Sainavle

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 B^2-.
Answer: Y²⁺ < X- < A- < B^2-

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 SO₃^2-, NH⁺⁴, and PO^4-₃ 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?

• 2p_z– 2p_z
• 2p_s + 2p_z
• 2_z + 2p_z
• 2p_y + 2p_y

Answer: 2p_x– 2p_X;2p_x + 2p_y

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?

• 2P_z > 2P_z
• 2s > 2P_y
• ls, 2s
• 2P_x, 2P_x

Answer: 2s > 2P_y ;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 π_2px 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 Cl^– is 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 N^3– 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 Cl^– ions 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.

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

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.

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.

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.

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 C_2-2.
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:

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⁺, O^2-
Answer: Al³⁺ < Mg²⁺ < F^– < O^2-

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.

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

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.

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.

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, CO^2-, 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 CO^2-₃ 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 S^2-; 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 CO^2-₃ 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 (CO^2-₃) can be represented by a combination of the following 3 resonating structures in which CO^2-₃ 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.

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.

• 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

Answer: 3. 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.

Question 5. The number of acid protons in H³PO³ is

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

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

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

1. 2 unpaired electrons π_b MO
2. 2 unpaired electrons in π* MO
3. 2 unpaired electrons in σ* MO
4. 2 unpaired electrons in σ_b 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

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 NO^–₂is—

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 \(\mathrm{NO}_2<\mathrm{NO}_2^{-}<\mathrm{NO}_2^{+}\)

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

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

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

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

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.

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

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

Answer: 

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—

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

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

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.

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—

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

Answer: 3. 3>2>1

Question 23. Thenumberoflone pairs of electrons on the central atoms of H₂O, SnCl₂ PCl_3, 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 CO₂ 

1. HgCl₂
2. SnCl₂
3. C₂H₂
4. NO₂

Answer: 1. HgCl₂

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

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<Li^–₂<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 I^–₂ 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-

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-

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, NO^–₂, NO^–₃, NH^–₂, NH^–₄, SCN^–

1. NO^–₂ and NO^–₃
2. NH⁺₄ and NO^–₃
3. SCN^– and NH^–₂
4. NO^–₂ and NH^–₂

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. O^–₂
3. O₂^2-
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

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.

Question 62. Which contains no n -bond—

1. SO₂
2. NO₂
3. CO₂
4. H₂O

Answer: 4. H₂O

Question 63. Which ofthe following lanthanoid ions is diamagnetic (Atnos. Ce=58, Sm=62,Eu=63, Yb70)-

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

O^–₂

Answer: 4. O^–₂

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

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 O^2-(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 O^2- in the gas phase is unfavorable even though O^2- 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 O^2- 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

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

Both NH₃ and PH₃ have pyramidal geometry.

XeF₄ has sp³d² hybridisation while XeO₄ has sp³ hybridisation.

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 CO²–₃ 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:

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:

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

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.

back 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

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

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

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.

⇒ \(\stackrel{(+1)}{\mathrm{NaCl}}<\stackrel{(+2)}{\mathrm{MgCl}_2}<\stackrel{(+3)}{\mathrm{AlCl}_2}\)

⇒ \(\text { Bond order } \propto \frac{1}{\text { Bond length }}\)

\(\text { Bond order }=\frac{\text { Bond order of each } \mathrm{C}-\mathrm{O} \text { bond }}{\text { Total no. of resonating structures }}\)

⇒ \(\begin{aligned}
& \mathrm{CO} \rightarrow \mathrm{C} \equiv \mathrm{O} \Rightarrow \frac{2+1}{1}=3.0 \\
& \mathrm{CO}_2 \rightarrow \mathrm{O}=\mathrm{C}=\mathrm{O} \Rightarrow \frac{2+2}{2}=2.0
\end{aligned}\)

Hence, the decreasing order of bond length is, CO2-₃ > HCO₂> CO₂ > CO

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

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. B₂–₂ 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):

⇒ \(K K\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}\right)^2\left(\pi_{2 p_x}\right)^2\left(\pi_{2 p_y}\right)^2 ; \text { B.O. }=\frac{1}{2}(8-4)=2.0\).

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.

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.

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.

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]

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.

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.

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, \(r_{\mathrm{F}^{-}}<r_{\mathrm{Cl}^{-}}<r_{\mathrm{Br}^{-}}<r_{1^{-}}\)

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¯

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.

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.

\(\text { Atomic volume }=\frac{\text { Gram-atomic mass }}{\text { Density }\left(\text { in } \mathrm{g} \cdot \mathrm{cm}^{-3}\right)}\)

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

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.

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.

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.

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

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