 atoms ofthe gas= 32 g
[since the gas is diatomic]

Thus, the mass of 1 atom of the given gas

⇒ \(=\frac{32}{2 \times 6.022 \times 10^{23}}=2.656 \times 10^{-23} \mathrm{~g} .\)

Question 11. For complete combustion, 24g of a solid element requires 44.8L of 02 at STP. The gaseous oxide produced in combustion occupies a volume of 4.8 L at STP. What is the molecular mass of the produced gaseous oxide?
Answer: Molar volume of any gas at STP = 22.4L.

∴ 22.4L of 02 at STP =1 gram-molecule of 02 = 32 g 02

∴ Mass of 44.8L of 02 at STP \(=\frac{32}{22.4}\) 44.8 = 64 g 02

Now, 24 g of solid element reacts completely with 64 g of
02 to produce a gaseous oxide.

In this case, the total mass ofthe reactants = (24 + 64) =88 g.

Now According to the law of constant proportion (in the case) an oxide is only formed) or the law of conservation ofmass, the total mass ofthe product will be 88 g.

∴ Mass of 44.8 L ofthe gaseous oxide formed at STP = 88 g.

∴ Mass of22.4L ofthe gaseous oxide at STP = y = 44 g.

Hence, the gram-molecular mass of the gaseous oxide formed = 44 g

∴ The molecular mass ofthe gaseous oxide formed = 44.

Question 12. A sample of hard water contains 20 mg of Ca2+ ions per liter. How many millimoles of Na2C03 would be required to soften the L of the sample? Also, calculate the mass of.Na2C03
Answer: Reaction \(\mathrm{Ca}^{2+}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{CaCO}_3+2 \mathrm{Na}^{+}\)

Number of moles ofNa2C03 molecules

= Number of moles of Ca2+ ion \(=\frac{20 \times 10^{-3}}{40}=5 \times 10^{-4}\)

∴ Number of millimoles of Na2C03 required to soften 1L of the sample = 5 x 10-4 x 103 = 0.5

∴ Mass of Na2C03 =5xl0-4xl06 =0.05g

Valency

Valency Definition: The capacity with which an atom of an element combines chemically with the atom(s) of another element is called the combining capacity or valency of the element.

It is determined by the number of hydrogen atoms that combine with an atom of the element or are displaced from a hydrogenated compound by one atom of that element.

There are some elements that do not combine directly with hydrogen or can’t displace hydrogen from any hydrogenated compound.

In such cases, the valency of an element can be determined by the number of atoms ofthe element that combine with an element of known valency.

Elements with different valencies are mentioned in the following table

Class 11 Chemistry Some Basic Concepts Of Chemistry Elements With Different Valencies

Valency of radicals: Like an element, the valency of a radical can also be explained in the same way.

Valency of radicals Definition: Valency of a radical refers to the number of hydrogen atoms with which a radical can combine.

Thus, the valency of nitrate radical (NOJ) is because combines with 1 atom of hydrogen to produce an HN03 molecule. Similarly, sulfate (SO1-) and phosphate (PO1-) radicals exhibit di and tri valencies respectively.

This is because they combine with 2 and 3 atoms of hydrogen to form H2S04 and H3P04 molecules respectively.

Except NH+, other radicals behave like nonmetals. NH+ behaves like a monovalent metal. Like elements, radicals may also be monovalent, divalent, trivalent, etc.

Some Radicals And Their Valencies

Class 11 Chemistry Some Basic Concepts Of Chemistry Some Radicals And Their Valencies

In the case of a binary compound, the ratio of the number of atoms of the constituent elements is the inverse of the ratio of their valencies.

Let, x atoms of element A combine with y atoms of element B to form a compound and the valencies of A and B are a and b respectively.

So, the total valency of A in the compound =xa [For x atoms] and that often that compound =yb [For y -atoms

Now, in the case of a compound composed of two elements, the total valency of one element is equal to that ofthe other.

⇒ \(x a=y b \quad \text { or, } \frac{x}{y}=\frac{b}{a}\)

Thus, the tint formula of the compound formed by the elements A and I will be A;(Hrt.

So, It In evident that In a molecule of a binary compound, the ratio of (lie number of atoms of the constituent elements is Inverse of the ratio of their valencies.

Equivalent Weight Or Equivalent Mass Or Chemical Equivalent Of An Element

Equivalent Weight Definition: The equivalent weight or mass of an element Is the number of parts by mass of the element which combines with 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 ports by mass of chlorine or can displace the same amount of hydrogen, oxygen or chlorine respectively from their compounds.

Thus, the equivalent mass of an element

⇒ \(\begin{aligned}
& =\frac{\text { mass of the element }}{\text { mass of hydrogen combined or displaced }} \times 1.008 \\
& =\frac{\text { mass of the element }}{\text { mass of oxygen combined or displaced }} \times 8 \\
& =\frac{\text { mass of the element }}{\text { mass of chlorine combined or displaced }} \times 35.5
\end{aligned}\)

The equivalent mass is a ratio of two masses. So it is a Pure number and has no unit.

Alternative Definition: The equivalent weight of an element may be defined as the number of parts by weight of the element which combines with 11.2L of hydrogen or 5.6L of oxygen or 11.2L of chlorine (at STP) or displaces the above-mentioned volume of hydrogen or oxygen or chlorine (at STP) from any compound.

Hence, the Equivalent mass of an element

⇒ \(\begin{aligned}
& =\frac{\text { mass of the element }(\text { in gram })}{\text { volume of hydrogen combined }} \times 11.2 \\
& \text { or displaced (in L) at STP } \\
& =\frac{\text { mass of the element }(\text { in gram })}{\text { volume of oxygen combined }} \times 5.6 \\
& \text { or displaced (in L) at STP } \\
& =\frac{\text { mass of the element }(\text { in gram })}{\text { volume of chlorine combined }} \times 11.2 \\
& \text { or displaced ( In L) at STP } \\
&
\end{aligned}\)

Gram-equivalent mass (or weight) and gram-equivalent: The equivalent mass of a substance (element, radical, or compound) expressed in gram is called gram-equivalent mass. This particular amount represents 1 gram equivalent of the corresponding substance (element, radical, or compound)

Gram-equivalent mass (or weight) and gram-equivalent Example: The equivalent masses of Na and Mg are 23 and 12 respectively, so their gram-equivalent masses are 23 g and 12 g respectively. Again, 1 gram-equivalent of Na= 23g and 1 gram-equivalent of Mg= 24g.

The number of grams- the equivalent of an element

⇒ \(=\frac{\text { mass of the element (in gram) }}{\text { gram-equivalent mass of the element }}\)

Law of equivalent proportions: Elements combine with one another or displace the other from their compounds in the ratio of their respective equivalent mass or in simple multiples of their equivalent masses.

The law of equivalent proportions Is a direct corollary ofthe type of reciprocal proportions: Both calcium and chlorine combine separately with oxygen to form calcium oxide (CaO) and chlorine monoxide (C120) respectively.

In calcium oxide (CaO), 16 parts by mass of O combine with 40 parts by mass of Ca.

∴ 8 parts by mass of O combine with 20 parts by mass of a. So, equivalent mass of Ca= 20 In chlorine monoxide (C120), 16 parts by mass of O combines with (2 X 35.5) parts by mass of Cl.

∴ 8 parts by mass of nO combines with 35.5 parts by mass of Cl.

So, the equivalent mass ofchlorine = 35.5 From the above data, it is clear that 8 parts by mass of oxygen combine separately with 20 parts by mass of Ca and 35.5 parts by mass of Cl.

So according to the law of reciprocal proportion, if the elements Ca and Cl combine together, the ratio of their masses in the resulting compound will be either 20: 35.5 or any simple multiple of it.

Again according to the law of equivalent proportion, Ca and Cl will combine with each other in the ratio of their equivalent masses i.e., in the ratio of 20: 35.5 (because their equivalent masses are 20 and 35.5 respectively) In practice, it is also found that Ca and Cl combine with each other in the mass-ratio of 20: 35.5 to form calcium chloride (CaCl2).

Class 11 Chemistry Some Basic Concepts Of Chemistry Law of Equivalent Proportions Is a direct corollary

It is thus seen that by using the law of reciprocal proportions and the law of equivalent proportions arrive at the same conclusion regarding the chemical combination of two or more elements. So, these two laws are different versions ofthe same proposition.

Relation Between Equivalent Mass And Atomic Mass Of An Element

Let, the atomic mass of an element =A, its equivalent mass

=E and its valency = V.

We know that the valency of an element indicates the number of hydrogen atoms that combine with one atom of that particular element.

∴ V atoms of hydrogen combine with the atom ofthe element. So, V X 1.008 parts by mass of hydrogen combine with A parts by mass of that element.

∴ 1.008 parts by mass of hydrogen combined with

⇒ \(\frac{A \times 1.008}{V \times 1.008}=\frac{A}{V}\) parts by mass of that element.

Thus, according to the definition of equivalent mass, A/v stands for the equivalent mass ofthe element.

⇒ \(E=\frac{A}{V} \quad \text { or, } \quad A=E \times V\)

The atomic mass of an element = equivalent mass of that element x its valency

If the valency of an element (V) =1, then A = E.

∴ For monovalent elements, atomic mass and equivalent mass are equal.

Example: Atomic mass of Na = 23. As Na is monovalent, the value of its equivalent mass will also be 23.

The equivalent mass of an element can never be more than that of its atomic mass: From equation number (1), A = E=a/v From this equation, it is clear that if the value of E exceeds that of A then the value of V will be less than 1. But valency is always a whole number and its value can never be less than 1.

∴ The equivalent mass of an element can never be more than its atomic mass.

The equivalent mass of an element can never be zero:

  1. From equation no. (1), E = A/v
  2. Now, E will be zero only when A = 0.
  3. But the atomic mass of an element can never become zero.
  4. Hence, the equivalent mass of an element can never be zero.

The equivalent mass of an element is inversely proportional to its valency:

⇒ \(=\frac{\text { atomic mass of the element }(A)}{\text { its valency }(V)}\)

But, for a given element, A = constant.

⇒ \(E=\frac{\text { constant }}{V} \text { or, } E \propto \frac{1}{V}\) Therefore Equivalent mass of an element varies inversely with its valency.

Equivalent Mass Of An Element May Vary

Equivalent mass of an element= \(=\frac{\text { atomic mass of the element }}{\text { valency of that element }}\)

The atomic mass of an element has a fixed value. However, the element may have variable valencies. In such cases, the equivalent mass ofthe elements may vary.

Thus, while mentioning the equivalent mass of an element having variable valencies, the compound that contains the element or the reaction in which the element participates must be mentioned.

Examples: Copper exhibits more than one valency; For Example in cuprous compounds (e.g., Cu20), the valency of Cu is 1 while in cupric compounds (For example CuO), the valency of Cu is 2.

Therefore, the equivalent mass of copper in cuprous compounds

⇒  \(=\frac{\text { atomic mass of } \mathrm{Cu}}{\text { valency }}=\frac{63.5}{1}=63.5\)

And the equivalent mass of copper in cupric compounds

⇒ \(=\frac{\text { atomic mass of } \mathrm{Cu}}{\text { valency }}=\frac{63.5}{2}=31.75\)

Similarly, the valency of metallic iron in ferrous compounds such as FeO, FeCl2, etc., is 2 while in ferric compounds such as Fe203, FeCl3, etc., is 3.

∴ Equivalent mass of Fein ferrous compounds \(=\frac{\text { atomic mass of } \mathrm{Fe}}{\text { valency }}=\frac{55.85}{2}=27.925\)

And the equivalent mass of Fein ferric compounds \(=\frac{\text { atomic mass of Fe }}{\text { valency }}=\frac{55.85}{3}=18.616\)

The equivalent mass of the elements that exhibit variable equivalent masses can be determined if the reactions in which they participate are known.

For example, in the reaction of Fe with HC1, ferrous chloride and H2 gas are produced.

⇒ \(\begin{aligned}
& \mathrm{Fe}+2 \mathrm{HCl} \longrightarrow \mathrm{FeCl}_2+\underset{2 \times 1.008 \text { parts by mass }}{+\mathrm{H}_2} \\
& 55.85 \text { parts by mass } \\
&\text { Equivalent mass of } \mathrm{Fe}=\frac{55.85 \times 1.008}{2 \times 1.008}=27.925
\end{aligned}\)

Again, red-hot Fe reacts directly with Cl2 to form FeCl3.

⇒ \(\begin{aligned}
& 2 \mathrm{Fe}+3 \mathrm{Cl}_2 \quad \rightarrow \quad 2 \mathrm{FeCl}_3 \\
& 2 \times 55.85 \text { parts by mass } \quad 3 \times 71 \text { parts by mass } \\
& \text { Here, equivalent mass of } \mathrm{Fe}=\frac{2 \times 55.85 \times 35.5}{3 \times 71}=18.616 \\
&
\end{aligned}\)

Examples of some other elements having variable equivalent mass are Sn, Pb, Hg, Cr, Mn, As, etc. Elements like Na, K, Mg, Ca, etc., are not characterized by variable valencies, and hence their equivalent masses are always fixed.

Thus while mentioning the equivalent masses of such elements, it is not necessary to mention the compound containing the element or the reactionin which the element participates.

Equivalent mass of radicals, acids, bases, salts, oxidants, and reductants

Equivalent mass of radicals: Equivalent mass of a radical denotes the number of parts by mass of a radical which combine with 1.008 parts by mass ofhydrogen, 8 parts by mass of oxygen, or 35.5 parts by mass of chlorine, or one equivalent of any other element or a radical.

Examples:

In HN03, 1.008 parts by mass of hydrogen combine with 62 parts by mass of nitrate (N03) radical, and hence the equivalent mass of nitrate (N03 ) radical = 62. 0 In H2S04, 2 x 1.008 parts by mass of hydrogen remain associated with 96 parts by mass of sulfate (SO2-4) radical.

Hence the equivalent mass of sulphate (SO2-4) radical= =\(\frac{96 \times 1.008}{2 \times 1.008}=48\)

⇒ \(\text { Equivalent mass of a radical }=\frac{\text { formula mass of the radical }}{\text { valency of the radical }}\)

Equivalent mass of an acid: The equivalent mass of an acid is defined as the number of parts by mass of the acid which contains 1.008 parts by mass irreplaceable hydrogen.

⇒ \(\begin{aligned}
\text { Equivalent mass of an acid } & =\frac{\text { molecular mass of the acid }}{\begin{array}{l}
\text { no. of replaceable } \mathrm{H} \text {-atoms } \\
\text { present per molecule of the acid }
\end{array}} \\
& =\frac{\text { molecular mass of the acid }}{\text { basicity of the acid }}
\end{aligned}\)

Examples:

Equivalent mass of HN03 = 63/1 = 63

Equivalent mass of H2S04 =98/2 = 49

Basicity of HN03 and H2S04 are1 and 2 respectively]

Equivalent mass of a base (or alkali): The equivalent mass of a base (or alkali) is defined as the number of parts by mass of the base (or alkali) which requires one equivalent mass of an acid for complete neutralization.

Equivalent mass of a base (or alkali)

⇒ \(=\frac{\text { molecular mass or formula mass of the base }}{\text { acidity of the base }}\)

Examples: Equivalent mass of NaOH =40/1= 40

Equivalent mass of CaO =56/2=28

[∴ Acidity of  NaOH and CaO are1 and 2 respectively]

Equivalent mass of salt: The equivalent mass of a normal salt denotes the number of parts by mass of that salt which contains one equivalent mass of active cation or anion.

The equivalent mass of a slat

⇒ \(\begin{aligned}
& =\frac{\text { molecular mass or formula mass of the salt }}{\text { (number of cations or anions present per molecule }} \\
& \text { of the salt } \times \text { valency of that cation or anion) } \\
& =\frac{\text { molecular mass or formula mass of the salt }}{\text { total valency of the cation or anion in one molecule of the salt }} \\
& =\frac{\text { molecular mass or formula mass of the salt }}{\text { total charge of the cation or anion in one molecule of the salt }}
\end{aligned}\)

Example: In A12(S04)3, the valency of Al3+ =3

∴ Equivalent mass of A12(S04)3 \(=\frac{342}{2 \times 3}=57\)

Similarly, in A12(S04)3, valency of SO2-4 = 2

∴ Equivalent mass of A12(S04)3 \(=\frac{342}{3 \times 2}=57\)

The equivalent mass of salt is also defined as the stun of the equivalent masses ofthe radicals present in that salt

∴ Equivalent mass of ofsalt= equivalent mass of cation + equivalent mass opinion

Examples: 1. Equivalent mass of Na2S04 = Equivalent mass of Na+ + Equivalent mass of SO2-4 =23 + 48 =71

Equivalent mass of A12(S04)3 = equivalent mass of Al3+ +Equivalent mass of SO2-4 =9 + 48 =57

Equivalent mass of Ca3(P04)2

= Equivalent mass of Ca2+ + Equivalent mass ofPO3-4

= 20 + 31.67 =51.67

Equivalent maw of oxidant and reductant: Equivalent muss of oxidant turd reductant tilth In determined by the following two methods. These are:

Electronic method:

⇒ \(\begin{aligned}
& \text { Equivalent mass of an oxidant } \\
& =\frac{\text { molecular mass or formula mass of the oxldant }}{\text { mumber of electron( } \mathrm{s}) \text { gained per molecule }}
\end{aligned}\)

Examples: Equivalent mass of KMnO4:0 Acidic medium, KMn04 Is reduced by suitable reductant to manganous (Mn2+) salt.

Reactions \(\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\)

∴ In an acidic medium, the equivalent mass of KMn04

⇒ \(=\frac{\text { molecular mass of } \mathrm{KMnO}_4}{\text { number of electrons gained }}=\frac{158.1}{5}=31.6\)

In a neutral medium, KMn04 Is reduced by a suitable reductant to Mn02.

Reaction: \(\mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O}+3 e \rightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{-}\)

∴ In a neutral medium, the equivalent mass of KMn04 Is reduced by a suitable reductant to Mn02.

In a strongly alkaline medium, KMn04 is reduced to K2Mn04.

Reaction: \(\mathrm{MnO}_4^{-}+2 \mathrm{H}_2 \mathrm{O}+3 e \rightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{-}\)

Therefore In a neutral medium, the equivalent mass of KMn04

⇒ \(=\frac{158.1}{3}=52.7\)

In a strongly alkaline medium, KMn04 is reduced to K2Mn04.

Reaction: \(\mathrm{MnO}_4^{-}+e \rightarrow \mathrm{MnO}_4^{2-}\)

In strongly alkaline medium,

Equivalent mass of KMn04 \(=\frac{158}{1}=158\)

Equivalent mass of K2Cr207: K2Cr207 on reduction by a suitable reductant produces chromic salt (Cr3+).

Reaction: \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)

Equivalent mass of K2Cr207 \(=\frac{294.18}{6}=49.03\)

Equivalent Mass Of A Reductant

⇒ \(=\frac{\text { molecular or formula mass of the reductant }}{\text { number of electron(s) lost per molecule }}\)

Examples: In the presence of a suitable oxidizing agent, FeS04 gets oxidized to Fe2(S04)3.

Reaction: \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+e\)

Equivalent mass of FeS04 \(=\frac{151.85}{1}=151.85\)

In the presence of a suitable oxidizing agent, oxalic acid (C2H204 2H20) is oxidized to C02.

Reaction: \(\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{CO}_2+2 e\)

Here, the molecular mass of C2H204.2H20 = 126 and number of electrons lost = 2

∴ Equivalent mass of oxalic acid \(=\frac{126}{2}=63\)

Oxidation Mass of oxidant \(=\frac{\text { molecular or formula mass of the oxidant }}{\text { change in oxidation number }}\)

Examples: In the presence of d .H2S04, KMnCO4 Is reduced to manganous sulplutto (MnS04).

∴ \(\stackrel{+7}{\mathrm{KMnO}_4 \rightarrow} \stackrel{+2}{\mathrm{MnSO}_4}\)

Here, change In oxidation number of Mn in KMn04 = 7-2 = 5 units and molecular mass or formula mass of KMn04 =150.

Equivalent mass of KMn04 \(=\frac{158}{5}=31.6\)

In presence of dil. H2S04, K2Cr207 is reduced by a suitable reductant to Cr2(S04)3.

\(\mathrm{K}_2 \stackrel{+6}{\mathrm{Cr}_2 \mathrm{O}_7} \stackrel{+3}{\mathrm{Cr}_2}\left(\mathrm{SO}_4\right)_3\)

Here, change in oxidation number of 2 Cr -atoms in K2Cr2O7 =2x(+6)-2x(+3)=6 units.

Since the Equivalent mass of k2crO7 = 294/6=49

Equivalent mass of a reductant \(=\frac{\text { molecular or formula mass of the reductant }}{\text { change in oxidation number }}\)

Examples: 1 FeS04 is oxidized by an oxidant to Fe2(S04)3.

⇒ \(\stackrel{+2}{\mathrm{FeSO}_4} \stackrel{+3}{\rightarrow} \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\)

Here, the change in oxidation number of Fe in FeS04 = 3-2 = 1 unit

∴ Equivalent mass of FeS04 \(=\frac{151.85}{1}=151.85\)

Sodium thiosulphate (Na2S203-5H20) is oxidized by an oxidant to sodium tetrathionate (Na2S406)

⇒ \(2 \mathrm{Na}_2 \stackrel{+2}{\mathrm{~S}_2} \mathrm{O}_3 \rightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{Na}^{+}\)

Here, changein oxidation number of4 S-atoms in Na2S203 = 4 X (+2.5)- 4 x (+2) = 18- 8 = 2 units.

∴ Changes in oxidation number for two S-atoms =1 unit Now, molecular mass of Na2S203.5H20 = 248

Equivalent mass of Na2S203-5H20 \(=\frac{248}{1}=248\)

In any chemical reaction, the number of equivalent or gram-equi-valent of reactants and products are always equal: In the chemical reaction, the reacting substances unite together in accordance with their equivalent masses. By analyzing the chemical reaction, it has been observed that the number of equivalents or gram-equivalents of constant(s) and produces) are always equal.

Example 1 \(\begin{aligned}
& \text { (1) } \underset{56 \mathrm{~g}}{\mathrm{Fe}}+\underset{2 \times 36.5 \mathrm{~g}}{2 \mathrm{HCl}} \longrightarrow \underset{129 \mathrm{~g}}{\mathrm{FeCl}_2}+\underset{2 \mathrm{~g}}{\mathrm{H}_2} \\
& 2 \text { gram-eqv. } 2 \text { gram-eqv. } 2 \text { gram-eqv. } 2 \text { gram-eqv. } \\
&
\end{aligned}\)

Generally, if in the case of the reaction, A + B-+C +D, the gram-equivalent of A reacts with .v gram-equivalent of B then, the gram-equivalent of C and .v gram-equivalent of D will be produced.

Methods for the determination of the atomic mass of an element:

Determination of atomic mass by Dulong and Petit’s law: At ordinary temperature, the atomic heat of all the solid elements (except C, B, Si, and Be) is always the same and numerically equal to 6.4 (approximately). The product of atomic mass and specific heat is known as atomic heat.

Hence, atomicmass x specific heat = 6.4 (approx) Approximate atomic mass \(=\frac{6.4}{\text { specific heat }}\)

The approximate atomic mass so obtained is divided by the equivalent mass ofthe element to get its valency. Since the atomic mass is not accurate, the value of valency may be fractional.

So, the nearest whole number should be considered as its valency which on multiplication with equivalent mass gives accurate atomic mass.

Determination of atomic mass of an element by applying Mitscherlich’s law of isomorphism: Let, a salt of an element E be isomorphous with K2S04, and the salt contains a % of E.

The atomic mass of E is to be determined. Let, the atomic mass of E be r. Since the salt of E is isomorphous with K2S04, the formula of the salt, according to Mitscherlich’s law, would be K2E04. Now, the molecular mass of K2E04

= 2×39+x + 4x 16 = 78+X + 64 = 142 + x

∴ Percentage of E in the salt \(=\frac{x}{142+x} \times 100\)

By the above assumption \(\frac{100 x}{142+x}=a\)

As ‘a’ is known, x can easily be calculated.

An alternative method (hy Isomorphous replacement): Experimental results show that different key dementia (l.a, the elements which are different) forming an Isomorphous compound can replace each other, atom for atom, without changing the crystal structure.

Let A and li be the key elements (having atomic masses a and respectively) In a pair of isomorphous compounds.

If W g of 1 is replaced hy W2 g of, then according to the law oflsomorphlsm.

⇒ \(W_1 / a=W_2 / b \text { or, } W_1 / W_2=a / b\)

⇒ \(\text { i.e., } \frac{\text { mass of the substituted element, } A}{\text { mass of the substituting element, } B}=\frac{\text { atomic mass of } A}{\text { atomic mass of } B}\)

This relation can be used to determine the atomic mass of A if that of B is known or vice versa.

Two compounds having the following characteristics are said to be isomorphous crystals:

  1. Both have similar external crystalline structures.
  2. They together can form mixed crystals.
  3. One can form overgrowth on the other if a small crystal of the latter is placed in the saturated solution of the former.

The property by virtue of which isomorphous crystals are formed is called isomorphism (iso: similar, morph: form).

Examples of isomorphous crystals:

White vitriol (FeS04-7H20), Epsom salt (MgS04- 7H20), green vitriol (FeS04-7H20)

Potassium permanganate (KMn04), potassium perchlorate (KC104)

Potash alum [K2S04-A12(S04)3-24H20]; chrome alum [K2S04-Cr2(S04)3-24H20]

It is observed from the formulae of isomorphous crystals that the total number of atoms in their molecules is the same.

Mitscherlich’s law of isomorphism: The same number of atoms combine in the same way to produce isomorphous crystals.

The crystalline form of these crystals depends only on the number of atoms in their molecules and the way by which they combine but is independent of their chemical properties.

Question 1. The atomic mass & equivalent weight of an element are 27 & 9 respectively. Find the formula of its chloride
Answer: Valency of element \(=\frac{\text { atomic mass }}{\text { equivalent weight }}=\frac{27}{9}=3\)

∴ Formula of its chloride = MC13 [since valency of Cl =1]

Question 2. x gram of an element forms y gram of its chloride. Calculate the equivalent weight of the element.
Answer: Mass ofthe chloride =y g and mass of element = x g

Mass of chlorine =(y- x) g

So, (y- x) g of chlorine combines with x g of the element.

∴ 35.5 g ofchlorine combines with \(\frac{35.5 \times x}{y-x} g\)

∴ Equivalent weight ofthe element \(=\frac{35.5 x}{y-x}\)

Example 3. Calculate the relative equivalent weight (£) of copper in cuprous oxide.
Answer: Formula of cuprous oxide = Cu20

∴ 16 g of oxygen combined with 2 x 63.5 g of Cu

So, 8 g ofoxygen combines with 63.5 g of Cu

∴ Equivalent weight (E) = 63.5

Example 4. A metallic oxide contains 60% metal. Calculate the equivalent weight of the metal.
Answer: 100 g of metallic oxide contains 60 g of metal. 100 g of metallic oxide contains (100 – 60) = 40 g of oxygen So, 40 g of oxygen combines with 60 g of metal.

8 g of oxygen combines with \(\frac{60 \times 8}{40}\)

Therefore, the equivalent weight of the metal = 12

Question 5. 0.56 g of metallic oxide contains 0.16 g of oxygen. Determine the equivalent weight of that metal.
Answer: Mass of metallic oxide = 0.56 g and mass of oxygen = 0.16 g

So, mass ofmetalin the metallic oxide = (0.56- 0.16) g = 0.40 g

Equivalent weight of the metal

⇒ \(=\frac{\text { mass of the metal }}{\text { mass of combined oxygen }} \times 8=\frac{0.40 \times 8}{0.16}=20\)

Question 6. Determine the equivalent weight of O carbonate radical Ferrous Sulphate [Fe = 56]
Answer: Formula mass of carbonate radical =12 + 3×16 = 60

Therefore Equivalent weight of carbonate radical

⇒ \(=\frac{\text { formula mass }}{\text { valency }}=\frac{60}{2}=30\)

Formula mass of ferrous sulphate =56 + 32 + 64 = 152

Therefore Equivalent weight formula of ferrous mass sulphate

⇒ \(\begin{aligned}
& =\frac{\text { formula mass }}{\text { total valency of cation or anion per molecule }} \\
& =\frac{152}{2}=76
\end{aligned}\)

Question 7. When 0.3 g of a metal is dissolved in dilute IIC1 the volume of H2 gas liberated is 1 10mL at 17°C and 755 35.5 g ofchlorine combined with g ofthe element Hg of pressure. [Aqueous tension at 17°C = 14.4 Hg] Determine the equivalent mass of the metal.
Answer: Pressure of dry hydrogen gas= 755.14.4

If the volume of H2 gas produced at STP is VmL, then

⇒ \(\frac{110 \times 740.6}{(273+17)}=\frac{V \times 760}{273}\)

So, \(V=\frac{110 \times 740.6 \times 273}{290 \times 760} \mathrm{~mL}=100.91 \mathrm{~mL}\)

Now,mass of22400 mL of H2 at STP = 2x 1.008g

Mass of100.91 mL of H,2 at STP \(=\frac{2 \times 1.008 \times 100.91}{22400} \mathrm{~g}\)

=0.00908g

So, 0.00908 g of H2 is displaced by 0.3 g of metal.

1.008 g of H2 is displaced by \(\frac{0.3 \times 1.008}{0.00908}\) =33.3 g of mental

Therefore Equivalent mass of the mental=33.3

Question 8. The equivalent mass of a metal is 11.6. When 0.177g of that metal is allowed to react completely with dilute HC1, what will be the volume of H2 gas liberated at 12°C and 766mm Hg pressure?
Answer: The equivalent mass ofthe metal is 11.6.

∴ 11.6 g metal displaces 11200 mL of H2 at STP

0.177 g metal displaces \(\frac{11200 \times 0.177}{11.6}\)=170.89 ml of H2

Let volume by V at 12°C and 760 mm Hg pressure

∴ \(\frac{V \times 766}{(12+273)}=\frac{170.89 \times 760}{273}\)

Or, \(V=\frac{170.89 \times 760}{273} \times \frac{285}{766}=177.0 \mathrm{~mL}\)

Question 9. 20 g of a metal reacts with dilute H2S04 to liberate 0.504 g of H2 gas. Calculate the amount of metal oxide formed from 2.0 g of the metal.
Answer: 0.504 g of H2 is liberated by 20 g of the metal

∴ 1.008 g H2 is liberated by \(\frac{20 \times 1.008}{0.504}\)

So, the equivalent mass ofthe metal = 40

∴ 40 g of metal combined with 8 g of oxygen

So, 2.0 g metal combines with \(=\frac{8 \times 2}{40} g=\) 0.4 of oxygen

∴ Amount of metal oxide = mass of metal + mass of oxygen

= (2.0 4-0.4) = 2.4 g

Question 10. 3.26 g of zinc reacts with acid to liberate 1.12L of hydrogen gas (H2) at STP. Calculate the relative equivalent weight of zinc.
Answer: 1-12L of H2 (at STP) is liberated by 3.26 g of zinc, therefore, 11.2L Of H2 is Liberated By \(\frac{3.26 \times 11.2}{1.12}\) =32.6g of Zinc.

So, the equivalent weight of zinc = 32.6

Question 11. 15 g of an element reacts completely with 30 g of We know, A =Ex V A = 9.02 x V another element. Calculate the specific equivalent weight of A if that of B is 60.
Answer: We know that two elements A and B will react with each other in the ratio of their equivalent weight (EA and EB).

So, in the reaction between A and B Or, 15/30 = Ea/Eb

∴ Ea=1/2 X Eb =1/2 X 60 =30

So, the equivalent weight of A = 30

Question 12. 0.362g of metal is added to an aqueous solution of AgNOs. Consequently, 3.225 g of silver is precipitated. What is the equivalent mass of the metal? [Atomic mass of Ag = 108, valency = 1]
Answer: Equivalent mass of Ag \(=\frac{\text { mass of the metal }}{\text { mass of } \mathrm{Ag}}\) =108/1 =108

Now, the displacement of Ag by the metal from a compound of the metal occurs in proportion to their equivalent masses.

So, \(\frac{\text { equivalent mass of the metal }}{\text { equivalent mass of } \mathrm{Ag}}=\frac{\text { mass of the metal }}{\text { mass of } \mathrm{Ag}}\)

Let the equivalent mass of the metal be E

According to the problem \(\frac{E}{108}=\frac{0.362}{3.225}\)

Or, E=12.12

∴ Equivalent mass ofthe metal =12.12

Question 13. A metallic oxide contains 53% neutral. The vapor density of the chloride of the metal is 66. Find the atomic weight of the metal.
Answer: The metallic oxide contains 53% metal.

∴ The metallic oxide contains (100-53) = 47 % oxygen.

Hence, the equivalent weight (£) of the metal.

⇒\(=\frac{\text { mass of the metal }}{\text { mass of oxygen combined }} \times 8=\frac{53 \times 8}{47}=9.02\)

As given, vapour density of the metallic chloride = 66 Molecular weight ofthe chloride = 2 x 66 =132

since M=2xD

Let, the molecular formula of the metallic chloride be MCIv [where, V = valency of the metal]

∴ Molecular weight of MClv = A 4- 35.5 V [where, A = atomic weight of the metal]

We know, A =Ex V A = 9.02 x V

Hence, 9.02 X V4- 35.5 x V = 132 or, V \(=\frac{132}{44.52} \approx 3\)

So, atomic weight ofthe metal =ExV = 9.02 X 3 = 27.06

Question 14. A metallic chloride contains 20.2% by mass of metal (M). If the atomic mass of the metal is 27. What is the molecular formula of the metallic chloride?
Answer: Metal content in metallic chloride = 20.2 %

∴ Amount ofchlorine in the metallic chloride=(100- 20.2) = 79.8%

So, 79.8 parts by mass of chlorine combined with 20.2 parts by mass of the metal (M)

∴ 35.5 parts by mass of chlorine combined with \(\frac{20.2 \times 35.5}{79.8}\) = 8.986 Parts by mass of the mental.

Hence, the Equivalent mass of mental (M)=8.986

\(\quad V=\frac{A}{E}=\frac{27}{8.986}=3 \text { (approx.) }\)

So, the valency of the metal in the metallic chloride = 3

∴ The molecular formula of the metallic chloride = MC13

[Therefore Valency of chlorine =1]

Question 15. 8.08g of metallic oxide on being reduced by H2, produces 1.8g of water. Find the quantity of Oz In the above oxide and the equivalent mass of the metal.
Answer: Amount of oxygen present in 18 g of water = 16 g

Amount of oxygen present 1.8 g of water = 1.6 g

Oxygen content in 8.08 g ofthe metallic oxide = 1.6 g

[since all of this oxygen comes from the metallic oxide]

Hence, the amount ofthe metal present in that oxide =(8.08-1.6)=6.48g

∴ 1.6g of O2 COmbine with 6.48g of mental

Or, 8g Of O2 Combine with \(\left(\frac{6.48}{1.6} \times 8\right)\) g of the mental

Therefore, the equivalent mass of the metal = \(=\frac{6.48}{1.6} \times 8=32.4\)

Question 16. In the following reaction, determine the equivalent
weight of H3P04.
\(\begin{array}{r} \mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_3 \mathrm{PO}_4 \rightarrow \mathrm{CaHPO}_4+2 \mathrm{H}_2 \mathrm{O} \\{[\mathrm{Ca}=40 ; \mathrm{P}=31] \text { [II 82] }} \end{array}\)
Answer: In this acid-base neutralization reaction, 2 hydrogen atoms of H3P04 are replaced by atoms of calcium.

Equivalent weight of H3P04

⇒ \(\begin{aligned}
& =\frac{\text { molecular mass of } \mathrm{H}_3 \mathrm{PO}_4}{\text { number of displaced hydrogen atoms }} \\
& =\frac{3 \times 1.008+31+4 \times 16}{2}=49.012
\end{aligned}\)

Question 17. A metallic bromide weighing 1.878 g is heated in a current of hydrogen chloride. Consequently, 1.0 g of metallic chloride is obtained. What is the equivalent mass of the metal?
Answer:

Let the equivalent mass ofthe metal be the Equivalent mass ofthe metallic bromide =E+ 80

[∴ Equivalentmass ofbromine = 80 ] Equivalent mass ofmetallic chloride =E + 35.5

[∴ Equivalent mass ofchlorine = 35.5]

So, the number of gram-equivalent of metallic bromide and metallic chloride is \(\frac{1.878}{E+80} \text { and } \frac{1.0}{E+35.5}\) Respectively.

Now, in any chemical reaction, the gram-equivalents of the reactant and the product are equal.

So, in the reaction, the gram-equivalent of metallic bromide is equal to the gram-equivalent of metallic chloride.

∴ \(\frac{1.878}{E+80}=\frac{1.0}{E+35.5}\)

or, E(1.878-1) = 80-35.5X 1.78 =80-66.669

or, 0.878 x £ = 13.331 or, E= 15.18

∴ Equivalent Mass of mental = 15.18

Question 18. The hydride of element A contains 25% of hydrogen by mass. The percentages of oxygen in two oxides of that element are 57.1 and 72.7 respectively. If the atomic mass of the element is 12 determine the formula of the hydride and the two oxides.
Answer: The quantity of the element A in the hydride of element
A =(100-25)% = 75%

So, in the hydride of element A, 25 parts by mass ofhydrogen combine with 75 parts by mass element A.

∴ 1.008 parts by mass of hydrogen combined with

75/25 X 1.008 = 3.024 Parts by mass of element A

∴ Equivalent mass of ‘A’ anhydride = 3.024
∴ Quantity of A in the first oxide = (100- 57.1) % = 42.9 %

and in the second oxide = (100- 72.7) % = 27.3 %

So, in the first oxide, 57.1 parts by mass ofoxygen combine with

42.9 parts by mass of A

In the first oxide, 8 parts by mass of oxygen combine with

⇒ \(\frac{42.9}{57.1} \times 8\)=60.1 parts of mass of A.

Similarly, in the second oxide, 8 parts by mass of oxygen react with \(\frac{27.3 \times 8}{72.7}=\) = 3 parts by mass of A.

∴ The valency of A in the first hydride.

⇒ \(=\frac{\text { atomic mass of } \mathrm{A}}{\text { equivalent mass of } \mathrm{A} \text { in the hydride }}=\frac{12}{3.024}=4\)

Formula of the hydride = AH4

Valency of A in the first oxide

⇒ \(=\frac{\text { atomic mass of } \mathrm{A}}{\text { equivalent mass of } \mathrm{A} \text { in the first oxide }}\)

=12/6.01 =2

Formula of first oxide = AO [Therefore Valency of oxygen =2]

Valency of A in the second oxide

⇒ \(=\frac{\text { atomic mass of } \mathrm{A}}{\text { equivalent mass of } \mathrm{A} \text { in the second oxide }}\)

=12/3=4

therefore Formula ofthe second oxide = AO2

Question 19. The specific equivalent weight of a solid element is 17.8 and its specific heat is 0.124 cal .K-1. g-1. Find its valency and correct specific atomic mass.
Answer: As perDoulong and Petit’s law, the approximate atomic mass
ofthe solid element = \(\frac{6.4}{\text { specific heat }}=\frac{6.4}{0.124}=51.61\)

therefore Valency of element = \(=\frac{\text { atomic mass (approx) }}{\text { equivalent weight }}=\frac{51.61}{17.8} \approx 3\)

Therefore Correct atomic mass ofthe element

= equivalent weight X valency = 17.8 X 3 = 53.4

Question 20. The specific heat of magnesium is 0.262 and magnesium chloride contains 25.5% by weight of magnesium. Determine the atomic mass, valency of magnesium, and also the formula of magnesium chloride.
Answer: Percentage of magnesium chloride= 25.5

∴ Percentage of chlorine in the salt = 74.5

∴ Equivalent weight of magnesium

⇒ \(=\frac{\text { mass of magnesium }}{\text { mass of chlorine }} \times 35.5\)

⇒ \(=\frac{25.5}{74.5} \times 35.5=12.15\)

According to Dulong and Petit’s law,

Atomic mass ofthe metal x specific heat = 6.4 (approx)

So, approximate atomic mass of magnesium \(=\frac{6.4}{0.62}=24.42\)

∴ Valency of Mg \(=\frac{\text { approximate atomic mass }}{\text { equivalent weight }}\)

Therefore, the atomic mass of magnesium = equivalent weight x valency = 12.15×2 = 24.30, and the formula of magnesium chloride is MgCl2.

Question 21. An aqueous solution contains 0.22 g of metallic chloride. 0.51 g of AgNOa is required for the complete precipitation of chloride from that solution. If the specific heat of the metal is 0.057 then what will be the correct atomic mass of that metal? What is the formula of that metallic chloride?
Answer: S. Let the equivalent mass ofthe metal be E.
Equivalent mass of the metallic chloride =E + 35.5 [ v Equivalent mass ofchlorine = 35.5 ]

Equivalentmass of AgN03

= Equivalent mass of Ag+ + Equivalent mass of N03

= 108 + 62 =170.

In any chemical reaction, the reactants combine together in the proportion of their equivalent masses.

⇒ \(\text { So, } \quad \frac{E+35.5}{170}=\frac{0.22}{0.51} \quad \text { or, } E=37.83\)

Determination of atomic mass ofthe metal: According to Dulong and Petit’s law, the approximate atomic mass of metal

⇒ \(=\frac{6.4}{\text { specific heat }}=\frac{6.4}{0.057}=112.28\)

Therefore Valency of the metal \(=\frac{\text { approx. atomic mass of the metal }}{\text { equivalent mass }}\)

⇒ \(=\frac{112.28}{37.83}=2.968 \approx 3\)

Therefore, the correct atomic mass of M =37.83×3 = 113.49

So, the molecular formula of chloride salt = MC13.

Question 22. The percentage by weight of chromium in green-colored chromium oxide (chromic oxide) is 68.43 and it is isomorphous with ferric oxide. Estimate the = 74.5— X35.5 = 12.15 correct atomic weight of chromium.
Answer: Chromium oxide is isomorphous with Fe203. So, according to Mitscherlich’s law of isomorphism, the formula chromium oxide will be Cr203.

Since the valency of Fe in Fe203 is 3 the valency of Cr in Cr203 will also be 3.

Now, the quantity of chromium in chromium oxide = 68.43 % Quantity of oxygen =(100-68.43) = 31.57 % Equivalent weight of chromium

⇒ \(=\frac{\text { weight of chromium }}{\text { weight of combined oxygen }} \times 8=\frac{68.43}{31.57} \times 8=17.34\)

Hence, the correct atomic weight chromium

= equivalent weight x valency = 17.34 x 3 = 52.02

Question 23. Potassium manganate and potassium chromate (K2Cr04) Percentage of manganese by mass present in potassium manganate is 27.86. Determine the atomic mass of manganese.
Answer: Potassium chromate (K2Cr04) is isomorphous with potassium manganate. So according to Mitscherlich’s law of isomorphism, the formula of potassium manganate should be K2Mn04.

Suppose, the atomic mass of Mn = a

Molecular mass of K2Mn04 = 2×39 + a + 4xl6 = a + 142

Therefore Percentage of Mn in K2Mn04 \(=\frac{a}{a+142} \times 100\)

As given in the question \(\frac{a \times 100}{a+142}=\)

Therefore Atomic mass of manganese = 54.84

Question 24. An element (X) reacts with KOH to form a salt. The salt is isomorphous with potassium permanganate (KMn04). The oxide of the element X contains 61.2% of oxygen. Determine the formula of the oxide and atomic mass of X.
Answer: Percentage of oxygen in the oxide of X = 61.2

Therefore Percentage of X in the oxide of X = (100- 61.2) = 38.8

Therefore \(\text { Equivalent mass of } \mathrm{X}=\frac{\text { mass of } \mathrm{X}}{\text { mass of combined oxygen }} \times 8\)

⇒ \(=\frac{38.8}{61.2} \times 8=5.07\)

In reaction with KOH, the element X produces a salt that is isomorphous with KMn04. Since the valency of Mn in KMn04 is 7, according to Mitscherlich’s law, a formula of the salt produced will be KX04 where the valency of X is 7.

Therefore Atomic mass of X = equivalent mass x valency

∴ = 5.07X7 = 35.49.

∴ The formula of the oxide would be X207

Question 25. Oxides of two metals A and B are isomorphous. The atomic weight of B is 43.5 and the vapor density of its chloride is 75. The amount of oxygen present in 1.02g ofthe oxide of A is 0.48g. Determine the atomic weight of A.
Answer: Determination of equivalent weight of A: Quantity of oxygen present 1.02g of the oxide of A = 0.48 g

So, the quantity of A in that oxide = (1.02- 0.48) = 0.54 g

Equivalent weight of the metal A

\(=\frac{\text { mass of } \mathrm{A} \times 8}{\text { mass of combined oxygen }}=\frac{0.54 \times 8}{0.48}=9\)

Determination of valency of B: Vapour density ofthe chloride metal B = 75

Molecular mass ofthe chloride of metal B = 2 x 75 = 150 If the valency of B is x, then the formula of the chloride

of B = BC1X

Molecular mass ofthe compound BC1X = 43.5 + 35.5 X x

Therefore 43.5 + 35.5 X x = 150 or, x = 3

Therefore Valency of B = 3 and the formula of its oxide is B203

Determination of valency of A: Oxides of A and B are isomorphous. So according to the law of isomorphism, the formula ofthe oxide of A is A203 and the valency of A is 3. Hence, the atomic mass of A.

= equivalent weight x valency = 9×3 = 27

Quantity of water associated with 1 11 g or 1 mol of the anhydrous salt \(=\frac{1.48 \times 111}{1.52}=108.07 \mathrm{~g}\)

Therefore, the quantity of water associated with 1 mol of anhydrous CaCl2 is 108.07g.

Percentage Composition Empirical And Molecular Formula

Percentage Composition

The percentage composition of a compound means the parts by mass of each of the constituent elements in every 100 parts by mass of that compound.

Chemical analysis of water reveals that in every 100 parts by mass of water, 11.1 parts by mass of hydrogen, and 88.9 parts M1 Gram-formula mass of Ca3(P04)2 by mass of oxygen is present the percentage composition of water is: hydrogen (H) = 11.1 % and oxygen (O) = 88.9%.

Hence, it is clear that the percentage composition of the = [3 X 40 + 2(31 + 64)] = 310g
Ca3(P04)2 can be regarded as constituent elements in a compound independent of the Gram-molecular mass of P205 = (2 X 31 + 5 x 16) = 142g quantity of the compound taken.

For example, irrespective of the quantity of water taken % of P2O5 in calcium phosphate = x 100 = 45.8 (say 4g or 18g or 50g, etc.), the percentage of hydrogen and
oxygen by mass will always be 11.1 and 88.9 respectively.

Percentage composition from the molecular formula:

  1. The molecular mass or formula mass of the compound is first calculated from its molecular formula.
  2. The mass of each element or radical present in 1 gram ofthe compound is then calculated separately.
  3. Finally, the percentage of the mass of each element present in the compound is computed separately.

% of an element or radical present in the compound

\(\begin{aligned}
& \text { mass of the element or radical in } 1 \text { gram-mole } \\
& =\frac{\text { of the compound }(\mathrm{g})}{\text { gram-molecular mass or gram-formula mass }} \times 100 \\
& \text { of the compound (g) } \\
&
\end{aligned}\)

Question 1. 3 g of hydrated calcium chloride yields 1.52 g of the anhydrous salt on heating. What is the percentage of water present in the hydrated salt? Find the quantity of water associated per mole ofthe anhydrous salt.
Answer: Amount of water present in 3g of hydrated calcium chloride =(3- 1.52) =1.48g

∴ Percentage of water in hydrated salt= \(\frac{1.48}{3} \times 100\) 100 = 49.33 

Gram-formula mass of any. CaCl2 =(40 + 2 x 35.5) = 392 = 111g

Quantity of water associated with 1. 52g anhydrous =1.48g

∴ Quantity of water associated with 1 11 g or 1 mol of the anhydrous salt \(=\frac{1.48 \times 111}{1.52}=108.07 \mathrm{~g}\)

∴ The quality of water associated with 1 month 1 mol of anhydrous CaCl2 is 108.07g

Question 2. What is the percentage of P205 in calcium phosphate [Ca3(PO4)2]?
Answer: Gram-formula mass of Ca3(P04)2 = [3 X 40 + 2(31 + 64)] = 310g

Ca3(P04)2 can be regarded as [3CaO P205]

Gram-molecular mass of P205 = (2 X 31 + 5 x 16) = 142g

∴ % of P2O5 in calcium phosphate \(=\frac{142}{310} \times 100=45.8\)

Question 3. A compound contains 28% of nitrogen and 72% of metal mass. In the compound, 3 atoms of the metal remain combined with two atoms of nitrogen. Find the atomic mass of the metal.
Answer: The formula of the compound is M3N2. [metal =M]

∴ Molar mass of M3N2 =3a + 28 [a = atomic mass ofM]

So, percentage of metal in the compound \(=\frac{3 a}{3 a+28} \times 100\)

∴ \(\left(\frac{3 a}{3 a+28}\right) \times 100=72 \text {, or } a=24\)

Therefore, the atomic mass of the metal = 24

Question 4. Give the percentages of ammonium and sulfate radicals in Mohr salt [(NH4)2S04 -FeS04 -6H20 ].
Answer: Molecular mass of Mohr salt = (2 X 18) + 96 + 56 + 96 + (6 x 18) =392

In each mole of Mohr salt, the amount of ammonium

⇒\(\left(\mathrm{NH}_4^{\oplus}\right)\)

radical=2×18 =36g and amount of sulphate \(\left(\mathrm{SO}_4^{2-}\right)\) radical

= 2×96 = 192g

∴ Percentage of ammonium \(\left(\mathrm{NH}_4^{\oplus}\right)\) radical in mohar salt.

\(=\frac{36 \times 100}{392}=9.18\)

Percentage of sulphate \(\left(\mathrm{SO}_4^{2-}\right)\) radical in Mohr Salt \(=\frac{192 \times 100}{392}=48.98\)

Question 5. Haematite (Fe203) contains some water in addition to Fe203. 4.0 kg of this mineral contains 2.5 kg of iron. Find the purity of Fe2Os in the mineral.
Answer: Gram-molecular mass of Fe203 =2×55.85 + 3×16 =159.7g

Now, (2 x 55.85) g Fe= 159.7 g Fe2O3

∴ 2500g Fe = \(\frac{159.7 \times 2500}{2 \times 55.85} \mathrm{~g} \mathrm{Fe}_2 \mathrm{O}_3=3574 \mathrm{~g} \mathrm{Fe}_2 \mathrm{O}_3\)

So, 4000 g mineral contains 3574 g of pure Fe203.

∴ Purity of Fe203 in haematite \(=\frac{3574 \times 100}{4000}=89.35 \%\)

Empirical And Molecular Formula

Empirical formula: The empirical formula of a compound indicates the number of atoms of different elements present in a molecule of the compound, expressed in a simple whole number ratio.

The empirical formula merely indicates the ratio of the atoms of elements constituting the molecule of a compound. Thus, this formula may not represent the actual number of atoms in the molecule of a compound.

Example: The empirical formula of glucose is CH2O. This shows that in the molecule of glucose, C, H, and O-atoms are present in the simplest ratio of 1:2: 1.

Molecular formula: The molecular formula of a compound indicates the actual number of atoms of various constituent elements present in a molecule of the compound.

The molecular formula represents the actual formula of the molecule of a compound as it gives the actual number of atoms constituting the molecule.

Example: The molecular formula of glucose is C6H1206. This shows that a molecule of glucose consists of six C-atoms, twelve H-atoms, and six O-atoms.

However, in certain cases, the empirical formula and the molecular formula may be identical. For example, formaldehyde has both empirical and molecular formula CH20.

Relation between empirical formula and molecular formula: The molecular formula of a compound is a whole number multiple ofthe empirical formula of that compound.

Molecular formula of a compound = (empirical formula)n

[where n = 1, 2, 3, etc. are simple whole numbers]

Example: The empirical formula of glucose is CH20 while its molecular formula is C6H1206 or CH12O6 When n — 1, both the empirical formula and the molecular formula will be the same.

In the case of formaldehyde, the empirical formula i.e., CH20 and the molecular formula [i.e., CH20 (HCHO)] are the same.

The sum of the atomic masses of the constituent elements present in the molecular formula of a compound indicates the molecular mass of that compound.

On the other hand, the sum of the atomic masses of the constituent elements present in the empirical formula of a compound indicates the empirical formula mass of that compound.

The molecular formula of a compound is either equal to or is a simple multiple of its empirical formula mass Molecular mass of a compound =n x empirical formula mass.

Now, if n is 1, i.e., molecular mass = empirical formula mass of a compound, then the empirical formula and the molecular formula ofthe compound would be identical.

Basis of determination of the empirical formula of a compound: Let the molecular mass of a compound, composed of two elements A and B, is M and the percentage composition by mass of A and B in that compound are x and y respectively.

Therefore Amount of A in M gram i.e., 1 gram-mole compound \(=\frac{x \times M}{100}\) gram. Similarly, the amount of B in Mgram i.e., 1 gram-mole ofthe compound gram.

Let, the atomic masses of A and B be a and b respectively.

∴ Number of gram-atoms of A in 1 gram-mole of the \(=\frac{x \times M}{100 \times a} .\)atoms of B in1 gram-mole of the compound \(=\frac{x \times M}{100 \times a}: \frac{y \times M}{100 \times b}=\frac{x}{a}: \frac{y}{b}\)

In 1 molecule of the compound, the ratio of the number of atoms of A and B

Hence, in a molecule of the compound, the ratio of the number of atoms of A to the number of atoms of B \(=\frac{\text { mass of } \mathrm{A} \text { in the compound }(\%)}{\text { atomic mass of } \mathrm{A}}: \frac{\text { mass of } \mathrm{B} \text { in the compound }(\%)}{\text { atomic mass of } \mathrm{B}}\)

Similarly, the number of grams of A in the compound(%). mass of B in the compound(%) atomic mass of A atomic mass of B

Therefore, the ratio of the quotients, obtained by dividing the percentage mass of each constituent element by its corresponding atomic mass, gives the ratio of the atoms of the elements present in a molecule of that compound. From this ratio, the empirical formula can be determined.

Determination of the empirical formula of a compound: The empirical formula ofany compound can be determined the following steps

The percentage by mass of each element in the compound is accurately evaluated by suitable methods.

The percentage by mass of each element is divided by its atomic mass in order to get the relative number of different types of atoms present in the molecule.

Each of the numbers obtained is divided by the smallest of these numbers to get the simplest ratio of atoms.

In determining the ‘simplest ratio of atoms; the ‘rule of approximation’ is followed. For example, if the quotient is 2.99 or 4.01, then the nearest whole numbers, i.e., 3 or 4 respectively are accepted as the required quotient.

But when the ‘rule approximation’ cannot be applied, e.g., if any quotient is 1.5 or 1.6, each of the quotients obtained is to be multiplied with a suitable factor so as to convert all the quotients into lowest whole numbers.

The ratio, thus obtained, expresses the ratio of the number of atoms of the elements constituting a molecule and thus gives the empirical formula of the compound.

Example: In an acetic acid molecule, constituent atoms are C, H, and O. The ratio of the number of atoms is 1:2:1 as determined by the preceding steps. Thus, the empirical formula acetic acid will be CH20.

Method of determination of molecular formula:

  1. The empirical formula of a compound is first determined. by the method described above.
  2. The molecular mass of the compound is determined experimentally.
  3. In the case of volatile compounds, the molecular mass is evaluated using the equation, M = 2D, where D = experimentally determined value of the vapor density of the compound.
  4. [It is to be noted that the equation, M = 2D is applicable only in the case of those stable compounds that do not undergo dissociation or decomposition reaction in the vapor state.
  5. From the empirical formula of the compound, the empirical formula mass is calculated.
  6. The empirical formula mass of any compound is the sum of the atomic masses of atoms of different elements represented by the empirical formula.
  7. The molecular mass of the compound. n x empirical formula mass
  8. So, in order to evaluate ‘n ’, the molecular mass of the compound is to be divided by its empirical formula
  9. Finally, the empirical formula ofthe compound multiplied by n ’, gives the molecular formula of the compound.

Numerical Examples 

Question 1. A compound of carbon, hydrogen, and oxygen contains 40% of carbon and 6.67% of hydrogen. The vapor density of the compound, when vapourised, is 2.813 times the vapor density of oxygen. Determine the empirical formula and molecular formula of the compound.
Answer: 

In the compound, C = 40%, H = 6.67% and thus O = [100 -(40 + 6.67)]% =53.33%

Now, the ratio of the number of atoms of C: H:O in the compound, \(\mathrm{C}: \mathrm{H}: \mathrm{O}=\frac{40}{12}: \frac{6.67}{1}: \frac{53.33}{16}\)

Therefore Atomic masses of C, H, and O are 12, 1, and 16 respectively]

= 3.33: 6.67: 3.33 as 1: 2: 1

[dividing by the lowest number 3.33 ]

Therefore Empirical formula ofthe compound = CH20.

Molecular formula =(CH20) n [where n is an integer]

∴ Molecular mass ofthe compound=(12 + 2 + 16)n =30 n.

Again, the vapor density (D) of the compound

= vapour density of oxygen x 2.813 = 16 x 2.813 = 45.008

∴ The molecular mass of the compound =2 x D

= 2×45.008 =90.016

Hence, 30n= 90.016 i.e., n=3

∴ Molecular formula =(CH20)3 = C3Hg03.

Question 2. A gaseous hydrocarbon contains 75% carbon by moss, 100 cm3 of this gas at STP weighs 0.072 g. What is the molecular formula of the hydrocarbon? [Weight of litre hydrogen at STP = 0.09 g]
Answer: Hydrocarbon consists ofhydrogen and carbon only.

Let, in the hydrocarbon, the mass of carbon = 75 g The mass ofhydrogen = ( 100- 75) g = 25 g Ratio of the number of carbon (C) atoms and hydrogen (H) atoms in the compound, C: H \(=\frac{75}{12}: \frac{25}{1}\)

or, C: H =6.25: 25 =1:4

∴ The empirical formula of the hydrocarbon = CH4

∴ Its molecular formula = (CH4)n

Molecular mass of hydrocarbon = (12 + 4)n = 16n

Vapor density (D) of hydrocarbon

\(=\frac{\text { mass of } 100 \mathrm{~cm}^3 \text { hydrocarbon at STP }}{\text { mass of } 100 \mathrm{~cm}^3 \text { hydrogen at STP }}\) \(=\frac{0.072}{0.009}=8\)

Molecular mass hydrocarbon =2×0 =2×8 =16

So, 16n = 16 or, n = 1

Molecular formula ofthe hydrocarbon = (CH4)1 = CH4

Question 3. A compound on analysis gives the following percentage composition: K=31.83, CI= 28.98, and 0=39.19. Find the molecular formula of the compound if its molecular mass is 122.5
Answer: Empirical formula of the compound = KC103, molecular formula =(KC103)n

Molecular mass = n X (39 + 35.5 + 48) = 122.5 x n 122.5 x n = 122.5 or, n = 1

So, the molecular formula ofthe compound = KC103.

Question 4. 0.93 g of a compound containing C, H, and N, on C : H: N = burning, produces 2.64 g C02 and 0.63 g HzO. In another experiment, 0.186 g of that compound yields 24.62 cm3 of nitrogen at 1 atmospheric pressure at 27°C. Molecular weight ofthe compounds 93. What is its molecular formula?
Answer: Molecular masses: C02 =44 and H20 = 18

Quantity of carbon in 44g of C02 = 12 g

Quantity ofcarbon in 2.64g of C02= \(=\frac{12 \times 2.64}{44}=0.72 \mathrm{~g}\)

Again, quantity of H2 in 18g of H20 =2g

Quantity of H2 in 0.63g of H2O \(=\frac{2}{18} \times 0.63=0.07 \mathrm{~g}\)

Therefore, 0.93g ofthe compound contains 0.72g carbon and 0.07g hydrogen.

In another experiment, it was found that the volume of N2 produced at 1 atm pressure and 27°C temperature from 0.186g of that compound is 24.62 cm3.

Let, the volume ofthe gas at STP be V2 cm3.

As given in the question,

P1 = 1 atm, Kj = 24.62cm3, P2 = 1 atm

T1 = (273 + 27) = 300 K, T2 = 273 K, V2 = ?

we know \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

∴ \(V_2=\frac{P_1 V_1 T_2}{P_2 T_1}=\frac{1 \times 24.62 \times 273}{1 \times 300}\) = 22.4 cm3

1 mol or 22400 cm3 of nitrogen at STP weighs 28 g

∴ 22.4 cm3 of nitrogen at STP weighs \(\frac{28 \times 22.4}{22400}\) = 0.08g

So, the mass of nitrogen obtained from 0.186 g of the compound = 0.028 g

∴ Mass of nitrogen obtained from 0.93 g of compound

⇒ \(=\frac{0.028 \times 0.93}{0.186}=0.14 \mathrm{~g}\)

Hence, the mass of C, H, and N in 0.93 g of the compound is 0.72, 0.07, and 0.14 g respectively.

In the given compound,

Percentage of C by mass =(0.72/0.93) x 100 =77.42,

percentage of by mass = (0.07/0.93) X 100 = 7.52,

percentage ofN by mass = (0.14/0.93) x 100 = 15.05 .

The ratio of the number of atoms of C, H, and N,

⇒ \(C: H: N=\frac{77.42}{12}: \frac{7.52}{1}: \frac{15.05}{14}\)

= 6.451:7.52:1.075 =6:7:1 [dividing by the lowest number 1.075 ]

∴ Empirical formula compound = C6H7N

∴ Molecular formula =(C6H2N)n [where n is an integer]

∴ Molecular mass = (6 X 12 + 7 x 1 + 1 x 14)/t = 93 n

According to the given condition, 93n = 93 n = 1

Therefore Molecular formula of compound =(C6H7N)1 = CgH7N

Question 5. A compound composed of carbon, hydrogen, and chlorine contains C = 10.04% and Cl = 89.12%. The vapor density of the compound is 59.75. Determineitsmolecular formula.
Answer: Amount of hydrogen in the compound = 100- (10.04 + 89.12) = 0.84% Ratio of number of atoms of carbon (C), hydrogen (H) and chlorine (Cl) in the compound,

\(\begin{aligned}
\mathrm{C}: \mathrm{H}: \mathrm{Cl} & =\frac{10.04}{12}: \frac{0.84}{1}: \frac{89.12}{35.5} \\
& =0.836: 0.84: 2.510 \approx 1: 1: 3
\end{aligned}\) dividing by the lowest number 0.836 ]

∴ Empirical formula compound = CHC13

∴ Molecular formula ofthe compound = (CHC13)n

Molecular mass ofthe compound = (12 +l + 3x 35.5)n = 119.5n

Again, vapour density (D) ofthe compound =59.75

Molecular mass (Af) = 2 x D = 2 x 59.75 = 119.5

Hence, 119.5n = 119.5

∴ n = 1

Molecular formula of compound = (CHC13)1 = CHC13

Question 6. lg of phosphorus on combustion produces 1.77g oxide. What is the empirical formula of the compound? If the vapour density ofthe compound is 110, what is its molecular formula?
Answer: Quantity of phosphorus (P) in 1.77g of oxide =1 g The percentage-mass of phosphorus in its oxide

⇒ \(=\frac{1}{1.77} \times 100\) = 46.49 and that of oxygen in the oxide

= 100-56.49 =43.51

The ratio of the number of atoms of P and O in 1 molecule of the oxide
\(P: O=\frac{56.49}{31}: \frac{43.51}{16}\)

=1.822: 2.719

=1: 1.492 [dividing by the lowest number]

=2:3 [to get the lowest whole number, the

the ratio is multiplied by 2]

1.24 X 100

= 40%

∴ The empirical formula of compound = (P203)n, where n is an integer.

So, molecular mass of compound =(2 x 31 + 3 X 16)n = 110 n

Now, the vapour density (D) ofthe oxide = 110

Therefore The molecular mass (M) ofthe oxide = 2 X D

= 2 x 110 = 220

So, HOn = 220

∴ n =2

Thus, molecular formula ofthe compound = (P203)2 or P4Og

Question 7. The empirical formula of a gaseous compound is CH2C1.0.12 g ofthe compound occupies 37.20 cm3 at 105°C 12 1 16 Empirical formula of compound = CH20 temperature and 768 mmHg pressure. Find the molecular formula of the compound.
Answer: The volume of 0.12 g of the compound is 37.20 cm3 at 105°C and 768 mm pressure. Let, the volume of 0.12 g ofthe compound be V cm3 at STP.

⇒ \(\text { So, } \frac{768 \times 37.20}{(273+105)}=\frac{760 \times V}{273} \quad \quad V=27.15 \mathrm{~cm}^3\)

Again, mass of 27.15 cm3 of gaseous compound at STP = 0.12 g

Therefore Mass of 22400 cm3 i.e., 1 mol ofthe compound at STP

⇒ \(=\frac{0.12 \times 22400}{27.15}=99 \mathrm{~g}\)

[Since the volume of 1 gram-mole of gas at STP = 22400 cm3 ] The molecular mass ofthe compound = 99 Now, the empirical formula ofthe compound = CH2C1.

The molecular formula of the compound = (CH2Cl)n.

[where n is a whole number]

Its molecular mass = (12+1×2 + 35.5)n = 49.5n

So, 49.5n = 99 or, n = 2

Molecular formula ofthe compound =(CH2C1)2 = C2H4C12

Question 8. An organic compound contains C, H, and O as its constituents. On heating in the absence of air, 3.10 g of this compound, produces 1.24g of carbon. But if 0.5 g of the compound is burnt in the presence of air, 0.3 g of H2O is formed. 0.05 gram-mole of the compound contains 4.8 g of oxygen. What is the molecular formula ofthe compound?
Answer: 3.10 g of the compound when heated in the absence of air, produces 1.24 g of carbon.

∴ Carbon content of the compound \(=\frac{1.24 \times 100}{3.1}=40 \%\)

Again, 0.5 g ofthe compound yields 0.3 g of H20

Now, amount of hydrogen in 18 g of H20 = 2 g

∴ 0.3 g of H2O contains,\(=\frac{2 \times 0.3}{18}\) = 0.033 g

∴ Hydrogen content ofthe compound \(=\frac{0.033 \times 100}{0.5}=6.6 \%\)

∴ Oxygen content ofthe compound = 100- (40 + 6.6) = 53.4%

The ratio of the number of atoms of C, H, and O in a molecule ofthe compound

⇒ \(\begin{aligned}
& =\frac{40}{12}: \frac{6.6}{1}: \frac{53.4}{16} \\
& =3.33: 6.6: 3.33=1: 2: 1
\end{aligned}\)

∴ The empirical formula of the compound = CH2O

Let, the molecular formula of the compound be (CH2O)n, where n is an integer.

So, molecular mass of the compound = (12 + 2 + 16)n =30n

∴ Its gram-molocular mass = 30n g

Mass of O2 present In 1 gram-mole ofthe compound = 16n g. Now, most of 02 In 0.05 gram-mole ofthe compound =4.0 g. Mass of 02 in 1 gram-mole of compound = 96 g.

Hence, 16 n – 96 i.e., n = 6.

∴ Molecular formula = (CH20)G = C6H1206.

Question 9. A hydrocarbon contains 10.5 g of carbon and lg of hydrogen. The weight of 1 liter of the hydrocarbon at 127°C and atm pressure is 2.8 g. Determine the molecular formula ofthe compound.
Answer: Let, V Lbe the volume of the gas at STP

Thus \(\frac{1 \times 1}{(273+127)}=\frac{V \times 1}{273}\)

therefore V=0.6825l

Now, the mass of 0.6825 L of hydrocarbon at STP = 2.8 g

Therefore Mass of 22.4 L of hydrocarbon at STP= \(\frac{2.8 \times 22.4}{0.6825}=91.9 \mathrm{~g}\)

Therefore Molecular mass ofthe compound = 91.9

The ratio of the number of atoms of C and H in a molecule of the compound \(=\frac{10.5}{12}: \frac{1}{1}\) = 0.875:1 =1:1.143 =7:8

[subsequently dividing by the smallest number and then multiplying by 7]

∴ The empirical formula ofthe compound = C7H8

∴ Molecular formula ofthe compound = (C7H8)n

∴ Molecular mass = (7 X 12 + 8 x l)n =92 n

Hence, 92n = 91.9 \(\text { or, } n=\frac{91.9}{92} \approx 1\)

Molecular formula of the compound =(C7H8)1 = C7H8

Question 10. Combustion of 0.2 g of a monobasic organic acid produces 0.505 g C02 and 0.0892 g HzO. 0.183 g of the above acid requires 15 cm3 of (N/10) NaOH solution for complete neutralization. Determine the molecular formula ofthe organic acid.
Answer: 15 cm3 0.1(N)NaOH= 0.183g of acid

Therefore 1000 cm3 1(N) NaOH \(\equiv \frac{0.183 \times 1000}{15 \times 0.1}\) =122 g of acid.

So, the equivalent mass of acid = 122

As the acid is monobasic, its molecular mass will be equal to its equivalent mass.

∴ The molecular mass ofthe acid =122

Again, 0.2 g of the acid on combustion produces 0.505 g of C02 and 0.0892 g of H20.

∴ 0.505 g of C02 contains = \(=\frac{12 \times 0.505}{44}\) = 0.1377g carbon

[ 44 g of C02 contains 12 g carbon] and the amount of hydrogen present in 0.0892 g of \(\mathrm{H}_2 \mathrm{O}=\frac{2 \times 0.0892}{18}\)

= 9.911 X 10-3 g [ V 18 g of H20 contains 2 g of H2].

Therefore 0.2 g of the acid contains 0.1377 g of carbon and 9.911 x 10-3 g of hydrogen.

Therefore In the acid, mass percent of C \(=\frac{0.1377 \times 100}{0.2}\) = 68.85

and mass percent of H \(=\frac{9.911 \times 10^{-3} \times 100}{0.2}=26.2\)

Therefore Percentage of O in the acid = 100- (68.25 + 4.95) = 26.2

∴ The Ratio of the number of atoms of C, H, and O in 1 molecule ofthe acid, C : H: O
\(=\frac{68.85}{12}: \frac{4.95}{1}: \frac{26.2}{16}\)

= 5.7375 : 4.95: 1.6375

= 3.5: 3.02: 1 a: 7: 6: 2

[Multiplication by 2 gives whole numberration]

Therefore Empirical formula ofthe acid = C7Hg02

Let the molecular formula of the acid be (C7H602)

Molecular mass ofthe acid =(7xl2 +6xl + 2x 16)n

= 122 n

Therefore 122n = 122/122 or, n = = 1

∴ Molecular formula of the acid = (C7H602)1 = C7H602

Chemical Calculations Based On Chemical Equations: Stoichiometry Calculations based on the quantitative relationship between the reactants [substance(s) participated in a chemical reaction] and products [substance(s) produced in the chemical reaction] in terms of their mole numbers, masses, and volumes in any chemical transformation, are known as stoichiometry.

Now, the quantity of reactants or products is expressed in terms of mass or volume (if gaseous). Thus in any chemical reaction, there exists three types of relationship between the reactants and the products, viz.

  1. Mass-mass
  2. Mass-volume and
  3. Volume volume.

Naturally, there can be three possible modes of calculations on the basis of chemical equations. These are

Calculations involving mass-mass relationship: In these calculations, the mass of product(s) formed from a given; mass of reactant(s) or the mass of reactant(s) required to produce a certain mass of product(s) can be determined. For example, carbon burns produce carbon dioxide gas.

⇒ \(\mathrm{C}(12 \mathrm{~g})+\mathrm{O}_2(32 \mathrm{~g}) \rightarrow \mathrm{CO}_2(44 \mathrm{~g})\)

Evidently, 44 parts by mass of C02 are produced from 12 parts by mass of C, or in other words, 12 parts by mass of C is to be burnt to obtain 44 parts by mass of C02.

Calculations involving mass-volume relationship: In these calculations, any unknown volume of product that is produced from a given mass of reactant or any unknown mass of reactant required to obtain a known volume of product can be determined. For example, when KC103 is heated, oxygen gas (O2) is liberated.

⇒ \(\underset{245 \mathrm{~g}}{2 \mathrm{KClO}_3} \longrightarrow 2 \mathrm{KCl}+\underset{3 \times 22.4 \mathrm{~L} \text { (at STP) }}{30_2}\)

So, 67.2 L of 02 at STP is produced from the decomposition of 245 g of KC103, or in other words, 245 g of KCIO3 is required to yield 67.2 L of Oz at STP.

Calculations involving volume-volume relationship: If both the reactants and products are in the gaseous state, such calculations are done.

For example, carbon monoxide on burning produces carbon dioxide.

⇒ \(\underset{2 \text { volume }}{2 \mathrm{CO}}+\underset{\text { 1 volume }}{\mathrm{O}_2} \longrightarrow \underset{2 \text { volume }}{2 \mathrm{CO}_2}\)

The above equation shows that under the same conditions of temperature and pressure, 2 volumes of CO combine with 1 volume of O2 to produce 2 volumes of CO2.

Some Significant Information Regarding Chemical Calculations

During chemical reactions, reactant molecules interact with each other in a simple ratio of whole numbers.

The product molecules also bear a simple whole-number ratio with the reactant molecules.

The whole numbers representing the moles of reactants and products involved in a chemical equation are called stoichiometric coefficients.

The simplest whole number ratio of moles reactants and products involved in the reaction is called stoichiometric ratio.

Density (d)= \(\frac{\text { mass of the substance }(\mathrm{m})}{\text { volume of the substance }(\mathrm{V})}\)

Relative Density \(=\frac{\text { mass of the substance }}{\text { mass of equal volume of water at } 4^{\circ} \mathrm{C}}\)

In the CGS unit, relative density or specific gravity of the substance = density of the substance So in the CGS unit, the mass of a substance

=relative density of the substance x the volume of the substance

Mass of 1 L hydrogen gas at STP = 0.089 g

Molecular mass (M) of any gas =2 X vapor density (D)

Mass of 1 L of any gas at STP

= vapour density ofthe gas x mass of1 L ofhydrogen gas

at STP = vapour density of the gas x 0.089 g

Percentage: The percentage of a constituent in a mixture refers to the amount of that constituent by parts present in 100 parts of the mixture.

For a solid mixture, percentage denotes percentage by mass. For example, an alloy of copper contains 70% copper. It means that 70 parts by mass of copper are present in 100 parts by mass ofthe alloy.

For a liquid mixture or solution, the percentage is expressed in terms of mass or volume. In other words, the percentage indicates parts by mass of the substance dissolved in 100 parts by mass or 100 cm3 of the liquid or solution. For example, 10% H2S04 by mass means.

10g of H2S04 is dissolved in 100 g of H2S04 solution. Again, 10% H2S04 by volume means 10g of H2S04 is dissolved in 100 cm3 of H2S04 solution.

For a gas mixture, percentage means the percentage by volume.

In all chemical calculations, similar types of units are used.

The general method of calculation on the basis of chemical equations involves the following steps:

A balanced chemical equation is to be written first.

The relative number of moles of the relative masses (gram-atomic or gram-molecular masses) of gaseous reactants and products are written below their formulae.

Relative volumes (in multiples of 22.4 L at STP) of gaseous reactants and products are written below their formulae.

Unitary methods applied for calculations.

Mass-mass calculations

At the time of calculations involving masses, the points mentioned below should be kept in mind.

A properly balanced equation of the corresponding chemical reaction should be written.

The respective relative masses (sometimes gram-mole or mole-number) of each of the reactants and products that appear in the equation should be written below their corresponding formulae.

The molecular mass of the reactant or product when multiplied by gram-mole or mole number, gives their relative mass.

For monoatomic substances, atomic mass when multiplied by the number of gram-atoms gives the relative mass.

From the relative masses or gram-moles of the reactants and the products in the equation, a quantitative relationship between the reactants and the products is obtained. From this relationship and the given data, the desired mass can be calculated.

Limiting reagent: Sometimes in chemical reactions, one ofthe participating reactants is taken in excess of the stoichiometric requirement according to the balanced chemical equation to ensure the completion of the reaction. In such cases, the amount ofthe product formed is divided by the reactant which is present in the least amount (i.e., the reactant which is consumed completely during the reaction).

This reagent is called the limiting reagent. It is so-called since it makes the participation of the other reactants and also the amount of products formed in the reaction limited. The reactants that are taken in excess are partially left behind at the end of the reaction.

Limiting reagent Definition: The Reagent that is present in the least proportionate amount (in the reaction mixture) and hence gets completely consumed in the chemical reaction under consideration, is called the limiting reagent.

On the basis of the amount of the limiting reagent, the product and the reactant left unreacted can be quantitatively estimated.

Example: N2 and H2 gases react to form NH3 gas.

The corresponding reactions: N2 + 3H2→NH3 According to the above equation, 1 mol of N2 reacts with 3 mol of H2 to produce 2 mol of NH3.

Now, let us consider that the reaction has been initiated by taking 2 mol of N2 and 4 mol of H2. According to the above equation, for 2 mol of N2, 6 mol of H2 is required.

Here, only 4 mol of H2 has been used. So, the limiting reagent, in this case, is H2.

Here, the amount ofthe product is to be calculated on the basis ofthe amount of H2. According to the equation, 4 mol of H2 given \(\left(\frac{2}{3} \times 4\right)=\frac{8}{3}\) mol of NH3.

Again, 4/3 mol of H2 combine with| mol of N2.

∴ At the end of the reaction, amount of N2 left unreacted \(=\left(2-\frac{4}{3}\right)=\frac{2}{3} \mathrm{~mol} \text {. }\)

Numerical Examples 

Question 1. How many grams of carbon is to be burnt to produce 33g of CO2?
Answer: Carbon burns in oxygen to produce C02 gas.

Reaction: C (12g) + O2 (32g)→CO2 (44g)

The reaction shows that for the production of44g of CO2,

Amount of carbon required = 12 g

∴ For the production of 33 g of CO2, the amount of carbon required

⇒ \(=\frac{12 \times 33}{44}=9 \mathrm{~g}\)

Question 2. Find the mass of CaO obtained by heating 100 g of a reactant which sample of CaC03 is 95%pure.
Answer: The thermal decomposition of CaCO3 into CaO and CO2 can be represented by the equation—

∴ \(\begin{array}{cc}
\mathrm{CaCO}_3(s) & \mathrm{CaO}(s) \\
(40+12+48) \mathrm{g} & (40+16) \mathrm{g} \\
=100 \mathrm{~g} & =56 \mathrm{~g}
\end{array}\)

According to the given condition, CaCO3 is 95% pure. So, 100 g of the sample contains 95 g of pure CaCO3.

The above reaction shows that the amount of CaO that can be obtained from 100 g of pure CaCO3 = 56 g

Amount of CaO that can be obtained from 95 g pure

⇒ \(\mathrm{CaCO}_3=\frac{56}{100} \times 95=53.2 \mathrm{~g}\)

∴ A sample of 100 g of CaC03 which is 95% pure, on heating, produces 53.2 g of CaO.

Question 3. Calculate the volume of C02 (at STP) that can be obtained from 2kg CaC03.
Answer: CaC03 decomposes as follows:

⇒ \(\begin{array}{cc}
\mathrm{CaCO}_3(s) \longrightarrow & \mathrm{CaO}(s)+\mathrm{CO}_2(\mathrm{~g}) \\
100 \mathrm{~g} & 22.4 \mathrm{~L}(\text { at STP) }
\end{array}\)

So at STP volume of C02(at STP) obtained from 2 kg or 2000 g of CaC03
\(=\frac{22.4}{100} \times 2000=448 \mathrm{~L}\)

Question 4. How much KC103 must be heated to produce as much oxygen as that would be obtained from 200g of HgO? [Hg= 200.5,K= 39]
Answer: HgO when heated produces O2

⇒ \(\underset{(2 \times 216.5) \mathrm{g}}{2 \mathrm{HgO} \longrightarrow 2 \mathrm{Hg}}+\underset{\mathrm{O}_2}{32 \mathrm{~g}}\)

Therefore Amount of 02 obtained from 200g of HgO \(=\frac{32 \times 100}{2 \times 216.5} \mathrm{~g}\)

Preparation of O2 from KC103 \(\begin{aligned}
& 2 \mathrm{KClO}_3 \\
& (2 \times 122.5) \mathrm{g}
\end{aligned} \longrightarrow 2 \mathrm{KCl}+{ }_{96 \mathrm{~g}}^{3 \mathrm{O}_2}\)

96 g 02 is obtained by heating 2 x 122.5 g of KC103

therefore \(\frac{32 \times 200}{2 \times 216.5} \text { g O }_2\) O2 Is obtained by heating

⇒ \(\frac{2 \times 122.5}{96} \times \frac{32 \times 200}{2 \times 216.5} \mathrm{~g}=\) 37.72g of KCLO3

Question 5. A solution of nitric acid contains 60% nitric acid. The specific gravity of the solution is 1.46. How many grams of HN03 solution will be required to dissolve 5 g of copper oxide?
Answer: The reaction between CuO and HN03 takes place as per the following equation—

⇒ \(\begin{aligned}
& \mathrm{CuO}+2 \mathrm{HNO}_3 \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \mathrm{O} \\
& (63.5+16)=79.5 \mathrm{~g} \quad 2(1+14+48)=126 \mathrm{~g} \\
&
\end{aligned}\)

∴ Amount of HN03 required to dissolve 79.5 g CuO = 126g

∴ Amount of HNO3 required to dissolve 5 g of CuO

⇒ \(=\frac{126}{79.5} \times 5=7.924 \mathrm{~g}\)

As given in the problem, 100 cm3 HNO is a solution containing 60 g of pure HN03.

The specific gravity of IIN03 = 1.46

Mass of100 cm3 HN03 solution = 100 x 1.46 = 146 g

Thus, 60 g of HN03 is present in 146 g IINO., solution.

∴ 7.924 g HN03 is present in \(\frac{146 \times 7.924}{60} \mathrm{~g}\) g of HNO3 solution /.e., 19.282 g ofHN03 solution.

19.282 g HN03 solution will be required for the dissolution of 5 g of CuO

Question 6. How much water will be produced when the electric spark is passed through a mixture of 20 g ofhydrogen and 200g of oxygen? What amount of oxygen will remain unreacted?
Answer: The corresponding reaction is:

⇒ \(2 \mathrm{H}_2+\mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}\)

In this reaction, 4 g of H2 produces 36 g of water.

⇒ 20 g of H2 will produce \(\frac{36 \times 20}{4}\)= 180 g of water

Again, 4 g of H2 reacts with 32 g of 02

20 g of H2 reacts with \(\frac{32 \times 20}{4}\) = 160 g of 02

⇒ Amount of O2 left unreacted = (200- 160)g = 40 g

Question 7. An astronaut needs 34 g of sucrose per hour to maintain his physical strength. What quantity of oxygen should he carry if he has to stay 1 day in the spacecraft?
Answer: The energy obtained due to the oxidation of sucrose serves as the source of energy for the astronaut. The oxidation reaction of sucrose is-

\(\begin{gathered}
\mathrm{CaCO}_3+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \\
2 \mathrm{~mol} \\
2(1+35.5)=73 \mathrm{~g}
\end{gathered}\)

Amount of sucrose required per day =34×24 = 816g

[since Requirement of sucrose per hour= 34 g]

According to the reaction, the amount of 02 required for the oxidation of 342 g of sucrose = 384 g Amount of 02 required for the oxidation of 816 g of sucrose
\(=\frac{384 \times 816}{342}=916.2 \mathrm{~g}\)

Therefore, the astronaut will have to carry 916.2 g of 02 if he has to stay in the spacecraft for 1 day.

Question 8. 25.4g of I2 and 14.2g of Cl2 react together to form a mixture of IC1 and IC13. What is the ratio of the number of moles of IC1 and IC13 in the product mixture?
Answer: 25.4 g of I2 \(=\frac{25.4}{254}=\) 0.1 mol and 14.2 g of Cl2 \(=\frac{14.2}{71}\)

= 0.2 mol [ v = 254, = 71 ]

The reaction between, and Cl2 is—

⇒ \(\underset{1 \mathrm{~mol}}{\mathrm{I}_2}+\underset{2 \mathrm{~mol}}{2 \mathrm{Cl}_2} \longrightarrow \underset{1 \mathrm{~mol}}{\mathrm{ICl}}+\underset{1 \mathrm{~mol}}{\mathrm{ICl}_3}\)

The reaction indicates that 1 mol of I2 and 2 mol of Cl2 react together to form 1 mol of each IC1 and ICI3.

So, 0.1 mol of, and 0.2 mol of Cl2, will react together to form 0.1 mol of IC1 and 0.1 mol of IC13.

The ratio of number of moles of IC1 and IC13 in the resultantmixture= 0.1: 0.1 =1:1

Question 9. 1L of a sample of hard water contains 1 mg of each of CaCÿ and MgCl2. Express the hardness of the sample in ppm, in terms of CaC03.
Answer:

⇒ \(\begin{aligned}
1 \mathrm{mg} \mathrm{CaCl}_2 & \equiv \frac{1 \times 10^{-3}}{111} \mathrm{~mol} \mathrm{CaCO}_3 \\
& =\frac{10^{-3} \times 100}{111} \mathrm{~g}=0.9009 \times 10^{-3} \mathrm{~g} \mathrm{CaCO}_3 \\
1 \mathrm{mg} \mathrm{MgCl}_2 & \equiv \frac{1 \times 10^{-3}}{95} \mathrm{~mol} \mathrm{CaCO}_3=\frac{10^{-3} \times 100}{95} \mathrm{~g}^2
\end{aligned}\)

∴ Amount of CaC03 equivalent to (1 mg of CaCl2 + 1 mg ofMgCl2) = (0.9009 + 1.052) x 10-3 g = 1.9529 X 10-3 g

Hardness of water \(=\frac{1.9529 \times 10^{-3}}{1000} \times 10^6=1.9529 \mathrm{ppm}\)

(1 Lwater = 1000 g water; density of water =lg-mL-1)

Question 10. What amount of calcium oxide will react with 852g of P4 O10?
Answer: The reaction of P4O10 with CaO is

⇒ \(\begin{aligned}
& 6 \mathrm{CaO}+\mathrm{P}_4 \mathrm{O}_{10} \longrightarrow 2 \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2 \\
& 6 \times 56 \mathrm{~g} \quad 284 \mathrm{~g}
\end{aligned}\)

284 g P4O10 reacts with 6 x 56 g of CaO

So, 852 g P4 O10 reacts with \(\frac{6 \times 56 \times 852}{284}\) =1008 g of CaO.

Question 11. What percent by mass of lead nitrate [Pb(N03)2] is reduced when heated strongly?
Answer:

⇒ \(\begin{array}{cl}
2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2 \longrightarrow & 2 \mathrm{PbO}+4 \mathrm{NO}_2 \uparrow+\mathrm{O}_2 \uparrow \\
2(207+2 \times 14+6 \times 16) & 2(207+16) \\
=662 \mathrm{~g} & =446 \mathrm{~g}
\end{array}\)

On strong heating, Pb(N03)2 loses its mass as N02 and 02 escape out as gases, and only PbO is left behind as solid residue (yellow colored).

The above equation shows that 446 g of PbO is left behind as residue when 662 g of Pb(N03)2 is strongly heated.

∴ Loss in mass = (662- 446) g = 216 g.

∴ Percentage of loss by mass \(=\frac{216}{662} \times 100=32.62\)

Question 12. When a mixture of KI and KC1 is heated repeatedly with H2S04, iodine escapes completely and K2S04 is produced quantitatively. In the case of such a mixture, it is observed that the mass of K2S04 is equal to the mass of the mixture of K3 and KC1 taken. What is the ratio ofthe masses of KI and KC1 in this mixture?
Answer: Let, the masses of KI and KC1 in the mixture be x g and y g respectively. Now, the reactions ofKI and KC1 with H2S04 are as follows

⇒ \(\begin{aligned}
& 2 \mathrm{KI}+2 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+\mathrm{I}_2+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \\
& 2 \times(39+127) \quad 174 \mathrm{~g} \\
& =332 \mathrm{~g} \\
&
\end{aligned}\)

Therefore \(x \mathrm{~g} \text { KI produces }\left(\frac{174}{332} \times x\right) \text { g of } \mathrm{K}_2 \mathrm{SO}_4\)

⇒ \(\begin{aligned}
& \underset{2 \times(39+35.5)}{2 \mathrm{KCl}}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \underset{2}{\mathrm{~K}_2 \mathrm{SO}_4+2 \mathrm{HCl} \uparrow} \\
& =149 \mathrm{~g}
\end{aligned}\)

∴ y g KC1 produces \(\left(\frac{174}{149} \times y\right)\)g of K2SO4

∴ Total amount of K2S04 produced from x g of KI and y g of KCl

\(=\frac{174 \times x}{332}+\frac{174 \times y}{149}\)

As given in the question \(\frac{174 x}{332}+\frac{174 y}{149}=x+y\)

\(\text { or, } \quad \frac{158}{332} x=\frac{25}{149} y \text { or, } \frac{x}{y}=\frac{25 \times 332}{158 \times 149} \text { or, } x: y=1: 2.836\)

∴ The ratio ofthe masses of KI and KCl in the mixture is 1: 2.836

Question 13. How much of 5% impure NaN03 and 98% H2S04 will be required to produce 7.5kg nitric acid by the chemical reaction between them?
Answer: The reaction between NaN03 and H2S04 is—

⇒ \(\begin{aligned}
& 2 \mathrm{NaNO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{HNO}_3 \\
& 2(23+14+48)=170 \mathrm{~g} \quad(2+32+64)=98 \mathrm{~g} \quad(2 \times 63)=126 \mathrm{~g} \\
&
\end{aligned}\)

7.5 kg or 7500 g HN03 is produced from

⇒ \(\frac{170}{126}\) X 7500 = 10119 g ofpure NaN03

Therefore Amount of impure NaN03 required to produce

7.5kg HN03 = 10119 =10651 g = 10.651 kg

7.5 kg or 7500 g HN03 is produced from

7500 =5833.33 g ofpure H2S04.

∴ Amount of 98% H2S04 required to produce 7.5 kg of HN03 =\(\frac{100}{98} \times\) 5833.33 g = 5952 g = 5.952 kg

Question 14. 3 g of HCl is present per liter of gastric juice produced in the human body. If a person produces 2.5 L gastric juice per day, then how many antacid tablets are required to neutralize HCI produced per day? [Assume that each tablet contains 400mg of Al(OH)3]
Answer: 1 L gastric juice contains 3 g of HCI Molecular mass of P4Og = 4 x 31 + 6 X 16

Therefore 2.5 L gastric juice contains (3 X 2.5)g = 7.5HCL

The reaction between Al(OH)3 and HCI is as follows—

⇒ \(\begin{array}{cc}
\mathrm{Al}(\mathrm{OH})_3 & +3 \mathrm{HCl} \\
\begin{array}{c}
1 \mathrm{~mol} \\
(27+3 \times 17)=78 \mathrm{~g}
\end{array} & \begin{array}{c}
3 \mathrm{~mol} \\
(3 \times 35.5)
\end{array}=106.5 \mathrm{~g}
\end{array}\)

Evidently, 106.5gofHCI is neutralised by78gof Al(OH)3

Therefore 7.5 g HCI is neutralised by \(\frac{78 \times 7.5}{106.5}\) = 5.493 g

= 5493 mg AL(OH)3

Now, 400 mg of Al(OH)3 is present in 1 tablet.

Therefore 5493 mg Al(OH)3 will be present in \(=\frac{1 \times 5493}{400}\)

=13.73 14 tables.

∴ 14 tablets are needed to neutralize HCI produced per day.

Question 15. (C2P4)n is a polymeric substance where n is a large number. It is prepared by polymerization in the presence of a sulfur catalyst. The final product is found to contain 0.012% of S. Find the value of n if the polymeric molecule contains three S-atoms.
Answer: Since 1 molecule of the polymer contains 3 atoms of S, 1 gram-mole of the polymer should contain 3 gram-atoms or 3 x 32 g of sulfur.

Now, 0.01 2 g of sulfur is present in 100 g of polymer.

therefore 3 x 32 g of sulphur is present in \(\frac{100 \times 3 \times 32}{0.012}\)

= 8 x 10s g of polymer

Therefore Molecular mass of the polymer = 8 x 105

Again, the molecular formula of the polymer = (C2F4)n

∴ Molecular mass =n(2 x 12 + 4 x 19) = 100n

or, 100n =8 x 105

∴ n = 8000

Question 16. Calculate the number of moles of NaOH required to neutralize the solution produced by dissolving l.lg P406 in water. Use the following reactions:
Answer:

⇒ \(\begin{aligned}
& \mathrm{P}_4 \mathrm{O}_6+6 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{H}_3 \mathrm{PO}_3 \\
& 2 \mathrm{NaOH}+\mathrm{H}_3 \mathrm{PO}_3 \rightarrow \mathrm{Na}_2 \mathrm{HPO}_3+2 \mathrm{H}_2 \mathrm{O}
\end{aligned}\)

Molecular mass of P4Og = 4 x 31 + 6 X 16 = 124 + 96 = 220 g-mol-1

Number of moles of P4Og \(=\frac{1.1 \mathrm{~g}}{220 \mathrm{~g} \cdot \mathrm{mol}^{-1}}=\frac{1}{200}\)

4 mol H3PO3 is produced by 1/200 mol P4O6

1 mol H3PO3 is produced by \(=4 \times \frac{1}{200}=\frac{1}{50} \mathrm{~mol} \mathrm{P}_4 \mathrm{O}_6\)

1/200 mol P4O6 Produce 4x 1/200 mol H3PO3

Also, 1 mol of H3PO3 requires 3 mol of NaOH.

1/50 mol H3PO3 requires = 2x 1/50 =1/25 = 0.04 mol NaOH

Mass-volume calculations

In a chemical reaction, if any substance (reactant or product or both) exists in the gaseous state, then the following procedure of calculation is considered.

Important points, relevant to this type of calculation, are given below—

A properly balanced equation representing the chemical reaction should be written first.

The relative mass (or mole number) ofthe solid reactant or product is to be written under each formula.

Molecule mass multiplied by mole number gives the relative mass. In the case of monoatomic substances, relative mass is the product of atomic mass and the number of gram-atoms.

The amount of a gaseous substance is generally expressed by its volume. If the volume at STP is not given, then the volume at STP can be calculated as follows-

⇒ \(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)

The volume of 1 gram-mole or molar volume of any gas at STP = 22.4L.

If the conditions of the gaseous reaction (i.e., temperature and pressure) are not mentioned, then the reaction is assumed to have occurred at STP.

Mass of 1 L of gas at STP = 0.089 g.

Mass of 1L of a gas at STP = vapor density x 0.089 g.

At any given temperature and pressure, the mass of a certain volume of gas, or the volume ofthe gas from its mass, can be calculated with the help ofthe equation:

Where w and M are the mass and molecular mass ofthe gas respectively. The relation between the volume or mass of the reactant and the volume or mass ofthe products can be determined from a chemical equation. The value of an unknown quantity can also be determined from this obtained data.

Numerical Examples 

Question 1. A balloon of 1000 L capacity is to be filled up with hydrogen gas at 30°C and 750mm pressure. What amount of iron will be required to generate the required volume of hydrogen?
Answer: Let 1000 L H2 gas which is required to fill up the balloon of 1000L capacity occupy V L at STP (at given condition).

∴ \(\frac{750 \times 1000}{(273+30)}=\frac{760 \times V}{273}\)

∴ \(V=\frac{750 \times 1000 \times 273}{760 \times 303}=889.13 \mathrm{~L}\)

Generally, H2 gas is produced by reacting iron with steam (H20). (Consumption of Fe will be more acid is used).

⇒ \(\begin{gathered}
3 \mathrm{Fe}+4 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{H}_2 \\
(3 \times 55.85)=167.55 \mathrm{~g}
\end{gathered}\)

The reaction shows 4×22.4L of H2 gas at STP is produced from 167.55 g of Fe.

889.13L of H2 gas at STPis produced from

⇒ \(\frac{167.5}{4 \times 22.4} \times 889.13=1662.15 \mathrm{~g} \text { of } \mathrm{Fe}\)

1662.15 g Fe is required to produce the desired volume of H2.

Question 2. The volume of oxygen liberated at 26°C and 714mm pressure due to the thermal decomposition of xg of KC103 and collected over water is 760mL. What is the value of x? [Given that aqueous tension at 26°C = 26mm; K = 39;Cl =35.5; O =16]
Answer: Actual pressure of 02 =(714- 26) = 688mm.

Let the volume ofthe given oxygen gas at STP be V L.

⇒ \(\quad \frac{688 \times 760}{(273+26)}=\frac{760 \times V}{273}\)

Or, \(V=\frac{688 \times 760 \times 273}{299 \times 760}\)

= 628.17mL = 0.62817L

The reaction of thermal decomposition of KC103 is—

⇒ \(\begin{array}{cc}
2 \mathrm{KClO}_3 \longrightarrow 2 \mathrm{KCl} & +3 \mathrm{O}_2 \\
2 \mathrm{~mol} & 3 \mathrm{~mol} \\
2(39+35.5+48)=245 \mathrm{~g} & 3 \times 22.4 \mathrm{~L}(\mathrm{STP})
\end{array}\)

Now, the mass of KC103 required to produce 3 x 22.4 L

of 02 at STP =245 g

⇒ At STP, KCIO3 is required to produce 0.62817 L of 02

⇒ \(=\frac{245 \times 0.62817}{3 \times 22.4}=2.29 \mathrm{~g}\)

⇒ x= 2.29

Question 3. What volume of gas will be formed at 523K and 1 atm pressure by the explosive decomposition of 5g of ammonium nitrate, according to the given equation? 2NH4N03(s) = 2N2(g) + O2(g) + 4H2Q(g)
Answer:

⇒  \(\begin{array}{cccc}
2 \mathrm{NH}_4 \mathrm{NO}_3(s) & 2 \mathrm{~N}_2(g) & \mathrm{O}_2(g) & +4 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\
2 \mathrm{~mol} & 2 \mathrm{~mol} & 1 \mathrm{~mol} & 4 \mathrm{~mol} \\
2(14+4+14+48) & 2 \times 22.4 \mathrm{~L} & 22.4 \mathrm{~L} & 4 \times 22.4 \mathrm{~L} \\
=160 \mathrm{~g} & \text { (at STP) } & \text { (at STP) } & \text { (at STP) }
\end{array}\)

⇒ Total volume ofthe gases produced by the decomposition Of 160g of NH4NO3 = 7 X 22.4 L at STP.

Let, 7 x 22.4 L ofthe gas at STP occupy a volume of V L at 523K and 1 atm pressure.

⇒ \(\frac{1 \times(7 \times 22.4)}{273}=\frac{1 \times V}{523}\)

⇒ \(V=\frac{1 \times 7 \times 22.4 \times 523}{1 \times 273}=300.39 \mathrm{~L}\)

Thus, at 523K and 1 atm pressure, 160 g of NH4N03 produces 300.39L of gas.

At 523 K and1 atm pressure, 5 g ofNH4N03 produces

⇒ \(=\frac{300.39 \times 5}{160}=9.387 \mathrm{~L} \text { of gas. }\)

Question 4. Ignition of a wooden match stick involves the combustion of P4S3 in the oxygen of the air to produce a white smoke of P4O10 and gaseous sulfur dioxide (S09). Calculate the volume of SO., formed at 27°C and 770mm Hg pressure from the combustion of 0.0546 g of P4S3. [P= 31, S = 32, 0 = 16]
Answer: The equation for combustion reaction:

⇒ \(\mathrm{P}_4 \mathrm{~S}_3+8 \mathrm{O}_2 \rightarrow \mathrm{P}_4 \mathrm{O}_{10}+3 \mathrm{SO}_2\)
The molecular mass of P4S3 = 4×314-3×32 = 220.

Now, 220 g of P4S3 produces 3 x 22.4 L of S09 at STP.

At STP, 0.0546g of P4S3 produces \(\frac{3 \times 22.4}{220} \times 0.0546\)

= 0.0166L of S02

Let. 0.0166 L of S09 at STP occupy a volume of PL at 27°C and 770 mmHg pressure.

⇒ \(\frac{770 \times V}{(273+27)}=\frac{760 \times 0.0166}{273} \text { or, } V=\frac{760 \times 0.0166 \times 300}{770 \times 273}\)

V = 0.018 L

Therefore, the volume of S09 formed from the combustion of 0.0546 g of P4S3 at 27°C and 770 mm Hg is 0.018 L

Question 5.5 g of a pure sample of FeS reacts with dil. H9S04. H9S gas produced is then completely burnt in the air. Find the volume of SO, thus obtained, measured at 25°C and 750mm pressure of Hg.
Answer: FeS reacts with H9S04 to liberate HH2S gas—

⇒ \(\begin{array}{cc}
\mathrm{FeS}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{FeSO}_4+\mathrm{H}_2 \mathrm{~S} \\
1 \mathrm{~mol} \\
\begin{array}{c}
1 \mathrm{~mol} \\
(55.85+32)=87.85 \mathrm{~g}
\end{array} & 22.4 \mathrm{~L}(\text { at STP })
\end{array}\)

Now, 87.85 g of FeS produces 22.4 L of H2S at STP.

⇒ 5.5 g of FeS produces \(\frac{22.4 \times 5.5}{87.85}\) = 1.402 L of H2S at STP.

Reaction takes place when H and S burn in the air.

So at STP, 2 x 22.4 L of H2S produces 2 x 22.4 L of S02.

At STP, 1.402 L of H9S produces 1.402 L of S02.

Let, 1.402 L of S02 at STP occupies a volume of VL at 25°C and 750 mm pressure of Hg.

Thus, \(\frac{V \times 750}{(273+25)}=\frac{1.402 \times 760}{273}\)

Hence, at 25°C and 750 mm pressure, the volume of S02 produced = 1.550 L.

Question 6. 1.78 L of chlorine gas at STP is prepared by using 40% HC1 by weight according to the following reaction— Mn02 + 4HCl→MnCl2 + 2H20 + Cl2. Find the volume of hydrochloric acid and mass of Mn02 required to produce this amount of chlorine gas. [Specific gravity of the HC1 solution= 1.12]
Answer: The corresponding ingreaction is:

⇒ \(\begin{array}{ccr}
\mathrm{MnO}_2+4 \mathrm{HCl} \longrightarrow \mathrm{MnCl}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2 \\
1 \mathrm{~mol} & 4 \mathrm{~mol} & 1 \mathrm{~mol} \\
(54.94+32) & 4 \times 36.5 & 22.4 \mathrm{~L} \\
=86.94 \mathrm{~g} & =146 \mathrm{~g} & \text { (at STP) }
\end{array}\)

According to the reaction, the amount of Mn09 required for the production of 22.4 L of Cl9 at STP = 86.94 g.

Since Amount of Mn02 required for production of 1.78 L of Cl2 at STP \(=\frac{86.94 \times 1.78}{22.4}\)

Similarly, the amount of HC1 required for the preparation of 22.4L of Cl9 gas at STP = 146 g.

⇒ Amount of HC1 required for the preparation of 1.78 L of Cl9 gas at STP \(=\frac{146 \times 1.78}{22.4}=11.6 \mathrm{~g}\)

Now, 100 g HC1 solution contains 40 g of HC1.

∴ 11.6 g HC1 solution contains \(\frac{100 \times 11.6}{40}=29 \mathrm{~g} \text { of } \mathrm{HCl}\)

The specific gravity of the HC1 solution =1.12

Required volume ofthe HC1 solution

⇒ \(=\frac{\text { mass of acid solution }}{\text { specific gravity }}=\frac{29}{1.12}=25.89 \mathrm{~mL}\)

Question 7. If a particular HCl solution contains 22% of the acid by weight, then how much quantity of this acid solution will be required to produce 1L C02 gas at 27°C & 760mmpressure from pure CaC03?
Answer: Reaction between CaC03 and HCI is given by:

⇒ \(\begin{gathered}
\mathrm{CaCO}_3+2 \mathrm{HCl} \longrightarrow \mathrm{CaCl}_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \\
2 \mathrm{~mol} \\
2(1+35.5)=73 \mathrm{~g}
\end{gathered}\)

If the volume of 1L of C02 at27°C and 760 mm pressure is VLatSTP, the

⇒ \(\frac{760 \times 1}{(273+27)}=\frac{760 \times V}{273}; V=\frac{760 \times 273}{760 \times 300}=0.91 \mathrm{~L}\)

Now at STP, 22.4L of C02 is produced by 73 g of

At STP, 0.91L C02 is produced from \(\frac{73 \times 0.91}{22.4}=2.965 \mathrm{~g}\) of HCI solution.

Now, 22gHCI is present in lOOg of the given HCI solution

2.965 g ofHCI will be present in \(\frac{100 \times 2.965}{22}\) =13.48 g of HCL. solution.

So, 13.48 g ofthe given acid solution will be required.

Question 8. 1.5g of a mixture of CaC03 and MgCOs, ignition produces 360 mL of C02 at STP. Calculate the percentage composition of the mixture.
Answer: Let the mixture contains xg of CaC03 and (1.5- x) g of MgC03

The decomposition reactions involved are—

⇒ \(\underset{100 \mathrm{~g}}{\mathrm{CaCO}_3(s)} \underset{22400 \mathrm{~mL}(\mathrm{STP})}{\longrightarrow} \mathrm{CaO}(s)+\mathrm{CO}_2(g)\)

⇒ \(\begin{gathered}
\mathrm{MgCO}_3(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{CO}_2(g) \\
84 \mathrm{~g} \\
22400 \mathrm{~mL} \text { (at STP) }
\end{gathered}\)

Volume of CO2 (at STP) Produced From xg OF CaCO3 \(=\frac{22400}{100} \times x=224 x \mathrm{~mL}\)

The volume of COz (at STP) produced from (1.5- x) g of

⇒ \(\mathrm{MgCO}_3=\frac{22400}{84}(1.5-x)=266.66(1.5-x)\)

Now according to the given data,

224x + 266.66(1.5- x) = 360 or, x = 0.9376

So, the given mixture contains: CaC03 = 0.9376 g

and MgC03 = 1.5- 0.9376 = 0.5624 g

In the mixture,

Percentage of CaC03 \(=\frac{0.9376}{1.5} \times 100=62.5 \text { and }\)

Percentage of MgCO3 \(=\frac{0.5624}{1.5} \times 100=37.5\)

Question 9. Air contains 21% of oxygen by volume. What volume of that air at 27°C and 750mm pressure of Hg will be required for the complete combustion of 60 g of a candle? The candle contains 80% of carbon and 20% of hydrogen by mass.
Answer: Amount of Cin 60 g of a candle \(=\frac{80}{100} \times 60=48 \mathrm{~g}\)

Therefore Amount of H \(=\frac{20}{100} \times 60=12 \mathrm{~g}\)

During the combustion of the candle, C and present are oxidized to give C02 and H20. The relevant reactions are

⇒ \(\begin{aligned}
& \mathrm{C}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2 ; \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O} \\
& 12 \mathrm{~g} \begin{array}{c}
22.4 \mathrm{~L} \\
\text { (at STP) }
\end{array} \quad 2 \mathrm{~g} \quad \frac{1}{2} \times 22.4 \mathrm{~L} \text { (at STP) } \\
&
\end{aligned}\)

Therefore At STP, the volume of 02 required for combustion of 48 g of c \(=\frac{22.4 \times 48}{12}=4 \times 22.4 \mathrm{~L}\)
and volume of oxygen required for combustion of 12 g of hydrogen = \(\frac{22.4 \times 12}{2 \times 2}=3 \times 22.4 \mathrm{~L}\)

The volume of 02 at STP required for combustion of 60 g of candle = (4 x 22.4 + 3 x 22.4) = 156.8L.

Let, 156.8 L of 02 at STP occupy V volume at 27°C and 750mm pressure.

⇒ \(\quad \frac{V \times 750}{300}=\frac{156.8 \times 760}{273} \text { or, } \quad V=174.6 \mathrm{~L}\)

Volumetrically air contains 21% of 02.

Therefore Volume of air containing 174.60 of O2 =\(=\frac{100}{21} \times 174.60\)

Hence, for complete combustion of 60 g of a candle, 831.42L of air will be required at 27°C and 750mm pressure.

Volume-volume calculations

This type of calculation finds application in gaseous reactions. The reaction in which both the reactants and the products are in the gaseous state is called a gaseous reaction.

Eudiometry: A calculation involving the volumes of gases participating in a reaction is called eudiometry.

Eudiometry furnishes the following information:

  1. Volumetric relationship between the reactants and the products in any gaseous reaction.
  2. Volumetric composition of different constituents of the gas mixture produced in a gaseous reaction.
  3. The molecular formula of gaseous substances, particularly that of hydrocarbons.

Principle Of eudiometry: The principle of eudiometric calculations is based on Gay-Lussac’s law of gaseous volumes and Avogadro’s hypothesis. Now let us see how Gay-Lussac’s law of gaseous volumes and Avogadro’s hypothesis can be utilized in eudiometric calculations.

Example: In the reaction between hydrogen and chlorine gas, hydrogen chloride (HC1) gas is produced.

The reaction is:  H2 + Cl2→2HC1

According to the equation, 1 molecule of H2 and 1 molecule of Cl2 react to form 2 molecules of HC1 gas, or 1 gram-mole of H2 and 1 gram-mole of Cl2 combine together to produce 2 gram-moles of HC1 gas.

If the volumes are measured at STP, 2 x 22.4 L HC1 gas is produced by the combination of 22.4 L of H2 and 22.4 L of Cl2 [ v molar volume of any gas at STP = 22.4 L].

The reaction can be expressed in terms of molecule gram-mole or volume in the following way:

⇒ \(\begin{array}{ccc}
\mathrm{H}_2 & \mathrm{Cl}_2 & 2 \mathrm{HCl} \\
1 \text { molecule } & 1 \text { molecule } & 2 \text { molecules } \\
1 \text { gram-mole } & 1 \text { gram-mole } & 2 \text { gram-mole } \\
1 \text { volume }(22.4 \mathrm{~L}) & 1 \text { volume }(22.4 \mathrm{~L}) & 2 \text { volume }[2 \times 22.4 \mathrm{~L}(\mathrm{STP})]
\end{array}\)

The ratio ofthe number ofmolecules or gram-mole of H2, Cl2 and HC1 is given by: H2: Cl2: HC1 =1:1:2 At STP, the ratio of the volumes of H2, Cl2 and HC1 = 22.4:22.4:2×22.4 =1:1:2

Hence, in the gaseous reaction, the ratio of the number of gram-moles (or molecules) of the reactants and products is the same as the ratio of their volumes, measured under the same conditions of temperature and pressure.

In order to compare the volumes of different appearing in the equation of gaseous reaction, the volume of 1 gram-mole of gas is often taken volume.

The actual value of volume is 22.4L at STP.

Eudiometer:

An eudiometer is an apparatus that can be used to carry out gaseous reactions and to measure the volumes of gaseous reactants and their products.

This apparatus consists of a U-shaped glass tube one end of which is closed and the other end is kept open.

Two platinum wires are fused inside the closed end of the tube and these are used for electric sparking which brings about the required reaction between the gaseous reactants.

The reaction usually occurs with a little explosion. For measuring the volumes of gases, the arm with closed end is graduated.

Before starting the experiment, the tube is filled with mercury. Then the reacting gas is introduced into the closed end through the open end of the tube by the displacement of mercury.

Class 11 Chemistry Some Basic Concepts Of Chemistry Eudiometer

There is a stop-cock towards the lower part of the limb with an open end.

By removing mercury with the help of this stop-cock, the levels of mercury in both the limbs are made equal and the volume of the gas collected in the graduated closed limb is measured.

Similarly volume of each of the reacting gases, introduced one after another in the closed limb, is recorded.

Then these gases are made to react by passing an electric spark through the platinum wires. After the completion of the reaction, the tube is cooled down to room temperature.

If there is any contraction in volume due to cooling, the observed volume is recorded from the graduations in the closed limb. This contraction is known as the first contraction.

If there is a suitable absorbent for any gas in the gas mixture (For example KOH for C02, alkaline pyrogallate for 02, etc.), then it is introduced into the gas mixture.

As a result, the gas involved gets absorbed and the mixture again suffers contraction in volume. This is called a second contraction.

Tills contraction is recorded from the graduation of the From these experimental data regarding volume, audiometric calculation is done.

In this context, it is to be noted that all volumetric measurements are always made at the same temperature and pressure.

Some important points regarding calculations of audiometry:

A properly balanced equation of the gaseous reaction Is to be used.

Under similar conditions of temperature and pressure, the ratio of the volumes and number of gram-moles of gaseous reactant(s) and product(s) must be the same.

At STP, the volume of mole of any gas is equal to 22.4 L. At the time of comparing the volumes of gases, the volume of 1 gram-mole of any gas is considered to be 1 volume. At STP, ) volume = 22.4 L

The mass of 22.41. of a gas at STP expressed in grains is equal to the gram-molecular mass of the compound.

The volume of any solid or liquid reactant or product Involved in the gaseous reaction is considered to be zero.

In the absence of drastic conditions, nitrogen present in air generally does not take part in the gaseous reaction.

During eudiometric calculations, air Is considered to be a mixture of ()2 and N2 gases. (02 = 21% ; N2 = 79% )

In most cases, contraction in volume occurs on cooling the gas mixture produced In the eudiometer tube, to mom temperature. Expansion in volume may also occur ill some cases. Further, in some reactions, no perceptible.

limb, if there is any residual gas in the eudiometer rube after the first and second contraction, the volume of the gas left Is also recorded Change in volume is observed.

In the following table, some relevant observations have been recorded—

Class 11 Chemistry Some Basic Concepts Of Chemistry Combustion Of Hydrocarbon in ediometer

If the addition of an absorbent to u gas mixture produced in the gaseous reliction causes contraction In volume, then the gas can be Identified from the nature of the absorbent used and the volume of the gas may be determined from the magnitude of contact.

For this reason, prior to the addition of absorbent, the volume of the mixture is measured, if there is any contraction In volume after the addition of absorbent, then the volume of the gas mixture is again measured.

Ibe difference between the two readings gives the volume of the gas absorbed.

In this way, different gaseous constituents are removed one after another from the gas mixture by the successive addition of suitable absorbents and as a result, the contraction in volume in each case gives the measure of the volume of the absorbed gas concerned.

Numerical Examples 

Question 1. 60 mL of a mixture of CO and H2, mixed with 40 ml, of 02, are subjected to explosion in a eudiometer tube. On cooling the gas mixture after the end of the reaction, the volume is reduced to 30 mL.

Determine the composition of the gas mixture Initially taken. [All volumes arc measured at the same temperature and pressure.

Answer: Let the volume of CO in the initial mixture = x mL

Volume of H2 = (60- x) mL

Reactions taking place in the eudiometer tube—

⇒ \(\underset{\substack{2 \text { volume } \\ x \mathrm{~mL}}}{2 \mathrm{CO}}+\underset{\substack{1 \\ \text { volume } \\ \frac{x}{3} \mathrm{~mL}}}{\mathrm{O}_2} \longrightarrow \underset{x \mathrm{~mL}}{2 \text { volume }} \underset{x \mathrm{CO}_2}{\longrightarrow}\)

Therefore Contraction in volume \(=\left(x+\frac{x}{2}-x\right)=\frac{x}{2} \mathrm{~mL}\)

⇒ \(\begin{array}{ccc}
2 \mathrm{H}_2 & +\mathrm{O}_2 & \underset{2}{2} \mathrm{H}_2 \mathrm{O} \\
2 \text { volume } & 1 \text { volume } & 2 \text { volume } \\
(60-x) \mathrm{mL} & \frac{1}{2}(60-x) \mathrm{mL} & 0
\end{array}\)

[as water vapor on cooling gets condensed, its volume is taken as zero]

Contraction in volume \(\begin{aligned}
& =\left[(60-x)+\frac{1}{2}(60-x)-0\right] \\
& =\frac{3}{2}(60-x) \mathrm{mL}
\end{aligned}\)

Before the reaction, the volume of the mixture = volume of the mixture of CO and Il2 + volume of 02 = (60 + 40) = 100 mL Volume of the mixture after the reaction = 30 mL.

Therefore Contraction in volume in the reaction =100- 30 = 70 mL

From reactions 1 and 2, the total contraction in volume.

⇒ \(\left[\frac{x}{2}+\frac{3}{2}(60-x)\right]=(90-x) \mathrm{mL} \quad \text { or, } 90-x=70 x=20 \mathrm{~mL}\)

In the initial mixture, volume of CO = 20 mL and volume of H2 =(60-20) = 40mL.

Question 2. 30 mL of a mixture of CO and C02, mixed with 10 mL of oxygen, was exploded by an electric spark. The gas mixture produced was mixed with KOH solution and thoroughly shaken.

5mL of oxygen was left behind. What was the composition of the original mixture? [Volume was measured at STP.
Answer: Let. the volume of CO in 30 ml, in the mixture = xmL.

∴ Volume of carbon dioxide (C02) = (30 – x) ml.

⇒ \(\begin{aligned}
& \text { Reaction involved- } \mathrm{CO}+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{CO}_2 \\
& 1 \text { volume } \frac{1}{2} \text { volume } 1 \text { volume } \\
& x \mathrm{~mL} . \quad \frac{x}{2} \mathrm{~mL} . \quad x \mathrm{~mL} . \\
&
\end{aligned}\)

The reaction shows that the volume of ()., used \(=\frac{x}{2}\) mL

Volume of O2 used = volume of O2 taken – volume of O2 unused =(10-5) = 5ml..

Therefore \(\frac{x}{2}=5\)

Hence, the volume of CO in the original mixture = 10 ml.

Volume of C02 in the mixture = (30- 10) = 20 ml.

Question 3. 1 L of a mixture of CO and C02, when passed through a red hot tube containing charcoal, tire volume becomes 1.6L. All volumes are measured under the same conditions of temperature and pressure. Find the Composition of the mixture.
Answer: Volume of the mixture of CO and CO., 11.

Let, the volume of CO= xl

∴ Volume of CO2= (l-x)I.

The reaction involved CO2 +C→2CO

The volume of the mixture after passing over carbon = 1.6 L

∴ Total volume of CO =[x + 2(1 – x) ) = (2- v)L

According to the given data, 2 -x = 1.6 or, x = 0.4

∴ Volume Of Co In1 L Mixture = 0.4 L = 400 Ml And Volume

Of CO2 In 1 L Mixture = 0.6 L = 600 Ml.

Question 4. A mixture contains CO, CH4, and nitrogen. 25 cm3 of the mixture on oxidation in the presence of excess oxygen, resulted in a decrease in volume by 16 cm3. A further contraction of 17 cm3 was observed when the residual gas was treated with a KOH solution. What was the composition of the original gaseous mixture of 25 cm3 volume? [All volumes are measured at the same temperature and pressure.]
Answer: In the oxidation reaction, CO2 is obtained from CO while CH4 produces CO2 and water vapor but N2 remains unaffected by the process.

Let, the volume of CO in the original mixture = x cm3 and the volume of CH4 in the original mixture =y cm3

∴ Volume of N2 = [25- (x + y)] cm3

The reactions involved in this case are

⇒ \(\begin{aligned}
& \mathrm{CO}+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{CO}_2 \\
& 1 \text { volume } \frac{1}{2} \text { volume } 1 \text { volume } \\
& x \mathrm{~cm}^3 \quad \frac{x}{2} \mathrm{~cm}^3 \quad x \mathrm{~cm}^3 \\
&
\end{aligned}\)

Therefore Contraction in volume \(=\left[\left(x+\frac{x}{2}\right)-x\right]=\frac{x}{2} \mathrm{~cm}^3\)

⇒ \(\begin{aligned}
& \mathrm{CH}_4+2 \mathrm{O}_2 \longrightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} \\
& 1 \text { volume } 2 \text { volume } 1 \text { volume } 0 \\
& y \mathrm{~cm}^3 \quad 2 y \mathrm{~cm}^3 \quad y \mathrm{~cm}^3 \quad 0 \\
&
\end{aligned}\)

[Water vapor condenses to liquid on cooling. So, its volume is taken as zero.]

Contraction in volume this reaction = (y + 2y- y) = 2y cm3 Total contraction in volume =\(=\left(\frac{x}{2}+2 y\right) \mathrm{cm}^3\)

This contraction is equal to the first contraction i.e., \(\frac{x}{2}+2 y=16\)

The second contraction resulted from treatment with a KOH

volume of CO2 produced by the oxidation process -gift

= 17cm3. The total volume of CO2 produced by the first and second reactions = (x + y)cm3 i.e., x + y = 17 By solving [1] & [2] we have, x = 12 and y = 5.

Volume of CO in 25 cm3 of the mixture = 12cm3 and

volume of CH4 = 5cm3

∴ Volume of N2 = [25- (12 + 5)] = 8cm3

Question 3. 100 cm3 of a mixture of CO, C2H6, and N2 is exploded in the presence of excess O2. On cooling, the observed contraction in volume and the volume of C02 formed are both equal to the volume of the original mixture. Find the volumetric composition of the original mixture.
Answer: Let, the volumes of CO, C2Hg, and N2 be x, y, and z mL respectively in the mixture.

Reactions due to explosion—

⇒ \(\begin{array}{ccc}
\mathrm{CO} & +\quad \frac{1}{2} \mathrm{O}_2 & \longrightarrow \mathrm{CO}_2 \\
\text { 1volume } & 1 / 2 \text { volume } & 1 \text { volume } \\
x \mathrm{~mL} & x / 2 \mathrm{~mL} & x \mathrm{~mL}
\end{array}\)

⇒ Contration in volume \(\left(x+\frac{x}{2}-x\right)=\frac{x}{2} \mathrm{~mL}\)

Contraction in volume in volume = \(\left(y+\frac{7 y}{2}-2 y\right)=\frac{5 y}{2} \mathrm{~mL}\)

N2 remains unaffected by the explosion.

⇒ total contraction in volume \(=\left(\frac{x}{2}+\frac{5 y}{2}\right) \mathrm{mL}\)

After the explosion, the volume of C02 produced = (x + 2y) mL. According to the given condition, observed contraction in volume after explosion = volume of the original mixture.

⇒ \(\text { i.e., } \frac{x}{2}+\frac{5 y}{2}=x+y+z\)

Volume of C02 produced due to explosion = volume of the original mixture i.e.,x + 2y = (x + y + z) or, y = z

Substituting y = z in'[l] we have,\(\frac{x}{2}+\frac{5 z}{2}\) (x + z + z)

or, x + 5z = (2x + 4z) or, x = z. Hence, x = y = z

⇒ In the mixture, vol. ofCO = vol. of C2H6 = vol. of N2.

⇒ % of each constituent in the mixture = 100/3 = 33.3

Question 6. A gaseous mixture contains 50% of H2, 40% of CH4, and 10% of 02. What additional volume of 02 at STP will be required to completely burn 200 cc of this -[2] gaseous mixture at 27°C and 750mm pressure?
Answer: In 200 cm3 of gaseous mixture, volume of H2 ,\(=\frac{50}{100} \times 200=100 \mathrm{~cm}^3 \text {; }\)

Volume of CH4 \(=\frac{40}{100} \times 200=80 \mathrm{~cm}^3 \text { and }\)

Volume Of O2 \(=\frac{10}{100} \times 200=20 \mathrm{~cm}^3\)

Now, relictions due to combustion—

⇒ \(\begin{aligned}
& 2 \mathrm{H}_2+\mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O} \text { and, } \\
& 2 \text { volume } \quad \begin{array}{l}
\text { I volume } \\
100 \mathrm{~cm}^3 \\
50 \mathrm{~cm}^3
\end{array}
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{CH}_4+2 \mathrm{O}_2 \longrightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} \\
& 1 \text { volume } 2 \text { volume } \\
& 80 \mathrm{~cm}^3 \quad 160 \mathrm{~cm}^3 \\
&
\end{aligned}\)

Volume of the 02 required for complete burning of II2 and (111,| =(50+160) = 210cm3 Again, the volume of the 02 present in the mixture at 27C(! and 750 mm pressure = 20cm3.

Therefore, the additional volume of 02 required for complete combustion = (210- 20) = 190cm3.

Let, the volume ofadditional 02 he V cm3(atSTP)

⇒ \(\quad \frac{750 \times 190}{(273+27)}=\frac{760 \times V}{273}\)

Or, \(V=\frac{750 \times 190 \times 273}{300 \times 760}=170.625 \mathrm{~cm}^3\)

Therefore, the additional volume of 02 required at STP will he = 170.625cm3

Question 7. 25 ml, of a mixture containing nitrogen and nitric oxide is passed over heated copper. The volume of the gaseous mixture becomes 20mL. What is the percentage composition of the original mixture. [All volumes are measured at the same temperature and pressure.]
Answer: When the mixture of N2 and NO is passed over heated copper, NO is reduced to N2 but N2 remains unaffected in the reaction.

Let, the volume of NO in the original mixture =rmL

∴  The volume of N2 in the original mixture = (25- x) mL

⇒ \(\begin{array}{cc}
\mathrm{Cu}+ & \mathrm{NO} \longrightarrow \mathrm{CuO} \\
\begin{array}{c}
1 \text { volume } \\
x \mathrm{~mL}
\end{array} & \frac{1}{2} \mathrm{~N}_2 \\
& 1 / 2 \text { volume } \\
& x / 2 \mathrm{~mL}
\end{array}\)

Volume of the gas mixture at the end ofreduction = volume of N2 produced by reduction of NO + volume of N2 already present in the mixture \(=\left(\frac{x}{2}+25-x\right)=\left(25-\frac{x}{2}\right) \mathrm{mL} .\)

According to the given data, volume of the gaseous mixture at the end ofreduction =20 mL.

∴ \(\quad 25-\frac{x}{2}=20 \quad \text { or, } x=10\)

In the original mixture, the volume of NO = 10 ml, and the volume of N2 = (25- 1 0) =15 mL.

In the original mixture, percentage of NO \(=\frac{10}{25} \times 100=40\)

Therefore In the original mixture, percentage \(\mathrm{NO}=\frac{10}{25} \times 100=40\) and percentage of N2 \(=\frac{15}{25} \times 100=60 .\)

Question 8. Combustion of 1 volume of a compound (contains of C,H and N)In air produces 3 volumes of C02 and 4.5 volumes water vapour and 0.5 volume of N2. Calculate the molecular formula of the compound. [All volumes are measured at same temperature and pressure].
Answer: Let the formula of the compound be CxHyNz. It undergoes combustion according to the following equation

⇒ \(\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y \mathrm{~N}_z+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \longrightarrow x \mathrm{CO}_2+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O}+\frac{z}{2} \mathrm{~N}_2 \\
& 1 \text { volume }\left(x+\frac{y}{4}\right) \text { volume } x \text { volume } \quad \frac{y_2}{2} \text { volume } \frac{z}{2} \text { volume }
\end{aligned}\)

∴ x = 3,y=9,z=l

So, formula of the compound is: C3H9N

Determination of molecular formula of gaseous hydrocarbons from eudiometry

A known volume of the gaseous hydrocarbon is mixed with an excess of oxygen and exploded by sparking inside the closed limb of an eudiometer tube.

As a result, the hydrocarbon suffers oxidation to produce carbon dioxide (C02) and steam (H2O ).

As the combustion is completed, the gas mixture is cooled down to room temperature under atmospheric pressure.

As a result, steam condenses to water (liquid) but gaseous C02 and unused Oz remain unchanged.

At this stage, there usually occurs a contraction in volume of the gas mixture. This is called first contraction.

Such contraction in volume is due to two reasons—

On coolingdown the reaction mixture, steam condenses to water (having negligible volume).

Due to reaction between hydrocarbon and oxygen, the entire amount of the hydrocarbon and a majorportion of the oxygen disappear.

Then cone. KOH (or NaOH) solution is added to the reaction mixture. As a result, a second contraction in volume takes place because the whole amount of C02 is absorbed by the KOH solution.

∴ \(2 \mathrm{KOH}+\mathrm{CO}_2 \rightarrow \mathrm{K}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O} .\)

The second contraction in the volume is equal to the volume of C02 produced by the combustion of the hydrocarbon.

Now, the unused oxygen is the only gas left in the eudiometer. The volume of unused oxygen is measured by absorbing it with alkaline pyrogallatc solution.

If the above experiment is carried out with a compound composed of carbon, hydrogen and nitrogen instead ofa hydrocarbon, then the gas left in the eudiometer after second contraction consists ofnitrogen (N2) and unused oxygen (O2).

Volumes of all gases are measured at the same temperature and pressure.

The empirical formula of the hydrocarbon is determined

From the knowledge of the following quantities:

  1. initial volume of the gas mixture
  2. magnitude of first contraction and
  3. Magnitude Of Second Contraction.

It should be remembered that the volume of 02 consumed is equal to the volume of C02 produced.

Furthermore, the volume of hydrogen required to form water is twice the volume of 02 consumed. This hydrogen originates from die hydrocarbon.

Calculation: Volume of hydrocarbon + volume of 02 taken volume of C02 + volume of unused 02 + H20 (becomes liquid on cooling)

First contraction in volume =Total volume of gas mixture before electric sparking-total volume of gas mixture (under cold condition) after electric sparking.

= (Volume of hydrocarbon + volume of 02 taken) – (volume of C02 generated by oxidation + volume of unused oxygen) [volume ofwater (liquid) is taken as zero].

= Volume of hydrocarbon + (volume of 02 takenvolume ofunused oxygen)- volume of C02 generated by the oxidation = (Volume of hydrocarbon + volume of oxygen consumed) -volume of C02 generated.

Again, second contraction in volume = volume of C02 generated by oxidation.

So, first contraction = (volume of hydrocarbon + volume of 02 consumed)- second contraction.

1st contraction + 2nd contraction = volume of hydrocarbon + volume of 02 consumed or, volume of 02 consumed = 1st contraction + 2nd contraction- volume ofhydrocarbon.

Example Let the formula of a hydrocarbon is Cxliy. The oxidation reaction ofthe hydrocarbon is represented as—
Answer:

⇒ \(\begin{array}{ccc}
\mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 & x \mathrm{CO}_2+\frac{y_{\mathrm{H}_2} \mathrm{O}}{2} \\
\text { 1 volume } \quad\left(x+\frac{y}{4}\right) \mathrm{mol} & x \mathrm{~mol} & \frac{y}{2} \mathrm{~mol} \\
\text { 1volume } \quad\left(x+\frac{y}{4}\right) \text { volume } & x \text { volume } & 0 \\
v \text { volume } v\left(x+\frac{y}{4}\right) \text { volume } & v x \text { volume } & 0
\end{array}\)

[The volume ofwaterin the liquid state = 0 ] Hence, volume of 02 required for complete combustion of v volume ofthe hydrocarbon \(=v\left(x+\frac{y}{4}\right)\)

Now, the volume of 02 used= (firstvolume contraction + second volume contraction)-volume ofthe hydrocarbon therefore \(v\left(x+\frac{y}{4}\right)\) = first volume contraction =second volume contraction)-V

Volume of the C02 produced by complete combustion of v volume of the hydrocarbon —vx= second contraction.

Therefore, if first volume contraction and second volume contraction as well as the volume of the gaseous hydrocarbon are known, x and y may be evaluated from equations 1 and 2 and hence the formula of the hydrocarbon (CxHy) can be determined easily.

Sometimes in the determination ofmolecular formula of a gaseous hydrocarbon by eudiometric method, the empirical formula ofthe compound is obtained from the experimental results.

In such a case, the molecular formulais determined by ascertaining the vapour density or molecular mass ofthat compound.

General equations for combustion reaction of different compounds.

The problems related to determination of molecular formula of gaseous hydrocarbons with the help of eudiometry are of three types. These are explained separately with examples.

Type-1: When first and second contractions and volume of mixed oxygen are known.

Question 1. 20cm3 of a hydrocarbon mixed with 66cm3 of oxygen is exploded. After cooling the gaseous mixture, the volume becomes 56cm3. The volume of this mixture when shaken with KOH solution reduces to 16cm3. Determine the formula of the hydrocarbon. [All volumes are measured at the same temperature andpressure.
Answer: Let, the formula ofthe hydrocarbonis. The reaction takes place due to explosion—

⇒ \(\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \longrightarrow x \mathrm{CO}_2+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O} \\
& \begin{array}{llll}
1 \text { volume } & \left(x+\frac{y}{4}\right) \text { volume } & x \text { volume } & 0 \\
20 \mathrm{~cm}^3 & 20\left(x+\frac{y}{4}\right) \mathrm{cm}^3 & 20 x \mathrm{~cm}^3 & 0
\end{array} \\
&
\end{aligned}\)

since the volume ofwaterin the liquid state is zero.

∴ First contractionin this reaction \(=\left[20+20\left(x+\frac{y}{4}\right)-20 x\right]\)

As given, first contraction = (20 + 66)- 56 = 30 cm3.

(20 + 5y) = 30 or, y = 2

Now, second contraction = (56- 16) = 40cm3 = volume of C02 produced. Therfore 20x = 40 or, x = 2

∴ The formula ofthe hydrocarbon is C2H2.

Type-2: When first and second contractions are known but volume ofmixed oxygen is unknown.

Question 2. 20cm3 ofa gaseoushydrocarbon mixed with excess of oxygen is exploded. A contraction in the volume of 30cm3 takesplace. On treating theproduced gaseous mixture with KOH solution, it suffers a further contraction of 40cm3. Determine the molecular formula of the hydrocarbon. [All volumes are measured at the same temperature andpressure.]
Answer: Let the formula ofthe gaseous hydrocarbon = Cxliy. The reaction involvingits oxidation

⇒ \(\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2+\frac{y_2}{\mathrm{H}_2 \mathrm{O}} \\
& \begin{array}{llll}
1 \text { volume } & \left(x+\frac{y}{4}\right) \text { volume } & x \text { volume } & 0 \\
20 \mathrm{~cm}^3 & 20\left(x+\frac{y}{4}\right) \mathrm{cm}^3 & 20 x \mathrm{~cm}^3 & 0
\end{array} \\
&
\end{aligned}\)

since water occupies negligiblevolume., i.e., zero]

In this reaction, contractionin volume.

⇒ \(\begin{aligned}
& =\left[20+20\left(x+\frac{y}{4}\right)-20 x\right] \\
& =(20+5 y) \mathrm{cm}=\text { first contraction }
\end{aligned}\)

According to the given data, first contraction = 30 cm3
20 + 5y = 30 or y=2

Again, second contraction = 40cm3 = volume of CO,. Now, from the equation, it can be seen that volume of C02 evolved = 20x cm3.

therefore 20x = 40

therefore x=2

Hence, formula ofthe hydrocarbon is C2H2.

Type-3: When first contraction and vapour density of hydrocarbon are knownbut second contraction and volume ofmixed oxygen areunknown.

Question 3. 3cm3 of a gaseous hydrocarbon is exploded with excess of oxygen. On cooling the mixture, the observed contraction is found to be 6cm3. Vapour density of the hydrocarbon is 14. What is the molecular formula ofthehydrocarbon?
Answer: Let the formula ofthe hydrocarbon be Reaction due to explosion—

⇒ \(\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O} \\
& \text { 1 volume }\left(x+\frac{y}{4}\right) \text { volume } x \text { volume } \quad 0 \\
& \begin{array}{llll}
3 \mathrm{~cm}^3 & 3\left(x+\frac{y}{4}\right) \mathrm{cm}^3 & 3 x \mathrm{~cm}^3 & 0
\end{array} \\
&
\end{aligned}\)

Therefore On cooling, water vapour is condensed to liquid (water) whose volume is assumed to be zero.]

∴ Contractionin this reaction = \(\begin{aligned}
& =\left[3+3\left(x+\frac{y}{4}\right)-3 x\right] \\
& =3+\frac{3 y}{4} \mathrm{~cm}^3=1 \text { st contraction }
\end{aligned}\)

∴ \(3+\frac{3 y}{4}=6\) or y=4

Its molecular mass = 2 x 14 =28

since V.D = 14

The molecular formula ofthe hydrocarbon = CXH4.

Its molecular mass =(12x+ 4)

Hence, 12x+ 4 = 28 or, x = 2

Therefore, formula of the hydrocarbon is C2H4.

Question 1. A gaseous hydrocarbon of volume 10mL at STP is mixed with 80mL of 02 and burnt completely. As a result, the volume of the gaseous mixture was reduced to 70mL. On treating the obtained gaseous mixture with KOH solution, the volume becomes 50mL. Determine the molecular formula of the hydrocarbon.
Answer: At STP, 10 mL ofa gaseous hydrocarbon mixed with 80 mL
of 02 is burnt completely. The following reaction takes place—

⇒ \(\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O} \\
& \text { 1 volume }\left(x+\frac{y}{4}\right) \text { volume } x \text { volume } \quad 0 \\
& 10 \mathrm{~mL} \quad 10\left(x+\frac{y}{4}\right) \mathrm{mL} \quad 10 x \mathrm{~mL} \quad 0 \\
&
\end{aligned}\)

since volume of water in the liquid state is assumed to be zero.

therefore volume contraction in this reaction

⇒ \(=10+10\left(x+\frac{y}{4}\right)-10 x=10+2.5 y \mathrm{~mL}=\text { first contraction }\)

In this case, first contraction =[(10 + 80)- 70] = 20 mL

∴ 10 + 2.5y = 20 or, y = 4.

Again, second contraction= volume of C02 produced.

So, second contraction = (70- 50) mL = 20 mL 10x = 20 i.e., x = 2 Hence, the formula ofthe hydrocarbonis C2H4.

Question 2. Volume of a gaseous hydrocarbon is 1.12L at STP. Whenitis completelyburntin air, 2.2 gof C02 & 1.8 g of water are formed. Find the volume of required at STP and also mass of the compound taken. Give the molecular formula of hydrocarbon.
Answer: Let the formula of the hydrocarbon be CzUy.

The concerned reaction of combustion is—

⇒ \(\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \longrightarrow x \mathrm{CO}_2+\underset{2}{\chi_2} \mathrm{H}_2 \mathrm{O} \\
& 1 \mathrm{~mol} \quad\left(x+\frac{y}{4}\right) \mathrm{mol} \quad x \mathrm{~mol} \quad y_2 \mathrm{~mol} \\
&
\end{aligned}\)

Now, the volume of the hydrocarbon = 1.12 L.

No. ofmoles ofhydrocarbon in 1.12L at STP \(P=\frac{1 \times 1.12}{22.4}=0.05\)

From reaction it is found that 1 mol of gaseous hydrocarbon produces x mole of C02 and y/2 mole of H20.

therefore 0.05mol ofhydrocarbon produces 0.05x mol of C02 and \(\frac{0.05 y}{2}\) mol of H2O. Now 0.05x mol of C02= 0.05* x 44g= 2.2xg of C02 and \(\frac{0.05 y}{2}\) X 18g = 0.45g mol H2O

According to the given data, 2.2x = 2.2 , 0.45y =1.8 y = 4 and x = 1; the formula ofthe hydrocarbon = CH4 ixnoi Now at STP, 1.12L ofhydrocarbon =0.05 mol

Mass ofthe hydrocarbon taken = (0.05 x 16) = 0.8 g [ v Molecular mass ofhydrocarbon, CH4 = 16]

Again, the amount of 02 required for the combustion of 1 mol of hydrocarbon \(=\left(x+\frac{y}{4}\right) \mathrm{mol}=\left(1+\frac{4}{4}\right) \mathrm{mol}=2 \mathrm{~mol}\)

O2 required for combustion of 0.05 mol ofhydrocarbon = (0.05 x 2) = O.lmol

Now, Volume of1 mol of 02 at STP = 22.4 L

Therefore Volume of 0.1 mol of 02 at STP = 2.24 L. So, the volume

of 02 required for combustion ofhydrocarbon is 2.24 L.

Question 3. 5 mL of a gas composed of hydrogen and carbon is mixed with 30mL of oxygen and exploded with electric sparking. The volume of the gas mixture, obtained by explosion is found to be 25 mL. KOH is then added to the mixture and as a result, its volume is reduced to 15 mL. The residual gas is purely oxygen. All volumes have been measured at STP. Whatis the molecular formula ofthe gas?
Answer: Assuming the formula ofthe gas to be CÿHÿ, the reaction occured during explosion is represented as

⇒ \(\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \longrightarrow x \mathrm{CO}_2+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O} \\
& \begin{array}{cccc}
1 \text { volume } & \left(x+\frac{y}{4}\right) \text { volume } & x \text { volume } & 0 \\
5 \mathrm{~mL} & 5\left(x+\frac{y}{4}\right) \mathrm{mL} & 5 x \mathrm{~mL} & 0
\end{array} \\
& {[\text { volume of water (liquid) }=0 \text { ] }} \\
&
\end{aligned}\)

since volume of water liquid)=0]

First contraction \(=\left[5+5\left(x+\frac{y}{4}\right)-5 x\right]=(5+1.25 y) \mathrm{mL}\)

As per given data, first contraction = (5 + 30)- 25 =10 mL

Therefore 5 + 1.25y = 10 or, 1.25y = 5

Second contraction = volume of the C02 formed = 5x mL According to the given condition, second contraction = (25- 15) = 10 mL.

Therefore, 5x = 10 or, x = 2

Thus, formula oftire gaseous compound = C2H

Question 4. 10 mL of a gaseous organic compound composed of carbon, hydrogen and oxygen is mixed with 100 mL of oxygen and subjected to explosion. Volume of tire mixtureproducedby explosion when cooledbecomes 90mL. On treatment with KOH, the volume is reduced by 20mL. Mass of1 L ofgaseous organic compound is 2.053 g at STP. Determine the molecular formula.
Answer: Let, the formula ofthe organic compound =

The mass of 1 L ofgaseous organic compound at STP = 2.053 g

Mass of 22.4L ofgaseous organic compound at STP = (2.053X22.4) =45.9872 g.

The reaction involved is as follows—

⇒ \(\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y \mathrm{O}_z+\left(x+\frac{y}{4}-\frac{z}{2}\right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O} \\
& 1 \text { volume. }\left(x+\frac{y}{4}-\frac{z}{2}\right) \text { volume } x \text { volume } 0 \\
& 10 \mathrm{~mL} \quad 10\left(x+\frac{y}{4}-\frac{z}{2}\right) \mathrm{mL} \quad 10 x \mathrm{~mL} \quad 0 \\
&
\end{aligned}\)

In the reaction, volume contraction= 10 + 10 \(\left(x+\frac{y}{4}-\frac{z}{2}\right)-10 x\)

= (10 + 2.5y- 5z) mL = first contraction x mL

In this case, first contraction = [(10 + 100)- 90] = 20 mL.

Therefore, 10 + 2.5y-5z = 20 or, 2.5y-5z =10 – [l]

Again, the second contraction in the mixture =20mL

= the volume of the C02 formed. From the equation, it is observed that the volume ofthe C02 produced = lOx.

Therefore, lOx = 20 or, x = 2

Now, the molecular mass of CxHy02 =(12x+ y+ 16z)

12.v+ y + 16z =45.9072 or, 12×2+y+16z =45.9872

or, y + 16z = 21.9072

Solving equations 1 & 3 we have, y = 6 and z = 1

Therefore The formula ofthe organic compound is C2H60.

Question 5. When 3 volume of a gaseous organic compound of carbon, hydrogen & sulphur mixed with excess oxygen is exploded, 3 volumes of carbon dioxide, 3 volumes of sulphur dioxide and 6 volumes of water vapour arc produced. What is the formula of the compound?
Answer: Let, the formula of the organic compound is On combustion, the following reaction takes place—

⇒ \( \mathrm{C}_x \mathrm{H}_y \mathrm{~S}_z+\left(x+\frac{y}{4}+z\right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O}+z \mathrm{SO}_2\)

⇒ \(1 \text { volume }\left(x+\frac{y}{4}+z\right) \text { volume } x \text { volume } \frac{y}{2} \text { volume } z \text { volume }\)

⇒ \(3 \text { volume } 3\left(x+\frac{y}{4}+z\right) \text { volume } 3 x \text { volume } \frac{3 y}{2} \text { volume } 3 z \text { volume }\)

According to the given condition, 3x = 3 or, x = 1; \(\frac{3 y}{2}\)

= 6 or, y = 4 and 3z = 3 or, z = 1

The formula ofthe organic compound = CH4S.

Question 6. When an acetylenic hydrocarbon, in presence of excess oxygen is exploded, it shows a contraction in volume by 50mL. A further contraction of 75mL is observed when the obtained gas mixture comes in contact with KOH solution. Determine the molecular formulaofthe compound.
Answer: Let the formula be CnH2 and its volume = x mL

Reaction of the acetylenic compound with O2

⇒ \(\begin{aligned}
& \mathrm{C}_n \mathrm{H}_{2 n-2}+\left(\frac{3 n-1}{2}\right) \mathrm{O}_2 \longrightarrow n \mathrm{CO}_2+(n-1) \mathrm{H}_2 \mathrm{O} \\
& 1 \text { volume }\left(\frac{3 n-1}{2}\right) \text { volume } n \text { volume } 0 \\
& x \mathrm{~mL} \quad x\left(\frac{3 n-1}{2}\right) \mathrm{mL} \quad n x \mathrm{~mL} \quad 0 \\
&
\end{aligned}\)

Contraction in volume in this reaction

⇒ \(=x+x\left(\frac{3 n-1}{2}\right)-n x=\frac{x+n x}{2}=\frac{x(1+n)}{2} \mathrm{~mL}\)

This volume contraction is equal to the first contraction of the mixture. Therefore, \(\frac{x(1+n)}{2}\) = 50 or, x+nx=100

As per given data, the second contraction of the mixture = 75 mL =the volumeofthe C02 formed. So, nx= 75

From equations 1 and 2 we have, x = 25 and from equation 2 putting x = 25 we have, n = 3.

Formula ofthe acetylenic compound = C3H6-2 = C3H4

Determination Of Molecular Formula Of Other Gaseous Compounds With The Help Of Eudiometry

Type-1: When two elementary gases of known atomicity react together to form a gaseous compound, the molecular formula of that compound may be ascertainedif the volumes of the reacting gases and the product at STP are known.
Let, 1 L of A2 gas (diatomic) and 1 L of B2 gas (diatomic) react to produce 2L of a gaseous compound X at STP. So,

⇒ \(\begin{array}{ccc}
\mathrm{A}_2 & \mathrm{~B}_2 & 2 \mathrm{X} \\
1 \mathrm{~L} & 1 \mathrm{~L} & 2 \mathrm{~L} \text { (at STP) } \\
n \text { molecules } & n \text { molecules } & 2 n \text { molecules } \\
1 \text { molecule } & 1 \text { molecule } & 2 \text { molecules } \\
1 \text { atom } & 1 \text { atom } & 1 \text { molecule } \\
& \text { [according to Avogadro’s hypothesis] }
\end{array}\)

Hence, in 1 molecule of X, 1 atom of each A2 and B2 are present.

∴ The molecular formula of the compound is AB.

Suppose, 1 L of X2 gas (diatomic) and 2L of Y2 gas (diatomic) react to produce1 L of Z gas at STP. So,

⇒ \(\begin{gathered}
\mathrm{X}_2 \\
1 \mathrm{~L}
\end{gathered}+\underset{2 \mathrm{~L}}{2 \mathrm{Y}_2} \longrightarrow \mathrm{Z} \underset{\mathrm{L}(\text { at STP) }}{\mathrm{Z}}\)

⇒ \(\begin{array}{ccc}
n \text { molecules } & 2 n \text { molecules } & n \text { molecule } \\
1 \text { molecule } & 2 \text { molecules } & 1 \text { molecule } \\
2 \text { atoms } & 4 \text { atoms } & 1 \text { molecule }
\end{array}\)

Therefore in 1 molecule of Z gas, 2 atoms of X2 and 4 atoms of Y2 are present.

Molecular formula of Z gas is X2Y4.

Type 2. If the atomicity of the elementary gases produced by the decomposition of any gaseous compound and also the volumes of the reactants and the products at STP are known, the molecular formula of the gaseous compound maybe determined.
Example: Suppose at STP, 2 L of a gaseous compound decomposes to form1 L of A2 gas and 3 L of B2 gas (both A2 and B2 are diatomic).

∴ \(\begin{gathered}
2 \mathrm{X} \\
2 \mathrm{~L}
\end{gathered} \longrightarrow \underset{1 \mathrm{~L}}{\mathrm{~A}_2}+\underset{3 \mathrm{~L}}{3 \mathrm{~B}_2}\)

∴ \(\begin{array}{ccc}
\text { 2n molecules } & \text { n molecules } & 3 \text { n molecules } \\
2 \text { molecules } & 1 \text { molecule } & 3 \text { molecules } \\
1 \text { molecule } & \frac{1}{2} \text { molecules } & \frac{3}{2} \text { molecules } \\
1 \text { molecule } & 1 \text { atom } & 3 \text { atoms }
\end{array}\)

since A2 and B2 are diatomic]

So,1 molecule of X gas contains1 atom of A2 and 3 atoms of B2 gas. Hence, the molecular formula of X gas is AB3.

Numerical Examples 

Question 1. At high temperature, gaseous compound, S4N4 decomposes to produce N2 and sulphur vapour. 1 vol. of S4N4 on decomposition gives 2.5 vol. of gaseous mixture at STP. Determine formula of sulphur.
Answer: Thermal decomposition: S4N4→2N2 + sulphur vapour 1 vol. of S4N4 produces 2.5 vol. of gaseous mixture.

From the equation,itis observed that 2 volumes of N2 are produced from 1 volume of S4N4 at STP. So, during thermal decomposition at STP, volume of the sulphur vapour produced = (2.5- 2) = 0.5 volumes. At STP, decomposition of 1 volume S4N4 results in evolution of2volumes N2 and 0.5 volumes sulphur vapour.

or, 1 molecule of S4N4 on decomposition gives 2 molecules of N2 and 0.5 molecules of sulphur.

1 molecule of S4N4 produces 4 N-atoms and 4 S-atoms.

∴ In 0.5 molecule ofsulphur, number of S-atoms = 4

∴ In 1 molecule of sulphur, number ofS-atoms = 8

Therefore Formula of sulphur = S8.

Question 2. When 100mL of ozonised oxygen is shaken with turpentine oil, volume decreases by 20mL. When lOOmL of the same sample is heated, the mixture occupies a volume of 110mL. Determine the molecular formula of ozone. [All volumes are measuredunder same temperature andpressure.]
Answer: Turpentine oil when added to ozonised oxygen, absorbs ozone and oxygen is leftbehindin the mixture.

∴ Volume of ozone in 100mL ofozonised oxygen = 20 mL.

∴ Volume of oxygenin 100 mL of ozonised oxygen = 80 mL

When Ozone Mixed With O2 is heated, it undergoes thermal decomposition to produce 02.

Volume of the gas obtained by heating 100mL of ozonised oxygen 110mL of this gas, (110-80) = 30 mL of 02 have been obtained by decomposing 20 mL of ozone.

Under the same conditions of temperature and pressure, 20mL ozone produces 30mL of 02.

i.e., 2 volumes of ozone produce 3 volumes of 02.

So, 2 molecules ofozone produce 3 molecules of  O2 .

∴ 1 molecule of ozone produces 3/2 molecule of oxygen or [ v oxygen is diatomic]

Hence,1 molecule ofozone contains 3 oxygen atoms.

∴ Molecular formula of ozone is 03.

Calculations Involving Reactions Occurring In Solution

Methods of expressing concentration of solutions

A solution is a homogeneous mixture of two or more substances which do not react chemically with each other.

A solution consisting of only two components is called a binary solution. Some important modes of expressing concentration ofsolutions are indicated below.

Percentage strength It is defined as the amount of solute in grams presentin lOOg of the solution.

⇒ \(\%(W / W)=\frac{\text { mass of solute }(\mathrm{g})}{\text { mass of solution }(\mathrm{g})} \times 100\)

Example: Percentage strength (W/W) of a glucose solution is 10 indicates that lOg glucose is presentin lOOg ofsolution.

Strength in gram per litre: The strength of a solution is defined as the amount of the solute in grams present per litre ofthe solution.

Unit: g/L or g/dm3, i.e., g-L-1 or g-dm-3.

Example: If 4.0g of sodium carbonate is dissolved in 1L of sodium carbonate solution, the strength of the solution willbe 4.0 g-L-1 or 4.0 g-dm-3 Temperature, strengthin g-L 1 depends on temperature.

Mole fraction: The mole fraction of any componentin the solution is equal to the number of moles of that component dividedby the total number ofmoles of all the components.

Let us consider, n2 moles of a solute dissolvedin n 4 moles of a solvent.

Mole fraction of the solute and mole fraction ofUie solvent in the solution respectively.

⇒ \(x_2=\frac{n_2}{n_1+n_2} \& x_1=\frac{n_1}{n_1+n_2}\)

⇒ \(\text { Now, } x_1+x_2=\frac{n_1}{n_1+n_2}+\frac{n_2}{n_1+n_2}=\frac{n_1+n_2}{n_1+n_2}=1\)

The sum of mole fractions ofthe components is equal to 1.

Therefore mole fraction ofthe solvent, xx = l-x2

mole fraction of the solute, x2 = 1- x1

Effect of temperature: As mole fraction is a number, it is independent oftemperature.

PPm (partspermillion): The concentration ofvery dilute solution is expressed in terms of parts of the solute by mass present in million parts by mass of the solution (or ppm), i.e., ppmx \(=\frac{\text { mass of } x}{\text { Mass of solution }} \times 10^6\)

Pollution of the atmosphere is also expressed in terms of ppm but in that case we use volumes in place of masses i.e., volume (in cm3) of the harmful gases (i.e., S02 etc.) presenti n 106 cm3 ofthe air.

Molarity (M): It is a defined as the number ofgram-moles ofthe solute presentin1 L (or, 1000 mL) of the solution.

⇒ \(\begin{aligned}
& \text { Molarity }(\mathrm{M}) \text { of solution }=\frac{\text { number of moles of solute }}{\text { volume of the solution (in } \mathrm{L})} \\
& =\frac{\text { mass of the solute }(\mathrm{g})}{\text { gram-molecular mass of solute } \times \text { volume of solution (in } \mathrm{L})}
\end{aligned}\)

Example: If three different solutions are such that1 L of each of these solutions contain 1 mol, 0.5 mol and 0.1 mol of a dissolved solute then the molarity of the solutions will be 1, 0.5 and 0.1 respectively. Similarly, if 5L of a glucose solution contains 4 gram-moles of glucose then molarity ofthe solution will be,

⇒ \(\frac{\text { number of moles of glucose }}{\text { volume of the solution }(\mathrm{L})}=\frac{4}{5}=0.8(\mathrm{M})\)

Units ofmolarity: moles perlitre (mol-L-1 ) or moles per cubic decimetre (mol-dm’3 ) [1L = 1000 cm3 = 1dm3 ]

Molar solution: A molar solution is defined as a solution, 1 L ofwhich contains 1 gram-mole ofthe solute.

Example: 1L of sulphuric acid (H2S04) solution containing 98 g (i.e., 1 gram-mole) of H2S04 is called 1(M) H2S04 solution.

Similarly, 1 L of1(M) Na2C03 solution contains 106 g (i.e., 1 gram-mole) ofsodium carbonate. Molar, semi-molar, deci-molar and centi-molar solutions are denoted by1(M

⇒ \(\left(\frac{\mathrm{M}}{2}\right),\left(\frac{\mathrm{M}}{10}\right) \text { and }\left(\frac{\mathrm{M}}{100}\right)\) respectively.

Effect of temperature on the molarity of a solution: The mass of solute is independent of temperature but the volume of solution is dependent on temperature and hence the molarity of a solution depends on temperature.

Formality (F): It is defined as the number of gramformula mass of the dissolved solute present per litre of the solution.

⇒ \(\begin{aligned}
& \text { Formality }(F)=\frac{\text { number of gram-formula mass of solute }}{\text { volume of the solution (in } \mathrm{L} \text { ) }} \\
& =\frac{\text { mass of the solute }(\mathrm{g})}{\text { gram-formula mass of solute } \times \text { volume of solution (in } \mathrm{L})}
\end{aligned}\)

Example: If1 L of a magnesium chloride solution contains 190 g of MgCl2 then this solution is said to be 2.0 formal solution (t.e., 2.0 F solution) because the formula mass of MgCl2 is 95.

Formal solution: It is defined as a solution, 1 L of which contains 1 gram-formula mass of the solute.

Example: 1 L of 1(F) solution of calcium chloride (CaCI2) contains lllg {i.e., 1 gram-formula mass) of calcium chloride.

Effect of temperature on the formality of solution: The mass of a solute is independent of temperature but the volume of solution is dependent on temperature and hence the formality of a solution depends on temperature.

Molality (m): The molality of a solution is defined as the number of gram-moles of the solute dissolved in 1000 g {i.e., 1 kg) ofthe solvent.

⇒ \(\begin{aligned}
& \text { Molality }(\mathrm{m}) \text { of solution } \\
& =\frac{\text { number of moles of dissolved solu }}{\text { mass of the solvent }(\mathrm{kg})} \\
& =\frac{\text { number of moles of solute } \times 10}{\text { mass of the solvent }(\mathrm{g})} \\
& =\frac{\text { mass of solute }(\mathrm{g}) \times 1000}{\text { gram-molecular mass of solute } \times \text { mass of solvent }(\mathrm{g})}
\end{aligned}\)

Example: If 40 g of NaOH [i.e., 1 g-mole NaOH) is dissolved in lOOOg of water, then the resulting solution is said to be 1 molal solution [i.e., l(m) solution] of NaOH. Again, dissolution of 4g ofNaOH {i.e., 0.1 g-mole NaOH) in 1000 g water gives 0.1 molal [i.e., 0.1 (m)] solution of NaOH.

Unit of molality: moles perkilogram ( mol-kg-1)

Effect of temperature on molality of a solution: As the masses of substances are independent of temperature, so the molality ofa solution does not depend on temperature.

Molal solution: A molal solution is defined as a solution, which is prepared by dissolving 1 gram-mole of solute in 1000 g of solvent.

Example: If 98 g(i.e., 1 gram-mole) ofsulphuric acid is dissolved in 1000 g of water, then a molal solution of H2S04 [i.e., l(m) H2S04 solution] is o btained.

Molal, semimolal, deci-molal and centi-molal solutions denoted by l(m)

⇒ \(\left(\frac{\mathrm{m}}{2}\right),\left(\frac{\mathrm{m}}{10}\right) \text { and }\left(\frac{\mathrm{m}}{100}\right)\)

Normality (N): Normality of a solution is defined as the number ofgram-equivalents of the solute presentin1 L (or 1000 mL) of the solution.

⇒ \(\begin{aligned}
& \text { Normality }(\mathrm{N}) \text { of a solution } \\
& =\frac{\text { number of gram-equivalents of solute }}{\text { volume of solution }(\mathrm{L})} \\
& =\frac{\text { mass of the solute }(\mathrm{g})}{\text { gram-equivalent mass of solute } \times \text { volume of solution (L) }}
\end{aligned}\)

If 1 L of H2S04 solution contains 2 gram-equivalent (i.e.,98 g) of pure H2S04 the strength of the solution will be 2(N).

Similarly, if 1 L of Na,C03 solution contains 0.1 gram-equivalent (i.e., 5.3 g) of Na2C03 the strength of the solution will be 0.1(N).

Effect of temperature on normality ofa solution: Mass of solute is independent of temperature but volume of a solution depends on it. So, normality of a solution is dependent on temperature.

Normal solution: A normal solution is defined solution, 1 L (or 1000 mL) of which contains 1 equivalent of the solute.

Examples: 1 L of a normal solution of H,S04 contains 1 gramequivalent or 49 g H2SO4 . 1 L of a normal solution of NaOH contains 1 gram-equivalent or 40 g NaOH.

Normality Equation

Let Vx mL of a solution of normality A/j be diluted to V2 mL so that its normality changes to N2. Since number of gram-equivalents of solute before and after dilution is ame, we can write, V1 X N1 =V2XN2. This is called normality equation.

Let V1 mL of a solution of compound ‘ A ’ having normality Nx reacts completely with V2 mL of a solution of compound ‘B’ having normality N2. Since the ompounds react with each other in equivalent amounts we can write, V1 X N1 = V2XN2 This is another normality equation.

Molarity equation

Let V1 mL of a solution of molarity M1 be diluted to V2 mL so that its molarity changes to M2.

Since number of gram-moles of solute before and after dilution is same, we can write, V1 X M1 = V2 X M2 This is molarity equation.

Let V1 mL of a solution of molarity be mixed with V2 mL ofanother solution (complex ofsame components) having molarity M2. If M be the molarity of the mixed solution, (V1 + V2 ) M=V1 x M1 + V2 X M2

This is a second molarity equation.

Let V1 mL of a solution of compound ‘A ‘ having molarity M2 reacts completely with V2 mL of another solution of compound having molarity M2 according to the equation: xA(aq) + yB(aq)pC(aq) + qD(aq)

It can be shown that \(\frac{V_1 \times M_1}{x}=\frac{V_2 \times M_2}{y}\)

This is a thirdmolarity equation.

Relationship between molarity and normality of the solution ofan acid ora base: Normality of solution of an acid =Molarity x Basicity ofthe acid.

Normality of a base = Molarity x Acidity ofthe base. e.g., 1 (M) H2S04 Solution = 2 (N) H2S04 solution 1 (M) Ca(OH)2 solution = 2 (N) Ca(OH)2 solution.

Normality= No. ofg- eq L-1 = Number ofmilli eq mL-1 Molarity of a solution = Number ofmoles perlitre = Number of millimoles per litre.

Numerical Examples

Question 1. NaOH is dissolved in water and the solution is madeupto 500mL. Find themolarity of solution.
Answer: Number ofmoles of NaOH = \(\frac{\text { given mass }}{\text { molecular mass }}=\frac{8}{40}=0.2\)

∴ Molarity(M) = \(=\frac{\text { number of moles of } \mathrm{NaOH}}{\text { volume of solution }(\mathrm{L})}=\frac{0.2}{0.5}=0.4\)

Question 2. The density of 3(M) solution ofNaCl is 1.25 g • cm 3. Calculate themolarityofthe solution.
Answer: Amount ofNaN03 in 1L solution=3 mol=3×58.5g

Mass of 1 L solution= 1000 x 1.25 g = 1250 g

Mass of the solvent [i.e., water) in 1L ofthe solution

= (1250-175.5)= 1074.5g

∴ Molality \(=\frac{\text { no. of moles of solute }}{\text { mass of solvent }(\mathrm{kg})}=\frac{3}{1.0745}=2.79(\mathrm{M})\)

Question 3. Calculate the volume of 0.5(M) H2S04 solution required to dissolve 0.5 g ofcopper(2) carbonate.
Answer: Reaction \(\underset{1 \mathrm{~mol}}{\mathrm{CuCO}_3}+\underset{1 \mathrm{~mol}}{\mathrm{H}_2 \mathrm{SO}_4} \rightarrow \mathrm{CuSO}_4+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}\)

∴ 1 mol CuC03 = 63.5 + 12 + 48 = 123.5 g

∴ 1 mole of H2S04 is required to dissolve 123.5 g of CuC03

∴ Amount of H2S04 required to dissolve 0.5 g of CuC03

⇒ \(=\frac{1 \times 0.5}{123.5}=0.004 \mathrm{~mol}\)

Now, 0.5 mol of H2S04 is present in 1000 mL solution.

∴ 0.004 mol H2SO4 is present \(\frac{1000 \times 0.004}{0.5}\) = 8ml. given H2S04 solution. So, volume of H2S04 required= 8 mL.

Question 4. Calculate the volume of H2 gas (at STP) liberated by thereaction ofexcessZnwith 500mL 0.5(N) H2S04.
Answer: Reaction involved \(\mathrm{Zn}+\underset{98 \mathrm{~g}}{\mathrm{H}_2 \mathrm{SO}_4} \rightarrow \underset{22400 \mathrm{~mL}}{\mathrm{ZnSO}_4}+\underset{\mathrm{H}_2}{\mathrm{H}_2}\)

H2S04 presentin 1000 mL of0.5(N) H2S04 =0.5 g-eqv.

∴ Amount of H2S04 present in 500 mL of 0.5(N) H2S0 solution \(=\frac{0.5 \times 500}{1000}\) g-equiv \(=\frac{0.5 \times 500 \times 49}{1000} \mathrm{~g}=12.25 \mathrm{~g}\)

98gH-,S04 reacts withZn to liberate 22400mLH, (STP).

∴ 12.25 g H,S04 reacts with zinc to liberate \(\frac{22400 \times 12.25}{98}=2800 \mathrm{~mL} \mathrm{H}_2 \text { (at STP) }\)

Density of 3 molal NaOH solution is 1.110 g mL-1. Calculate themolarity ofthe solution.

3 molal NaOH solution means 3 moles of NaOH are dissolvedin1 kg solvent. So, the mass ofsolution = 1000g solvent + 120g NaOH =1120g [v Molar mass of NaOH =(23 + 16 + 1) =40]

Volume of solution = \(=\frac{\text { Mass of solution }}{\text { Density of solution }}\)

or, Volume of solution \(=\frac{1120}{1.110 \mathrm{~g} \cdot \mathrm{mL}^{-1}}=1009 \mathrm{~mL}\)

∴ \(=\frac{\text { Mole of solute } \times 1000}{\text { Vol. of solution }(\mathrm{mL})}=\frac{3 \times 1000}{1009}=2.973(\mathrm{M})\)

Question 6. If 4 g of NaOH dissolvesin 36g of H2O, then calculate the mole fraction of each component in the solution. Also, determine the molarity of the solution (specific gravityofsolutionis1 g.mL-1 ).
Answer: \(n_{\mathrm{NaOH}}=\frac{4}{40}=0.1 \mathrm{~mol} ; n_{\mathrm{H}_2 \mathrm{O}}=\frac{36}{18}=2 \mathrm{~mol}\)

⇒ \(x_{\mathrm{NaOH}}=\frac{0.1}{0.1+2}=0.0476 ; x_{\mathrm{H}_2 \mathrm{O}}=\frac{2}{0.1+2}=0.9524\)

Total mass ofsolution= (4 + 36) = 40 g

Volume of solution \(=\frac{40 \mathrm{~g}}{1 \mathrm{~g} \cdot \mathrm{mL}^{-1}}=40 \mathrm{~mL}\)

Molarity \(=\frac{\text { Mole of solute } \times 1000}{\text { Vol. of solution }(\mathrm{mL})}\)

⇒ \(=\frac{0.1 \times 1000}{40}=2.5(\mathrm{M})\)

Question 7. 1L ofa (N/2) HC1 solution was heated in a beaker and it was observed that when the volume of solution got reduced to 600 mL, 3.25 g of HC1 was lost. Calculate the the normality of the resulting solution.
Answer: Normality \(\frac{\text { Mass of } \mathrm{HCl}}{\text { Equivalent mass } \times \text { Volume of solution (L) }}\)

⇒ \(\text { or, } \quad 0.5 \mathrm{eqv} \cdot \mathrm{L}^{-1}=\frac{\text { Mass of } \mathrm{HCl}}{36.5 \mathrm{~g} \cdot \mathrm{eqv}^{-1} \times 1 \mathrm{~L}}\)

Mass of HCl=0.5 eqv L-1 X 36.5 g – eqv-1 x 1L= 18.25g

Mass of HC1 left after heating = 18.25- 3.25 = 15.0 g

Volume of the resulting solution = 600ml \(=\frac{600}{1000}=0.6 \mathrm{~L}\)

therefore normality \(=\frac{15.0 \mathrm{~g}}{36.5 \mathrm{~g} \cdot \mathrm{eqv}^{-1} \times 0.6 \mathrm{~L}}=0.685\)

Long Question And Answers

Question 1. Two substances A are B are of equal masses and = 29.28 their molecular masses are in the proportion of 2:3. What is the ratio of the numbers of their molecules?
Answer:

From the question, MA: = 2:3; where MA and MB are the molecular masses of A and B respectively. Now, let the mass of A = mass of B = W g

Number of molecules present in gram-mole of any substance

= 6.022 x1023

Number of molecules in w g of A \(A=\frac{W}{M_A}\)= x 6.022 x 1023 and that in W g of B \(=\frac{W}{M_B}\)= x 6.022 x 1023

∴ \(\frac{\text { Number of molecules in } W \mathrm{~g} \text { of } A}{\text { Number of molecules in } W \mathrm{~g} \text { of } B}\)

\(=\frac{W}{M_A} \times 6.022 \times 10^{23} \div \frac{W}{M_B} \times 6.022 \times 10^{23}=\frac{M_B}{M_A}=\frac{3}{2}\)

Question 2. Show that, the ratio ofthe masses of equal volumes of two gases at the same temperature and pressure is directly proportional to the ratio of their molecular masses.
Answer:

Let the molecular masses of two gases, A and B be MA and MB respectively, and WA, and WB be their respective masses.

Now, number of moles of gas A = WA/MA

Number ofmoleculesin gas A = \(A=\frac{W_A}{M_A} \times\) x 6.022 x 1023

Again, number ofmoles ofgas B = WB/MB

Number of molecules in gas B = x 6.022 x 1023

Since the gases (having equal volumes) are at the same temperature and pressure, according to Avogadro’s hypothesis they will contain the same number of molecules.

∴ \(\frac{W_A}{M_A} \times 6.022 \times 10^{23}=\frac{W_B}{M_B} \times 6.022 \times 10^{23} \text { i.e., } \frac{W_A}{W_B}=\frac{M_A}{M_B}\)

Question 3. Which are heavier and lighter than air: O2, CO2, CH4, NH3, Cl2 ? [Volumetric composition of air N2 = 74%, O2 = 24%, CO2=2%
Answer:

Vapour density of a gas \(=\frac{\text { molecular mass of gas }}{\text { molecular mass of air }}\)

Volumetric composition of air: \(=\frac{(28 \times 74)+(32 \times 24)+(2 \times 44)}{100}=29.28\)

Therefore, concerning air, vapor densities of

⇒ \(\begin{aligned}
& \mathrm{O}_2=\frac{32}{29.28}>1 ; \mathrm{CO}_2=\frac{44}{29.28}>1 ; \mathrm{CH}_4=\frac{16}{29.28}<1 ; \\
& \mathrm{NH}_3=\frac{17}{29.28}<1 \text { and } \mathrm{Cl}_2=\frac{71}{29.28}>1
\end{aligned}\)

Therefore Gases lighter than air are CH4 and NH2 whereas gases heavier than air are O2, CO2, and CI2.

Question 4. The mass of a sulfur atom is twice that of an oxygen atom. Hence, the vapor density of sulfur will be twice that of the vapor density of oxygen—justify’.
Answer: The mass of an S-atom is twice that of an O-atom. So, the
atomic mass of sulfur is twice that of oxygen.

Now, Molar mass =n x atomic mass [n = atomicity ofmolecule] and vapor density

⇒ \(=\frac{\text { molecular mass }}{2}=\frac{n \times \text { atomic mass }}{2}\)

Therefore Vapour density of sulphur \(=\frac{n \times \text { atomic mass of sulphur }}{2}\)

Vapour density of oxygen \(=\frac{2 \times \text { atomic mass of oxygen }}{2}\)

since atomicity of oxygen molecule =2

⇒ \(\quad \frac{\text { Vapour density of sulphur }}{\text { Vapour density of oxygen }}\)

⇒ \(=\frac{n \times \text { atomic mass of sulphur }}{2 \times \text { atomic mass of oxygen }}=\frac{n}{2} \times 2=n\)

since the atomic mass of sulfur is twice of that of oxygen] or, Vapour density of sulfur =n x vapor density of oxygen.

Thus, it is quite logical to conclude that depending on the atomicity (n), the vapor density of sulfur maybe 2, 4, 6, or 8 times the vapor density of oxygen.

Question 5. How many significant figures are there in each of the following numbers:

  • 0.437
  • 935100
  • 2.158 x 104
  • 0.00839
  • 207.39
  • 17.00
  • 2.0100 x 104
  • 6.0 x 1023
  • 0.00070

Answer:

  1. 3 significant figures (4, 3, 7);
  2. 4 significant figures (9, 3, 5, 1);
  3. 4 significant figures (2, 1, 5, 8);
  4. 3 significant figures (8, 3, 9);
  5. 5 significant figures (2, 0 7, 3, 9);
  6. 4 significant figures (1, 7, 0, 0);
  7. 5 significant figures (2, 0, 1, 0, 0);
  8. 2 significant figures (6, 0);
  9. 2 significant figures (7,0)

Question 6. Express the following up to three Significant figures:

  • 4.309251
  • 49.793500
  • 0.005728
  • 7000
  • 2.67876 x 103
  • The decimal equivalent of 2/3,
  • one-millionth of one.

Answer: The digit in the second decimal place is increased by 1 unit and the remaining digits after that are dropped. Hence the number obtained is 4.31.

The digit in the first decimal place is increased by 1 unit and the remaining digits are dropped. The number obtained is 49.8.

The digit in the fifth decimal place is increased by 1 unit and the remaining digits are dropped. The number obtained is 0.00573.

The given no. is first expressed in the exponential form i.e. 7000 = 7.000 X 103. Then the third digit after the decimal point of the tire’s first factor is dropped. Thus the number obtained is 7.00 X 103.

The given number is in the exponential form. Thus the digit at the second decimal place is increased by 1 and all other digits of its right side are dropped. Thus the number obtained is 2.68 x 103.

The given fraction is first expressed in the decimal form. The digit at the third decimal place is increased by 1 and all other digits on its right side are dropped. This gives the number as 0.667.

The given number is first expressed in the exponential form. Then all the digits after the second decimal place are dropped. This gives the number 1.00 x 10~6, which has three significant.

Question 7. On heating of 4.9 g of KC103, 1.92g of 02 is evolved and 2.97 g of KC1 is obtained as a residue. Show that these data illustrate the law of conservation of mass.
Answer: ,\(\text { Reaction: } 2 \mathrm{KClO}_3 \rightarrow \quad \begin{gathered}
12 \mathrm{KCl}+3 \mathrm{O}_2 \uparrow \\
4.9 \mathrm{~g}
\end{gathered} \quad \begin{gathered}
2.97 \mathrm{~g} \quad 1.92 \mathrm{~g}
\end{gathered}\)

Total mass ofthe products 02 + KC1 = (1.92 + 2.97)g

= 4.89g

Difference in mass between the reactant and products

= (4.9 -4.89)g =0.01g.

This small difference is due to experimental error. Otherwise, these data illustrate the law of conservation of mass.

Question 8. 10.1 g o/HCl is mixed with 6.3 g of NaHC03. Calculate the mass ofCO2 released if the residual mixture is found to weigh 12.1 g
Answer: \(\text { Reaction: } \mathrm{HCl}+\mathrm{NaHCO}_3 \rightarrow \underset{10.1}{6.3 \mathrm{~g}} \rightarrow \underset{12.1 \mathrm{~g}}{\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}}+\mathrm{CO}_2\)

According to the law of conservation of mass, the mass of
(HC1 + NaHC03) = mass of(NaCl + H20 + C02)

or, (10.1 + 6.3)g = mass of(NaCl + H20 + C02)

or, 16.4g = 12.1g + mass of C02 /. massofC02 = 4.3g

Question 9. Calculate the number of gram-atom present in 2.1 g of nitrogen and 0.23 g of sodium.
Answer: One gram-atom of nitrogen means 14g of nitrogen.

No. of g-atom present in 2.1 g of nitrogen \(=\frac{2.1}{14}=0.15\)

Again, One g-atom of sodium means 23g of sodium

No. of g-atom present in 0.23g of Na \(=\frac{0.23}{23}=0.01\)

Question 10. Give an example of a tetra-atomic element and calculate its molecular mass.
Answer: Tetra-atomic element: P4 molecular mass of P4 = atomic mass of P x its atomicity

= 31 x4 = 124

Question 11. ‘1-mole oxygen = 2 gram-atom oxygen’—justify.
Answer: 1 g-atom of oxygen means 16g of oxygen

∴ 2 g-atom of oxygen = 32g

∴ 1 mol oxygen = 2 g-atom oxygen [proved].

Question 12. Which one is the volume of 2.8g of ethylene gas at STP?

  • 2.24L
  • 22.4L
  • 224 L
  • 0.224 L.

Answer:  Gram-molecular mass of ethylene gas = 28g

∴ At STP 28g of ethylene gas occupies a volume of 22.4 L

∴ At STP, 2.8g ethylene gas occupy \(\frac{22.4 \times 2.8}{28}=2.24 \mathrm{~L}\)

∴ The volume of 2.8g of ethylene gas at STP is 2.24L

Question 13. Which one is the volume of lg of oxygen gas at STP?

  1. 0.7 L
  2. 4.8 L
  3. 1.4L
  4. 1.2 LW.

Answer: Gram-molecular mass ofoxygen gas at STP is 32g

At STP, volume 32 g ofoxygen gas = 22.4L

At STP volume of1 g of oxygen gas \(=\frac{22.4}{32} \mathrm{~L}=0.7 \mathrm{~L}\)

Question 14. Calculate the number of O-atoms in 0.5 mol of S02.
Answer: No. ofmolecules presentin1 mol S02 = 6.022 x 1023.

Number of molecules present in 0.5 mol S02

= 0.5 x 6.022 x 1023 =3.011 x 1023

2 oxygen atoms are present in each S02 molecule.

∴ The number of oxygen atoms present in 0.5mol of S02

= 3.011 X 1023 X 2 = 6.022 X 1023

Question 15. Calculate the mass o/lmol of electrons if the mass of one electron be 9.11 x 10-31 kg.
Answer: 1 mol of electrons = 6.022 x 1023 number of electrons.

Mass of one electron = 9.11 x 10-31kg

therefore The mass of 1 mol of electrons

= 6.022 x 1023 x 9.11 x 10-31kg = 5.486 x 10-7kg

Question 16. At a temperature of 273 K and 1 atm pressure, 1 L of a gas weighs 2.054 g. Calculate its molecular mass.
Answer: 22.4L of any gas at STP contains 1 mol of that gas.

1L of a gas weighs 2.054g

∴ 22.4L ofthe gas weighs = 2.054 X 22.4g = 46g

∴ Molecular mass ofthe gas is 46

Question 17. At the same temperature & pressure, two flasks of equal volume contain NH3 & S02 gas respectively. Identify the flask having a greater number of molecules of gaseous substance with greater mass and a greater number of atoms.
Answer: According to Avogadro’s hypothesis, at temperature and pressure equal volume of all gases contains an equal number of molecules.

At the same temperature & pressure, two flasks having equal vol. contain equal no. of NH3 & S02 molecules.

(H) NH3 = 17 and S02 = 64.

The flask containing S02 gas has a greater mass.

NH3 is a tetra-atomic and S02 is a triatomic molecule.

So, the flask with NH3 contains a greater no of atoms.

Question 18. Choose the correct options: The vapor density of carbon dioxide is—

  1. 22
  2. 22g.cm3
  3. 22g.l-1
  4. 44

Answer: Vapour density of a gas =1 /2 x its molecular mass

∴ Vapour density of C02 =(1/2) x 44 =22

Question 19. The equivalent mass of an element can never be zero explained.
Answer: Vapour density ofoxygen (D) = 32/2 = 16.

∴ Vapour density ofthe gaseous element =5×16 = 80.

∴ Molecular mass ofthe gaseous element = 2 x 80 = 160.

Relative atomic mass ofthe element

⇒ \(=\frac{\text { molecular mass of the element }}{\text { its atomicity }}=\frac{160}{2}=80\)

Question 20. Calculate the percentage composition (by mass) of the constituent elements of sodium sulfate (Na2S04).
Answer: Gram-molecular mass of Na2SO4 = 142g

mass% of Na in Na2SO4 = \(\frac{2 \times 23}{142} \times 100=32.39\)

mass % of S in Na2SO4 =\(\frac{32}{142}\) x 100 = 22.54

∴ mass % of O in Na2SO4 = \(\frac{64}{142}\)x 100 = 45.07

Question 21. An oxide of iron is found to contain 69.9% iron and 30.1% dioxygen (02) by mass. Calculate its empirical formula.
Answer: Fe = 69.9% and 02 = 30.1%. Now, the ratio of the number of atoms of Fe and Oin the compound, \(\text { Fe }: O=\frac{69.9}{55.85}: \frac{30.1}{16}=1.25: 1.88=1: 1.5=2: 3\)

Its empirical formula is Fe203.

Question 22. Calculate the amount of sodium hydroxide present in 100 mL of 0.1(M) NaOH solution.
Answer: 1000 ml 1 (M) NaOH solution contains 40gNaOH

Therefore 100mL 0.1 (M) NaOH solution contains

⇒ \(\frac{40 \times 100 \times 0.1}{1000} \mathrm{~g} \text { of } \mathrm{NaOH}=0.4 \mathrm{~g} \text { of } \mathrm{NaOH}\)

Question 23. 100 mL of 3 (N) Na2C03 solution is diluted to 300 mL by adding water. Calculate the normality of this solution.
Answer: Let S be the normality of the diluted solution. As given in the question, 100 x 3 = 300 x S S = 1

Question 24. Find the volume (in mL) of 0.2 (N) NaOH solution required to neutralize 25mL of 0.2(N) H2S04 solution.
Answer: Volume of H2SO4 solution V1 =25mL Normality of H2S04 solution S1 =0.2(N) Volume of NaOH solution = xmL Normality of NaOH solution S2 = 0.2(N) Now, V1S1 = V2S2 or, 25 X 0.2 = x x 0.2

∴ x =25

Question 25. Give the relation between normality & molarity of a solution.
Answer: Normality of an acid = Molarity x Basicity ofthe acid.

  • Normality of a base = Molarity x Acidity of the base.
  • Solution Of Warm Up Exercises
  • Mixture (homogeneous);
  • Mixture (homogeneous);
  • Mixture (homogeneous);
  • Compound;
  • Mixture (heterogeneous);
  • Element;
  • Mixture (homogeneous);
  • Mixture (heterogeneous);
  • Compound;
  • Mixture (homogeneous);
  • Mixture (heterogenous)

Answer: Normality of an acid = Molarity x Basicity ofthe acid. Normality of a base = Molarity x Acidity of the base.

Question 26. The reactant which is entirely consumed in any reaction is known as the limiting reagent. In the reaction, 2A + 4B → 3C + 4D, if 5 moles of A react with 6 moles of B, then

  • Which is the limiting reagent?
  • Calculate the amount of C formed.

Answer: Reaction: 2A + 4B → 3C + 4D

  • Here, 2 moles of A reacts with 4 moles of B.
  • Therefore, 5 moles of A reacts with= \(\frac{4}{2} \times 5=10\) moles of B.
  • But, we have 6 moles of B participating in the reaction.
  • It means B is the limiting reagent.
  • 4 moles of produce 3 moles of C.
  • Hence, 6 moles ofB gives =(3/4) x 6 = 9/2 = 4.5 moles of C.

Question 27. A granulated sample of aircraft alloy (Al, Mg, Cu) weighing 8.72 g was first reacted with alkali and then with very dilute HC1, which left behind a residue. The residue after boiling with alkali weighed 2.10 g and the acid-insoluble residue weighed 0.69 g. What is the composition of the alloy?
Answer: Let us suppose, the amount of Al, Mg, and Cu in the sample box, y, and z g respectively.

Reactions:

⇒ \(2 \mathrm{Al}+2 \mathrm{NaOH} \stackrel{2 \mathrm{H}_2 \mathrm{O}}{\longrightarrow} 2 \mathrm{NaAlO}_2+3 \mathrm{H}_2\)

⇒ \(\mathrm{Mg}+2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_2+\mathrm{H}_2, \mathrm{Cu}+\mathrm{HCl} \rightarrow \text { No reaction }\)

i.e., only Al reacts with NaOH and only Mg reacts with HC1.

Therefore + y+ z = 8.72 andy+ z = 2.10 (Residue left after alkali treatment)
z = 0.69 (Residue left after acid treatment)

therefore x = 6.62 g and y = 2.10- 0.69 = 1.41 g

∴ \(\% \text { of } \mathrm{Al}=\frac{6.62}{8.72} \times 100=75.9\)

⇒ \(\% \text { of } \mathrm{Mg}=\frac{1.41}{8.72} \times 100=16.2\)

⇒ \(\% \text { of } \mathrm{Cu}=\frac{0.69}{8.72} \times 100=7.9\)

Question 28. The equivalent mass of metal M is E and the formula of its oxide is MxOy. Show that the valency and atomic mass of metal M are 2y/x and 2yE/x respectively.
Answer: Let the atomic mass ofthe metal M be a

So, the molecular mass ofthe compound, MxOy = ax + 16y

Therefore Equivalent mass (E) ofthe metal M in the compound

⇒ \(\mathrm{M}_x \mathrm{O}_y, \mathrm{E}=\frac{a x}{16 y} \times 8=\frac{a x}{2 y}\)

∴ Atomic mass of M (a) = 2yE/x

Hence, valency of M = \(\frac{\text { atomic mass }}{\text { equivalent mass }}\)

∴ \(=\frac{\frac{2 y E}{x}}{E}=\frac{2 y}{x}\)

The valency of an element in its oxide = 2 x the number of oxygen atoms combined with 1 the atom ofthe element.

In atoms of M combine with y no. of O-atoms.

With 1 atom of, no. of oxygen atoms combined = y/x

Valency of M = 2 x (y/x) = 2y/x

Question 29. Two gases, A and B having equal mass are kept in two separate vessels under identical conditions of temperature and pressure. If the ratio of their molecular masses is 2 : 3, find the ratio of the volumes of the vessels.
Answer: Let, MA &Mfi be the molasses of A Ik B respectively.

As per given data, MA: MB =2:3

therefore \(M_A=\frac{2}{3} \times M_B\)

Here, mass of A = mass of B = Wg (say)

Therefore In Wg of each A and B, the ratio number of moles,

⇒ \(n_A: n_B=\frac{W}{M_A}: \frac{W}{M_B}=\frac{M_B}{M_A}=\frac{M_B}{\frac{2}{3} \times M_B}=\frac{3}{2}=3: 2\)

Question 30. Since under the identical conditions of temperature and pressure, the volumes of gases are directly proportional to their number of moles, [because according to Avogadro’s law, V at the same temperature and pressure.] the ratio of volumes of two gases, i.e., the ratio of volumes of two vessels, \(V_A: V_B=n_A: n_B=3: 2.\)
Answer: Here, 1 L of X combines with 2L of Y to form 1L of a gaseous compound.

Let, n be the number of molecules present per liter ofthe gas under the same temperature and pressure.

So according to Avogadro’s hypothesis, n molecules of X react with 2n molecules of Y to form n molecules of gaseous product.

i.e., 1 molecule of X + 2 molecules of Y =1 molecule of the gaseous productor, 2 atoms of X + 4 atoms of Y =1 molecule of the gaseous compound [since the reactants are diatomic]

Molecular formula ofthe compounds X2Y4.

Question 31. The vapor density of a gas at 25°C is 25. What will be its vapor density at 50°C?
Answer: Vapour density of gas \(=\frac{\text { mass of certain volume of gas }}{\text { mass of equal volume of } \mathrm{H}_2 \text { gas }}\)

[under similar conditions of temperature and pressure]

Now, an increase in temperature will result in a proportionate increase in the volume of both hydrogen and the given gas as all gases expand equally due to an equal rise in temperature irrespective of their chemical nature.

So, the vapor density ofthe gas does change with any change in temperature.

Thus, the vapor density of the gas at 50° C will be equal to the vapor density measured at 25°C. So, the vapour density ofthe gas at 50°C will also be 25.

Question 32. Do 1 mol 02 and 1 mol O signify the same quantity?
Answer: 1 mol 02 and 1 mol O do not signify the same quantity.

  1. 1 mol 02 =1 gram-mole 02
  2. = 2 gram-atom O =2×16 = 32 g oxygen.
  3. 1 mol O =1 gram-atom O =1 x 16 = 16 g oxygen.

Question 33. The experimental values of the vapor density of either NHÿCl or PCI- is less than that obtained from the equation D = M/2. Explain.
Answer: At the experimental temperature, both NHÿCl and PC15 undergo thermal dissociation.

⇒ \(\begin{aligned}
\mathrm{NH}_4 \mathrm{Cl}(\text { vap }) & \rightleftharpoons \mathrm{NH}_3(g)+\mathrm{HCl}(g) \\
\mathrm{PCl}_5(\text { vap }) & \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)
\end{aligned}\)

Due to such dissociation, the number of molecules in NH4C1 vapor or PC15 vapor increases.

This causes an increase in volume because, at a certain temperature and pressure, the volume of a gas is proportional to die number of molecules. As die mass is fixed, an increase in volume causes a decrease in the value of vapor density measured experimentally.

Question 34. There are two natural isotopes of hydrogen ( 1H > 99 %; 2H < 1 ). Chlorine also has two natural isotopes (35C1 = 75%; 37C1 = 25%). How many different molecules of HC1 are possible? Arrange them in the decreasing order of their relative abundance.
Answer: The possible HC1 molecules are 1H35Cl,2H35C1, 1H37CI and 2H37CI. Since the abundance of the 1IT isotope of 35 hydrogens and ‘ Cl isotope of chlorine are maximum, the 1 35 1 abundance of IT Cl will also be Since the abundance of 2H and 37C1 isotopes is much less, 2I I37C1 will be the least abundant.

So, the order of different molecules of HC1 arranged in the decreasing order of their relative abundance will be—

\({ }^1 \mathrm{H}^{35} \mathrm{Cl}>{ }^1 \mathrm{H}^{37} \mathrm{Cl}>{ }^2 \mathrm{H}^{35} \mathrm{Cl}>{ }^2 \mathrm{H}^{37} \mathrm{Cl}\)

Question 35. At the same temperature and pressure, a gaseous hydride contains twice of Its own volume of hydrogen. The vapor density of the hydride is 14. What is its molecular formula?
Answer: Let, at the same temperature and pressure, 2 volume of hydrogen is present in 1 volume of hydride.

According to Avogadro’s hypothesis, at the same temperature and pressure, if no. of molecules is present in 1 volume of hydride, the no. of hydrogen molecules in this hydride is 2n.

Therefore 4 hydrogen atoms or 2 molecules of hydrogen are present in 1 hydride molecule,| v hydrogen is diatomic.

Let, the number of carbon atoms in a hydride molecule he x. Therefore, the molecular formula of the hydride is CX.H4.

Molecular mass of the hydride = 2 x 14 = 28

Now, the molecular mass of CVH,1 = 12x + 4

Therefore, 12x + 4 = 28 or,x=2

Therefore Molecular formula of the hydride C2H4.

Question 36. According to Avogadro’s hypothesis, at the same temperature and pressure equal volume of all guscs contains an equal number of molecules. Can it be concluded from the given statement that all molecules have equal volume”?
Answer: Avogadro’s hypothesis states that dial equal volumes of all gases under similar conditions of temperature and pressure, contain equal numbers of molecules. This does not signify that the actual volumes of one molecule of different gases are the same.

Experimental studies have shown that the molecular diameters of different gases differ from each other depending on the molecular composition of the gas.

in fact, the total volume of any gas (i.e., the volume of the container in which the gas is enclosed) is equal to the sum of the volume of all gas molecules and the volume of the empty space which is available for free movement of the gas molecules.

Question 37. A mixture of formic acid and oxalic acid is heated with concentrated H2SO4. The gas evolved is collected and treated with KOH solution. The volume of the solution decreases by 1/H th of Its original volume. Find the molar ratio of the two acids in the original mixture.
Answer: Let, the mixture of a moles of oxalic acid and b moles of formic acid be heated with concentrated H2S04.

The concerned reactions are as follows :

⇒ \(\begin{aligned}
& (\mathrm{COOH})_2 \stackrel{\mathrm{H}_2 \mathrm{SO}_4 / \text { heat }}{\longrightarrow} \mathrm{CO}(\mathrm{g})+\mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(I) \\
& a \mathrm{~mol} \quad a \mathrm{~mol} \quad a \mathrm{~mol} \\
&
\end{aligned}\)

⇒ \(\underset{b \mathrm{~mol}}{\mathrm{HCOOH}} \stackrel{\mathrm{H}_2 \mathrm{SO}_4 / \text { heat }}{\longrightarrow} \underset{b \mathrm{~mol}}{\mathrm{CO}}(g)+\mathrm{H}_2 \mathrm{O}(l)\)

Total number of moles ofthe gaseous mixture

= number of moles of CO + number of moles of C02
=(a + b) mol + a mol = (2a + b) mol

Now, KOH absorbs only moles of C02, and the volume

ofthe solution decreases by l/6th of its initial volume.

According to Avogadro’s law

⇒ \(\frac{\text { number of moles of } \mathrm{CO}_2}{\text { number of moles of gas mixture }}=\frac{a}{(2 a+b)}=\frac{1}{6}\)

or, 6a = 2a + b or, 4a – b or, b/a = 4

Therefore Molar ratio of formic acid and oxalic acid =4:1

Question 38. Taking N2 and 02 as the main components of air (79% N2, 21% 02 by volume), find the average molecular mass of air.
Answer: For a mixture of different gases, the average molecular mass ofthe mixture is taken as—

Average molecular mass = \(\Sigma x_{\mathrm{i}} M_{\mathrm{i}}=x_{\mathrm{N}_2} M_{\mathrm{N}_2}+x_{\mathrm{O}_2} M_{\mathrm{O}_2}\)

where xN and xQ are mole fractions of N2 and 02 and MNÿ and MQ are their molecular masses.

At the same temperature and pressure, equal volumes of all gases contain an equal number of moles, and their molar ratio is the same as the ratio of their volumes.

Therefore XN2 = 0.79,xO2 = 0.21, Also MN2 = 28u, MO2= 32u

ATherefore Average molecular mass ofair = (0.79 x 28 + 0.21 x 32)u
= (22.12 + 6.72)u = 28.84U

Question 39. A gaseous hydrocarbon needs 6 times more volume of oxygen (02) than itself for complete oxidation. It produces 4 times more C02 (by volume) than it’s own. What is the formula of the hydrocarbon?
Answer: Equation of combustion:

⇒ \(\mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O}\)

1 volume ofhydrocarbon reacts with \(\left(x+\frac{y}{4}\right)\) volumes of oxygen to produce 2x volumes of C02.

Now, x + y/4=6 or, 4x+y=24

Given, x = 4. Therefore, y = 8

∴ The formula of the hydrocarbon is C4H8.

Question 40. What is the equivalent mass of KH(I03)2 oxidant in the presence of 4.0 (N) HC1 when IC1 becomes the reduced form? [where K = 39.0 andI = 127.0]
Answer: 103 is present in KH(I03)2 as I03

Let, the oxidation state of i be x.

Therefore x-6 = -l or, x = +5. \(\begin{array}{cc}
+5 & +1 \\
\mathrm{KH}\left(\mathrm{IO}_3\right)_2 & 2 \mathrm{ICl}
\end{array}\)

Therefore Decreasein oxidation state =(10-2) = 8

Equivalent mass of KH(103)2 \(=\frac{\text { molecular mass }}{8}\)

= 390/8= 48.75

Question 41. Pb(NO2)2 on strong heating loses 32.6% of its mass. Calculate the relative atomic mass of Pb.
Answer:  Let, the relative atomic mass of Pb = x

⇒ \(\underset{\substack{x+2(14+3 \times 16) \mathrm{g} \\=(x+124) \mathrm{g}}}{\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \stackrel{\Delta}{\longrightarrow}} \underset{(x+16) \mathrm{g}}{\mathrm{PbO}}+2 \mathrm{NO}_2+\mathrm{O}_2\)

On dissociation of (x+124)g of Pb(N03)2, the weight decreases = (x+ 124- x- 16)g = 108g So, for lOOg Pb(N03)2, the weight decreases

⇒ \(=\left(\frac{108 \times 100}{x+124}\right) \mathrm{g} \text { As, } \frac{100 \times 108}{x+124}=32.6\)

Or, x= 207.3

Question 42. Find the molecular formula of the compound. [All volumes are measured under identical conditions of temperature and pressure] Fill in the blank: Equivalent weight of Min MC13 is E. Relative atomic mass of M is

  • 0.7l
  • 4.8l
  • 1.4l
  • 1.2l

Answer:  3E

Volume of lg of 02 at STP \(\frac{22.4}{32}\)= 0.7L

Question 43. What does 1 mol of electron signify?
Answer: 1 mol of electron signifies, 6.022 x 1023 number of electrons. The total charge of these electrons is 96500 C or F.

5. Find the number of neutrons in 5 x 10-4 mol of 14C isotope.

1.0 g of metal (A) displaces 1.134g of metal (B) from its salt. Determine the equivalent weights of(A) and (B). 2

Equivalent weight of A \(Equivalent weight of A \)

Equivalent weight of B \(=\frac{1.134 \times 28.006}{1}=31.7588\)

Question 44. 5 g of an impure sample of common salt on treatment with an excess of AgNO3 solution yields 9.812 g of AgCl. What is the percentage impurity of that sample?
Answer: NaCl(58.5g) + AgNO3 = NaNOg + AgCU(143.5g)

143.5g AgCl is obtained from 58.5gNaCl

Therefore 1 g ofAgCl is obtained from \(\frac{58.5}{143.5} \mathrm{~g} \mathrm{NaCl}\)

Therefore 9.812g of AgCl is obtained from \(\frac{58.5 \times 9.812}{143.5}=4 \mathrm{~g} \mathrm{NaCl}\)

Percentage of purity in the sample is \(=\left(\frac{4}{5} \times 100\right)=80 \%\)

Therefore So, percentage impurity = (100- 80)% = 20%

Find the number of neutrons in 5 x 10-4 mol of 14C isotope.

The chloride of a metal (A) contains 55.90% of chlorine.

10g of metal (A) displaces 1.134g of metal (B) from its salt. Determine the equivalent weights of(A) and (B). 2 6. 5 g of an impure sample of common salt on treatment with an excess of AgN03 solution yields 9.812 g of AgCl. What is the percentage impurity of that sample?

Question 45. A gaseous hydrocarbon contains 75% C by weight. 1L of this gas at STP weighs 0.72 g. What is the molecular formula of the hydrocarbon? [weight of 1L ofhydrogen at STP = 0.09 g]
Answer: D\(\frac{0.72}{0.09}\) 8; Molecular weight (M) = 2 x 8 = 16 = 3.0115 X 1021.

⇒ \(\mathrm{C}: \mathrm{H}=\frac{75}{12}: \frac{25}{1}=6.25: 25=1: 4\)

Empirical formula:CH4; molecular formula:(CH4)n.

Molecular weight = (12 + 4xl)n

As, 16n = 16

⇒ n = 1. So, the actual formulais CH4.

Question 46. 84 g of a mixture of.CaCO3 and MgCO. were heated to a constant weight. The constant weight of the residue was found to be 0.96 g. Calculate the percentage composition of the mixture. (Relative atomic masses of Ca and Mg arc 40 and 24 respectively)
Answer: Let, CaC03 = xg, MgC03 = (1.84 -x)g

∴ \(\mathrm{CaCO}_3(100 \mathrm{~g}) \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}(56 \mathrm{~g})+\mathrm{CO}_2\)

∴ \(x \mathrm{~g} \mathrm{CaCO}_3 \longrightarrow \frac{56 x}{100} \mathrm{~g} \mathrm{CaO}\)

∴ \(\mathrm{MgCO}_3(84 \mathrm{~g}) \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}(40 \mathrm{~g})+\mathrm{CO}_2\)

∴ \((1.84-x) \mathrm{g} \mathrm{MgCO}_3 \longrightarrow \frac{40}{84}(1.84-x) \mathrm{g} \mathrm{MgO}\)

∴ \(\text { As, } \frac{56 x}{100}+\frac{40}{84}(1.84-x)=0.96 \quad \text { or, } x=1 \text {. }\)

∴ \(\mathrm{CaCO}_3=\frac{1}{1.84} \times 100=54.35 \% ; \mathrm{MgCO}_3=45.65 \%\)

Question 47. 10 mL of a mixture of CH4, C2H4, and CO2 was exploded with excess oxygen. After the explosion, there was a contraction of 17 mL on cooling and there was a further contraction of 14 mL on treatment with KOH. Find out the composition of the mixture.
Answer: Let, CH4 = x mL, C2H4 = y mL, CO2 = z mL
∴ x+ y+ z = 10

⇒ \(\begin{aligned}
& \mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) \\
& \begin{array}{lccc}
1 \mathrm{~mL} & 2 \mathrm{~mL} & 1 \mathrm{~mL} & 0 \mathrm{~mL} \\
x \mathrm{~mL} & 2 x \mathrm{~mL} & x \mathrm{~mL}
\end{array} \quad \\
&
\end{aligned}\)

Contraction of volume =(x+2x-x)mL = 2xmL

Produced CO2 = xmL

⇒ \(\begin{array}{cccc}
\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g}) & +3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow & 2 \mathrm{CO}_2(\mathrm{~g}) & +2 \mathrm{H}_2 \mathrm{O}(l) \\
1 \mathrm{~mL} & 3 \mathrm{~mL} & 2 \mathrm{~mL} & 0 \mathrm{~mL} \\
y \mathrm{~mL} & 3 y \mathrm{~mL} & 2 y \mathrm{~mL} &
\end{array}\)

Contraction of volume = (y + 3y- 2y)mL = 2ymL Produced CO2 = 2ymL but the quantity of CO2 produced = z mL.

Contraction of total volume = (2x+ 2y+ 0)mL and total quantity of C02 = (x+ 2y+ z)mL

According to the problem, 2x+ 2y – 17 x+ 2y+ z = 14

From, [3]- [l] we get, y = 4mL

Putting y = 4 inequation [2], x = 4.5mL

Putting y = 4mL and x = 4.5mL we get z = 1.5mL  from equation [3].

So, CH4 = 4.5mL, C2H4 = 4mL and CO2 = 1.5mL.

Question 48. Which of the following will have the largest no. of atoms?

  1. 1 gAu (g)
  2. 1g Na(s)
  3. 1g li(s)
  4. 1g of cl2 (g)

Answer: We know that, the number of atoms of an element

= number of moles x NA x atomicity

⇒ \(=\frac{\text { mass of the element }(\mathrm{g})}{\text { gram-atomic mass of element }} \times N_A \times \text { atomicity }\)

  1. No. atoms present in lg of Au = \(\frac{1}{197}\) 6.022 x 1023 (since Gram-atomic mass of Au= 197)
  2. No. atoms presenting lg of Na \(\frac{1}{23} \times 6.022 \times 10^{23}\)
  3. No. atoms present in lg of Li \(=\frac{1}{7} \times 6.022 \times 10^{23}\)
  4. No. of atoms present in lg of Cl2

⇒ \(=\left(\frac{1}{71} \times 6.022 \times 10^{23}\right) \times 2=\frac{1}{35.5} \times 6.022 \times 10^{23}\)

(Since 2 cl atoms are present in 1 molecule of Cl2) Therefore, in lg ofLi will have the largest no. atoms.

Question 49. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one)
Answer: The number of moles of ethanol present in 1 L of aqueous
solution indicates the molarity of the solution.

Let, 1L water be present in 1 L ethanol solution [since the solution is dilute.

Therefore the amount of water present in 1 L of water

⇒ \(n_{\mathrm{H}_2 \mathrm{O}}=\frac{1000 \mathrm{~g}}{18 \mathrm{~g} \cdot \mathrm{mol}^{-1}}=55.55 \mathrm{~mol}\)

For binary solutions, the mole fraction of the first component  + the Mole fraction ofthe second component= 1.

Therefore, in this case, CH2o + xC2H5OH=1

Or, \(x_{\mathrm{H}_2 \mathrm{O}}=1-x_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \quad \text { or, } x_{\mathrm{H}_2 \mathrm{O}}=1-0.040=0.96\)

Again, XH20 \(=\frac{n}{n_{\mathrm{H}_2 \mathrm{O}}+n_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}} \text { or, } 0.96=\frac{55.55}{55.55+n_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}}\)

Or, 53.328 + 0.96nc2h5OH = 55.55

Therefore nC2H5OH \(=\frac{2.222}{0.96}\) =55.55

Therefore \(n_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}=\frac{2.222}{0.96}=2.3146 \mathrm{~mol}\)

Therefore Molarity ofthe solution = 203146 mol-L-1.

Question 50. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

Which law of chemical combination is obeyed by the above experimental data? Give its statement.

Fill in the blanks in the following conversions:

  1. 1km =……….mm=………..pm
  2. 1mg=…………kg = ……….ng
  3. 1ml= …………l=………….dm3

Answer: The ratio of different masses of dioxygen which combine separately with a fixed mass (28g) of dinitrogen.

=32: 64:32: 80 = 2:4:2:5 which is a simple whole numberratio.

Therefore, the obtained result supports the law of multiple proportions.

⇒ \(\begin{aligned}
1 \mathrm{~km} & =1 \mathrm{~km} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{100 \mathrm{~cm}}{1 \mathrm{~m}} \times \frac{10 \mathrm{~mm}}{1 \mathrm{~cm}} \\
& =10^6 \mathrm{~mm}
\end{aligned}\)

⇒ \(1 \mathrm{~km}=1 \mathrm{~km} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{1 \mathrm{pm}}{10^{-12} \mathrm{~m}}=10^{15} \mathrm{pm}\)

Therefore 1km = 106mm = 1015pm

⇒ \(1 \mathrm{mg}=1 \mathrm{mg} \times \frac{1 \mathrm{~g}}{1000 \mathrm{mg}} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}}=10^{-6} \mathrm{~kg}\)

⇒ \(1 \mathrm{mg}=1 \mathrm{mg} \times \frac{1 \mathrm{~g}}{1000 \mathrm{mg}} \times \frac{1 \mathrm{ng}}{10^{-9} \mathrm{~g}}=10^6 \mathrm{ng}\)

Therefore long = 10~6kg = 106ng

⇒ \(1 \mathrm{~mL}=1 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}=10^{-3} \mathrm{~L}\)

⇒ \(\begin{aligned}
1 \mathrm{~mL}=1 \mathrm{~cm}^3 & =1 \mathrm{~cm}^3 \times \frac{1 \mathrm{dm} \times 1 \mathrm{dm} \times 1 \mathrm{dm}}{10 \mathrm{~cm} \times 10 \mathrm{~cm} \times 10 \mathrm{~cm}} \\
& =10^{-3} \mathrm{dm}^3
\end{aligned}\)

Therefore 1ml = 10-3l =10-3dm3

Question 51 . In a reaction A + B2AB2. Identify the limiting reagent, if any, in the following reaction mixtures.

  • O 300 atoms of A + 200 molecules of B 0
  • 2 mol A + 3 mol B
  • 100 atoms of A + 100 molecules of B
  • 5 mol A + 2.5 mol B 0 2.5 mol A + 5 mol B

Answer:  According to the given reaction, 1 atom of A reacts

88K with 1 molecule of B. Therefore, 200 atoms of A react with 200 molecules of B, and 100 no. of A atoms remain unused. In this case, B is the limiting reagent, as A is present in excess.

According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A reacts with 2 mol of B. Thus, A is the limiting reagent and B will remain in excess.

As per the reaction, 1 atom of A reacts with 1 molecule of B. So, 100 atoms of A reacts with 100 molecules of B. Thus, in this case, there is no limiting reagent.

Here, 2.5 mol of A reacts with 2.5 mol of B. Therefore, B is the limiting reagent and A remains in excess.

As per the reaction, 2.5 mol A reacts with 2.5 mol B. Thus, A is the limiting reagent, and B remains in excess.

Question 52. N2 and H2 react to produce ammonia according to the equation: N2(g) + H2(g)→2NH3(g) Calculate the mass of ammonia produced if 2.00 x 103g N2 reacts with 1.00 X 103 g of H2. Will any of the two reactants remain unreacted? If yes, which one and what would be its mass?

⇒ \(\begin{gathered}
\text { Reaction: } \mathrm{N}_2(\mathrm{~g}) \\
1 \mathrm{~mol}=28 \mathrm{~g}
\end{gathered} \quad+\quad \begin{gathered}
3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow \\
2 \mathrm{~mol}=6 \mathrm{~g}
\end{gathered} \quad \begin{array}{r}
2 \mathrm{NH} \\
2 \mathrm{~mol}=34 \mathrm{~g}
\end{array}\)

28g of nitrogen reacts with 6g of hydrogen.

Therefore lg nitrogen reacts with (6/28)g of hydrogen.

Therefore 2000g of nitrogen reacts with \(\frac{6}{28} \times 2000\) =428.57g of hydrogen. therefore nitrogen is the limiting reagent and an excess amount ofhydrogen is present. 28g of nitrogen produces 34g of NH3

Therefore 2000g of N2 produces 34/28 x 2000 = 2428.57g of NH3

Yes, excess hydrogen does not take part in the reaction and remains unchanged.

Amount of hydrogen that does not take part in the reaction = (1000- 428.57) = 571.43g

Question 53. Calculate the number of atoms in each of the following

  • 52 moles of Ar
  • 52u of He
  • 52 g of He

Answer: 1 molAr = 6.022 x 1023 no. of-atoms

∴ 52mol Ar = 52 X 6.022 x 1023 Ar-atoms

= 3.131 x 1025 no Ar-atoms

Mass of 1 He-atom = 4u

Therefore, 4u = mass of1 He-atom

Therefore 52uHe = mass of \(\frac{52}{4}\) He-atoms = mass of13 He-atoms

1 mol He-atoms = 6.022 x 1023 He-atoms = 4gHe

Hence, 4g He = 6.022 x 1023 He-atoms

∴ 52gHe \(=\frac{6.022 \times 10^{23} \times 52}{4}=7.8286 \times 10^{24} \mathrm{He} \text {-atoms }\)

Question 54. A welding fuel gas contains carbon & hydrogen only. Burning a small sample of it in oxygen gives 3.38 g of C02, 0.690 g of H20, and no other products. A volume of 10.0L (at STP) of this welding gas is found to weigh 11.6g. Calculate the empirical formula, the molar mass of the gas, and the molecular formula.
Answer: Mass of carbon in 44g of C02 = 12g

∴ Mass of Cin 3.38g of C02 \(=\frac{12}{44} \times 3.38 \mathrm{~g}=0.9218 \mathrm{~g}\)

Similarly,18 g H2O = 2g hydrogen

Therefore, the mass of hydrogen in 0.690g of H20

=(2/18) x 0.690g = 0.0767g

therefore Total mass of fuel gas = (0.9218 + 0.0767) = 0.9985g

Therefore % amount ofC in the gas \(=\frac{0.9218}{0.9985} \times 100\) = 92.32

Therefore % amount of H in the gas \(=\frac{0.0767}{0.9985} \times 100=7.68\)

Determination of the empirical formula of the fuel gas

Therefore, the empirical formula ofthe fuel gas = CH

As given the question,

Mass of 10L gas at STP = 1 1.6g

∴ Mass of22.4L gas at STP = \(\frac{11.6 \times 22.4}{10}=25.984 \mathrm{~g}\)

Therefore Molecular mass = 25.984= 26

Mass of the empirical formula (CH) = 12+1 = 13.

Let, the molecular formula of the compound be (CH)n

Again, n = \(\frac{\text { molecular mass }}{\text { mass of empirical formula }}\)

∴ n = 26/13 =2

Therefore formula of fuel gas = (CH)n=(CH)2=C2H2

∴ n = 26/13=2

∴ Formula Of Fuel gas = (CH)n = (CH)2 = C2H2

Question 55. Calcium carbonate reacts with aqueous HC1 to give CaCl2 and CO2 according to the reaction. What mass of CaCO3 is required to react completely with 25 mL of 0.75(M) HC1?
Answer: Gram-molecular mass ofHC1 (solute) = (1 + 35.5) = 36.5g We know that the molarity ofthe solution (M)

⇒ \(=\frac{\text { mass of the solute (in gram) }}{\text { gram-molecular mass of solute } \times \text { vol. of solution }(\mathrm{L})}\)

Or, 0.75 \(\frac{\text { mass of } \mathrm{HCl}}{36.5 \times 25 \times 10^{-3}}\) So, mass of HCl = 0.6844g

⇒ \(\underset{100 \mathrm{~g}}{\mathrm{CaCO}_3(s)+2 \mathrm{HCL}(a q)} \underset{2 \times 36.5=73 \mathrm{~g}}{\rightarrow \mathrm{CaCl}_2(a q)+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)}\)

73g ofHC1 reacts completely with 100g of CaCO3

Therefore 0.6844g of HC1 reacts completely with \(\frac{100 \times 0.6844}{73} \mathrm{~g}\)

= 0.9375g of CaCO3

Question 56. Chlorine is prepared in the laboratory by treating manganese dioxide (MnOz) with aqueous HC1 according to the reaction.

⇒ \(4 \mathrm{HCl}(a q)+\mathrm{MnO}_2(s)-\mathrm{2H}_2 \mathrm{O}(l)+\mathrm{MnCl}_2(a q)+\mathrm{Cl}_2(g)\) How many grams of HC1 react with 5.0g of MnO2?

Answer: The given reaction,

⇒ \(4 \mathrm{HCl}(a q)+\mathrm{MnO}_2(s) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{MnCl}_2(a q)+\mathrm{Cl}_2(g)\)

(4 x 36.5)g (55 + 32) = 87 g

According to the equation,

87g of Mn02 reacts with (4 x 36.5)g of HCl

therefore 5g of mnO2 reacts with ,\(\frac{4 \times 36.5 \times 5}{87}=8.39 \mathrm{~g} \text { of } \mathrm{HCl}\)

Question 57. Classify the following as pure substance, homogeneous mixture, heterogeneous mixture, element, and compound:

  • Milk
  • Air
  • Petrol
  • Distilled water
  • Common salt
  • Graphite
  • Tap-water
  • Smoke
  • Dry ice
  • Cold drinks
  • Gun powder.

Answer:

  1. Mixture (homogeneous);
  2. Mixture (homogeneous);
  3. Mixture (homogeneous);
  4. Compound;
  5. Mixture (heterogeneous);
  6. Element;
  7. Mixture (homogeneous);
  8. Mixture (heterogeneous);
  9. Compound; Mixture (homogeneous);
  10. Mixture (heterogenous

 Short Question And Answers

Question 1. Is the statement correct or incorrect— ‘35.5g of chlorine contains 6.022 x 1023 molecules’?
Answer: The statement is incorrect. 35.5g (lg-atom) of chlorine  contains 6.022 x 1023 no. of atoms.

Question 2. Calculate the number of molecules in 0.52g of acetylene.
Answer: Gram-molecular mass of acetylene = 26g.

Number of molecules present in 0.52g of acetylene

∴ \(=\frac{6.022 \times 10^{23} \times 0.52}{26}=1.2044 \times 10^{22}\)

Question 3. Find the number of molecules in l milli mole of C02.
Answer:  milli mol of C02 molecule = 10-3 mol of C02 Hence, number of molecules presentin1 millimole of

\(=10^{-3} \times 6.022 \times 10^{23}=6.022 \times 10^{20}\)

Question 4. The equivalent mass of iron in ferrous oxide is 28. Calculate its equivalent mass in ferric oxide.
Answer: Considering ferrous oxide (where valency of iron = 2),

Atomic mass of iron, A = £xF=28×2 = 56

Now, considering ferric oxide (where valency of iron = 3 ),

equivalent mass ofiron, E = A/ V = 56/3 = 18.67.

Question 5. Differentiate between accuracy and precision.
Answer: Cu Precision refers to the closeness of the set of values obtained from identical measurements of a quantity.

  1. Whereas accuracy means how measurements obtained experimentally agree with the exact value.
  2. A measurement may have good accuracy but poor precision as different measurements m