Classification Of Elements And Periodicity In Properties Long Answer Questions

Element Classification Techniques: Question 1. Three elements X, Y, and Z follow Dobereiner’s law of triads. If the atomic masses of X and Z are 7 and 39 respectively, then determine the atomic mass of Y.
Answer: Atomic mass of Y = arithmetic mean of atomic masses of X and Z. Therefore, the atomic mass of Y 7 + 39/2 = 23.

Question 2. For which elements, Newlands’ law of octaves is not applicable?
Answer: Heavier elements beyond Ca

Question 3. Which properties of the elements are dependent on their electronic configurations and which are not
Answer: The chemical properties and some physical properties of elements are dependent on their electronic configurations whereas nuclear properties (like radioactivity) do not depend on their electronic configurations.

Question 4. How many periods and groups are there in the modem version of Mendeleev’s table?
Answer: There are 7 periods and 9 groups of a modern version of Mendeleev’s periodic table.

Question 5. Elements of the 4th, 5th, and 6th periods of Mendeleev’s periodic table were divided into even and odd series—why?
Answer: Due to the difference in properties of each pair of elements belonging to the same period and group, the elements of the 4th, 5th, and 6th periods of Mendeleev’s periodic table were divided into even and odd series.

Question 6. Which periods in Mendeleev’s periodic table (modified form) do not contain subgroups?
Answer: Periods 1, 2 and 3

Question 7. Name the elements with which even and odd series of 4th, 5th, and 6th periods begin.
Answer: Even series begins with K, Rb, and Cs. The odd series begins with Cu, Ag, and Au.

Question 8. Identify the elements eka-aluminium and eka-silicon. What was the reason for such naming?
Answer: Ga and Ge. The elements were so named by Mendeleev because he predicted that the properties of ekaaluminium and eka-silicon would be similar to those of aluminium and silicon respectively.

Question 9. How many periods and groups are there in the present form (i.e., long form) of the periodic table?
Answer: 7 periods and 18 groups

Question 9. Which group ofthe long form of the periodic table contains solid, liquid, and gaseous elements?
Answer: Group-17.

Question 10. State the reason for the repetition of properties ofthe elements after certain regular intervals of their atomic numbers.
Answer: After certain regular intervals of their atomic numbers, elements show similar electronic configurations in their valence shell. For this reason, repetition of properties of the elements is observed.

Question 11. What are rare-earth elements? Why are they so called?
Answer: The 14 elements from cerium (58Ce) to lutetium (71Lu) ofthe periodic table are called rare-earth elements. These elements are so named because most of these elements occur in very small amounts in the earth’s crust.

Question 12. Write down the names ofthe coinage metals and indicate their positions in the long form ofthe periodic table.
Answer: Cu, Ag, Au (periods 4, 5, & 6in group 11).

Question 13. Give one example of each of the following and indicate their positions in the periodic table (long form) (1) liquid non-metal (11) liquid metal (ill) radioactive halogen (lv) radioactive inert gas (v) radioactive alkali-metal.
Answer:

1. Liquid non-metal: Bromine (Br), 17th group in 4th period,
2. Liquid metal: Mercury (Hg), 12th group in 6th period,
3. Radioactive halogen: Astatine (At), 17th group in 6th period,
4. Radioactive inert gas: Radon (Rn), 18th group in 6th period,
5. Radioactive alkali metal: Francium (Fr), 1st group in 7th period.

Question 14. Indicate the period that contains the first series of transition elements.
Answer: 4th period.

Question 15. Mention the names of the first member of each of the first, second, and third series of transition elements.
Answer: Scandium (Sc), Yttrium (Y), and Lanthanum (La)

Question 16. Mention the names of the first and last members of the actinide series.
Answer: Thorium (Th) and Lawrencium (Lr).

Question 17. What are pnictogens and chalcogens?
Answer: Gr-15 elements: pnictogens; Gr-16 elements: chalcogens

Question 18. Elements of the same group exhibit similar chemical properties—why?
Answer: Because they have similar outer electronic configurations.

Question 19. What are d-block elements? Give their general electronic configuration.
Answer: The reasons behind placing Na and K in the same group of the periodic table are— The valence shell electronic configurations of and are the same Na: ls22s22p63s1 K: ls22s22p63s23p64sl.

Question 20. Why are sodium (Na) and potassium (K) placed in the same group of the periodic table? Give any two reasons.
Answer: (n- l)d1-10 ns1-2f (n-2)f-14(n-l)d0-1ns2

Question 21. Give general electronic configurations of 1 transition elements and 2 inner-transition elements
Answer: Due to the presence of unpaired electrons in (n-l)d subshell

Question 22. What is the reason for the strong reducing character of s-block elements?
Answer: Since they have low ionization energy.

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Question 23. Which block in the periodic table contains metals, nonmetals, and metalloids? Give three examples of metalloids.
Answer: p -block; Si, Ge, As

Question 24. Cu in the +1 oxidation state and Hg in the +2 oxidation state resemble each other in their properties. Explain.
Answer: Valence shell configurations of Cu+([Ar]3d10) and Hg2+([Xe]5d10) are similar. So they resemble each other in their properties.

Question 25. Mention the names of two noble metals and indicate their positions in the periodic table.
Answer: Pt and Au. In the periodic table, they are present in the 6th period in the 10th and 11th groups respectively.

Question 26. Which one of the following exhibits paramagnetism? Sc3+, Cr3+, Cu+, Zn2.
Answer: Cr3+ : [Ar]3d3. Due to the presence of unpaired electrons, it exhibits paramagnetism.

Question 27. Locate the position of an element in the long form of the periodic table.
Answer: Electronic configuration: ls22s22p63s23p63d104s24p3.
Period = 4th, Group No. = 10+ total no. of electrons present in 4s and 4p orbitals =10 + 2 + 3 = 15

Question 28. Find the atomic number of an element that belongs to the third period and group 17 in the periodic table.
Answer: The atomic number ofthe given element = 17

Question 29. An element belongs to the third period of p -block. It has five valence electrons. Predict its group.
Answer: Group ofthe element= 10 + no. of valence electrons = 10 + 5 = 15

Question 30. Write the IUPAC name and symbol of the element with atomic number 135.
Answer: IUPAC name: Untripentium; Symbol: Utp

Question 31. How many elements beyond actinides have been given official names in the last period of the periodic table? Write the IUPAC name and symbol of the last element.
Answer: 11 elements beyond actinides have been given official names in the last period of the periodic table. IUPAC name ofthe last element is Ununoctium and its symbol is Uuo.

Question 32. Arrange the following elements in decreasing order of their atomic radius: Na, H, Si, S, P, Cl
Answer: Na > Si > P > S > Cl > H

Question 33. The atomic radius of elements in a period decreases with an increase in atomic number but for inert gases, it increases. Why?
Answer: Atoms in inert gases are held together by weak van der Waals forces, not by covalent bonds. Thus, atomic radii and van der Waals radii of inert gases are equal.

However, as the van der Waals radius is greater than the covalent radius, the atomic radius of elements across a period decreases with an increase in atomic number but for inert gases increases.

Question 34. Indicate the largest and smallest species among the following: Mg, Al, Mg2+, Al3+
Answer: Largest: Mg; Smallest: Al3+

Question 35. Consider the set of ions (Na+, N3-, Mg2+, 02-, F_, and Al3+) and answer the following questions: What is the common factor associated with the species? Arrange the ions in order of increasing radii.
Answer: All are isoelectronic species.

Al3+ < Mg2+ < Na+ < F- < O2- < N3

Question 36. The number of electrons in Sr2+ and Br- are the same. Justify whether the atomic radii ofthese two ions will be the same or not.
Answer: Although Sr2+ and Br- have the same number of electrons, their proton number are different Number of protons in Sr2+ is greater than that of Br-‘;

Thus, the nuclear pull on the electrons is greater in Sr2+ than in Br-. So atomic radius or Sr2-1, is less them that of 8r-.

Question 38. Write the names ofthe smallest cation and anion.
Answer: H+ and H-

Question 39. Calculate the atomic volume of sodium (atomic mass – 23) if its density is 0.972 g-cm-3.
Answer: Atomic volume of sodium \(=\frac{23 \mathrm{~g}}{0.972 \mathrm{~g} \cdot \mathrm{cm}^{-3}}=23.66 \mathrm{~cm}^3\)

Question 40. Based on atomic number and position in the periodic table arrange the following elements in decreasing order of their metallic character: Si, Na, Mg, P, Be.
Answer: Na > Mg > Be > Si > P

Question 41. Arrange in increasing order of oxidizing power: F, Br, Cl, I
Answer: I < Br < Cl < F

Question 42. The atom of an element has the electronic configuration ls22s22p63s23/?5. Identify a metal or non-metal.
Answer: Non-metal

Question 43. Which of the following oxides is the most acidic? A1203, S02, S03, P4O10 and CO.
Answer: Out of the elements Al, S, P, and C, S lies at the extreme right of the periodic table and so it is the most electronegative.

Again in S03 die oxidation state of S is the highest (+6). Hence, S03 is the most acidic compound. It reacts with water to give a strong acid.

Question 44. Arrange the elements according to the instructions given:

• Na, Cu, Zn (increasing order of electropositivity).
• Na, Cs, K, Rb, Li (increasing order of atomic volume)

Answer:

1. Cu < Zn < Na
2. Li < Na < K < Rb < Cs

Question 45. The first ionization enthalpy of Na is less than that of Mg, but the second ionization enthalpy o/Na is higher than that of Mg. Explain.
Answer: By losing an electron from its outermost shell. Na attains the stable electronic configuration of the nearest noble gas Ne. So, the first ionization enthalpy of Na is less than Mg.

On the other hand, the removal of one electron from the outermost shell of Mg+1 leads to the attainment of the stable electronic configuration of Ne. So the second ionization enthalpy of Mg is less than that of Na.

Question 46. Ionization enthalpy values ofSe, Br, Te andI are 869, 941, 1191, and 1142 kj. mol l-1. The values are not arranged in the correct order. Predict which element the ionization enthalpy value is 869kj. mol-1 and 1142kJ . mol-1 respectively.
Answer: Se and Te are found in the fourth and fifth periods of group 16 in the periodic table. On the other hand, Br and are found in the fourth and fifth periods of group 17.

On moving across a period from left to right, ionization enthalpy increases whereas on moving down a group, its value decreases.

So, the elements in increasing order of ionization enthalpy are Te <I < Se < Br. Therefore, ionization enthalpy values of Te and Br are 869 and 1142 kj-mol-1 respectively.

Question 47. Why is the ionization enthalpy of oxygen less than those of nitrogen and fluorine?
Answer: For the explanation ofthe ionization enthalpy of oxygen is less than nitrogen, the  Ionisation enthalpy of oxygen is less than fluorine because—

1. Nuclear charge increases from oxygen to fluorine,
2. The number of shells remains the same and the addition of differentiating electrons occurs in the same shell of fluorine,
3. The atomic size of oxygen is greater than fluorine.

Question 48. The outermost electronic configurations of the two elements are 2s2 and 2s22p1 respectively. Which has greater ionization enthalpy Which has the highest ionization enthalpy: N, O, Ar, P?
Answer: The element with outermost electron configuration 2s2
has greater ionization enthalpy.

Question 49. The ionization potential of hydrogen is 1312.0 kj mol-1. Express the value in eV atom-1. (leV = 1.6 x 10 19
Answer: Ionisation potential of hydrogen = 1312.0 kj-mol-1

\(\begin{aligned}
& =\frac{1312 \times 10^5}{6.023 \times 10^{23}} \mathrm{~J} \cdot \mathrm{atom}^{-1} \\
& =\frac{1312 \mathrm{w} \times 10^3}{6.023 \times 10^{23}} \times \frac{1}{1.6 \times 10^{-19}} \mathrm{eV} \cdot \mathrm{atom}^{-1} \\
& =13.61 \mathrm{eV} \cdot \mathrm{atom}^{-1}
\end{aligned}\)

Question 50. Electron-gain enthalpy of N is less than that of O. Explain.
Answer: Nitrogen (2s22p3) has stable outer electronic configu¬
ration with half-filled 2p -subshell. So, it is reluctant to accept an additional electron. Thus the electron-gain enthalpy of nitrogen is less than that of oxygen.

Question 51. What is the basic difference between electron-gain enthalpy and electronegativity of an element?
Answer: Electron-gain enthalpy means the enthalpy change involved when an electron is added to an isolated gaseous atom in its lowest energy state.

Whereas electronegativity means the tendency of an atom to attract the shared pairs of electrons toward its nucleus when the atom is covalently bonded in a molecule.

Question 52. Which one has the least electron-gain enthalpy: B, C, N, or O?
Answer: N.

Question 53. Electron-gain enthalpy value of Cu is negative while that of Zn is positive. Explain.
Answer: In the Cu atom, there is only a single electron in the 4s subshell of the valence shell. When it gains another electron, it attains a stable electronic configuration of 3d104s2. However, in Zn, the 4s -subshell of the valence shell is already filled with two electrons. So, the electron-gain enthalpy value of Cu is negative while that of Zn is positive.

Question 54. Indicate the similarity in the electron affinity/Mg and Explain the order of electronegativity: I < Br < Cl < F.
Answer: Both magnesium and nitrogen, have a +ve value of electron-gain enthalpy.

Question 55. Out of the following elements, which one has the least value electronegativity? Na, C, S, Br, Mg, Li
Answer: Out of the given elements Na has the least value of electronegativity.

Question 56. Arrangeinincreasingorderofelectronegativity: N, Na, Si, CL
Answer: Na < Si < N < Cl

Question 57. Which element in each of the following pairs has higher electronegativity: 1 K & Cu, 2 P & S, 3 C & Si
Answer:

1. Cu
2. S
3. C

Question 58. Justify the following statement—The electronegativity of the N atom in all nitrogen-containing compounds is the same.
Answer: The electronegativity of N-atom in all nitrogen-containing compounds is not the same because it depends on the state of hybridization ofthe element under consideration. In different hybridization states, the electronegativity of N follows the sequence: sp-N > sp2-N > sp3-N.

Question 59. Write the formula of the compound formed by the most electronegative and least electronegative elements.
Answer: The most electronegative element is F while the least electro-negative element is Cs. Formula ofthe compound is CsF. 85

Question 60. Which is mostacidic: S03, P205, ZnO, Na20?
Answer: S03 is the most acidic compound.

Question 61. Which is mostbasic: Si02, MgO, A1203, Na20 ?
Answer: Na20 is the most basic compound

Question 62. Arrange according to the instruction given: 4J A1203, P205, C1207, S03 (increasing order of acidity) ,{\() MgO, ZnO, CaO, Na20, CuO (increasingprderofbasicity).
Answer: A1203 < P205 < S03 < C1207

CuO < ZnO < MgO < CaO < Na20

Question 63. How does the basicity of the oxides of representative elements vary on moving a group in the periodic table? On moving across a period from left to right, how does die acidity of the oxides of representative elements vary?
Answer: On moving down a group in the periodic table, the basicity of the oxides of representative elements increases. On moving across a period from left to right, the acidity of the oxides ofthe representative elements increases.

Question 64. identify the following as acidic, basic, or amphoteric oxides: BeO, AlÿOy CaO, Si02.
Answer: Acidic oxide: Si02, basic Oxide: CaO, Amphoteric oxides: BeO, A1203.

Question 65. Between XaOH and CsOH, which one is more basic and why?
Answer: CsOH is more basic than NaOH because the electronegativity of Cs is more electropositive than Na.

Question 66. Formulas of oxide and chloride of an element M are M1O- & MCI3 respectively. State the group to which the element belongs to. Determine whether it is metal or non-metal.
Answer: The valency of the element M is 5 when oxygen Is considered the standard as the valency of M Is 3 when chlorine Is considered as the standard.

Since, the oxygon-butted valency of an element Is equal to Its group number, the group to which M belongs Is VA(I f>). lenient M Is a nonmetal as It Is located In group VA of the periodic table

Question 67. Arrange the following compounds in increasing order of their reducing: NH3, PH3, and AsH3.
Answer: Nh3 < ph3 < AsH3.

Question 68. Which products are liberated at the cathode and anode when molten ionic hydrides are electrolyzed?
Answer: When molten Ionic hydrides are electrolyzed, the metal Ions are discharged at the cathode while hydrogen gas is liberated at the anode.

Classification Of Elements And Periodicity In Properties Very Short Answer Type Questions

Question 1. Mention two laws (for the classification of elements), which were proposed before Mendeleev.
Answer: Dobereincr’s law of triads and Newland’s law of octaves.

Question 2. Write the IUPAC name and the symbol of the element with atomic number 108.
Answer: Unniloctium; Uno.

Question 3. What are the names given to eka-aluminum and cka -Aluminium predicted by Mendeleev?
Answer: The names of these elements are Gallium and Germanium respectively.

Question 4. Mention the number of periods and groups in the long form of the periodic table.
Answer: 7 periods and 18 groups

Question 5. Mention the name of the s -s-block element which is placed along with the p -block elements.
Answer: Helium (He, Is²).

Question 6. To which group of the long form of the periodic table do the chalcogens belong?
Answer: The chalcogens belong to group 16.

Question 7. Write the general electronic configuration of the inner-transition elements.
Answer: \((n-2) f^{1-14}(n-1) d^{0-1} n s^2 \text {. }\)

Question 8. Write the atomic number of the element placed just below cobalt (Z = 27) in the modern periodic table
Answer: The atomic number of the element is 27 + 18 = 45.

Question 9. Indicate the position of the element having electronic configuration Is²2s²2p⁶3s²3p⁶3d³4s² in the periodic table.
Answer: 4th period and 15th group in the periodic table

Question 10. What is the total number of elements incorporated
now in the periodic table? Write down the name and symbol of the last element.
Answer: The total number of elements incorporated in the periodic table is 118. The name of the last element is Ununoctium and its symbol is Uuo (Z = 118).

Question 11. Which group of the periodic table contains solid, liquid & gaseous elements? What are those elements?
Answer: Solid, liquid, and gaseous elements are present in group- 17(VIIA). These elements are fluorine (gas), chlorine (gas), bromine (liquid), iodine (solid) and astatine (solid).

Question 12. Which element is the most electronegative
Answer: Fluorine is the most electronegative of all the elements.

Question 13. Name one property that is not periodic
Answer: The radioactivity of an element is not a periodic property.

Question 14. Arrange according to the instructions given in the bracket:

• O, Te, Se, S (Increasing order of electronegativity)
• Na, Cu, Zn (Increasing order of electropositive character)
• I, F, Br, CI (Increasing order of metallic character)
• I, F, Br, Cl (Decreasing order of electron affinity)
• Na, K, F, Cl, Br (Increasing order of atomic radius)
• Mg, AI, Si, Na (Increasing order of ionization potential)
• PbO, MgO, ZnO (increasing order of basic character)
• Na+, Mg²⁺, Al³⁺ (Decreasing order of size)
• Cu, S, C (graphite) (Increasing order of electrical conductivity)
• Be, C, B, N, O (Increasing order of electron affinity)
• Cl, Mg, C, S (Increasing order of electronegativity)
• A1₂O₃, P₂O₅, CL₂O₇, SO₃ (Increasing order of acidic property)
• MgO, ZnO, CaO, Na20, CuO (Increasing order of basic property)
• Na⁺, F-, O^2-, Mg²⁺, N^3- (Increasing order of ionic radii)
• B —Cl, Ba—Cl, Br —Cl, Cl —Cl (Increasing order of bond polarity)
• Br, F, Cl, I (Increasing order of oxidising property)
• Na, Cs, K, Rb, Li (Increasing order of atomic volume)
• Sb₂O₃, N₂Og, AS₂O₃ (Increasing order of acidic property)

Answer:

• Te < Se < S < O
• Cu< Zn< Na
• F < Cl < Br <I
• Cl > F > Br >I
• F < Cl < Br < Na < K
• Na<Al<Mg<Si
• ZnO < PbO < MgO
• Na⁺ > Mg²⁺ > Al³⁺
• S < C(graphite) <Cu
• Be<N<B<C<0
• Mg < C = S < Cl
• A1₂O₃ < P₂O₅ < SO₃ < C1₂O₇
• CuO < ZnO < MgO < CaO < Na20
• Mg²⁺ < Na⁺ < F^– < O^2-– < N^3-
• Ba —Cl > B —Cl > Br —Cl > Cl —Cl
• I < Br < Cl < F
• Li < Na < K < Kb < Cs
• Sb₂O₃ < AS₂O₃ < N₂O₅

Question 15. Name the elements having the highest and lowest
ionization enthalpy.
Answer: Highest value: Helium (He), Lowest value: Cesium (Cs).

Question 16. What do you mean by the statement—the covalent radius of H-atom is 0.37 A?
Answer: It indicates that one-half of the internuclear distance between two bonded H-atoms in an H₂ molecule is 0.37 A.

Question 17. Indicate the similarity observed in the electron-gain enthalpy values of Mg and N.
Answer: Both Mg and N have positive electron-gain enthalpy values.

Question 18. Why is the size of F- smaller than that of O^2- ion? {
Answer: O^2- and F^– ions are isoelectronic, but their nuclei contain 8 and 9 protons respectively.

So, the nuclear attractive force acting on the electrons of the F^– ion is greater than that on the electrons of the O^2- ion. Consequently, the F^– ion is smaller than the O^2- ion.

Question 19. Compare the radii of K⁺ and Cl^– ions (each contain the same number of electrons)
Answer: The nuclei of the isoelectronic ions K⁺ and Cl^– contain 19 and 17 protons respectively. So the magnitude of nuclear attractive force acting on the electrons of K+ ions is greater than that on the electrons of Cl- ion. Consequently, the radius of the K+ ion is smaller than that of the Cl- ion.

Question 20. What do you understand by the negative value of electron-gain enthalpy of an element?
Answer: It signifies that energy is released when an isolated gaseous atom of the element under consideration accepts an electron to form a monovalent gaseous anion.

Question 21. The electronic configuration of the atom of an element is ls²2s²2p6²3s³3px. Locate its position in the periodic table. Is it a metal or non-metal? What is its valency?
Answer: The receives its last electron in 3p -orbital. So it belongs to p -p-block elements. Accordingly, its group number = 10 + no. of electrons in the valence shell = 10 + (2 + 1)= 13.

Again, the period of the element = several principal quantum numbers ofthe valence shell = 3.

It is metal because it belongs to the 13th group of the periodic table. Valency of the element = number of electrons in the valence shell =2+1 = 3.

Question 22. The atomic numbers of elements A, B, and C are 10, 13, & 17 respectively. Write their electronic configurations.
Which one of them will form a cation and which one an anion?0 Mention their valencies.
Answer: Electronic configuration of 10A: ls²2s²2p⁶ electronic configuration of 13B: ls²2s²2p⁶3s³3p¹ Electronic configuration of 17C: ls²2s²2p⁶3s²3p⁵ The element, A belonging to group 18, is an inert gas.

So it will form neither a cation nor an anion. The element B, belonging to group 13, is a metal. It will readily form a cation by the loss of 3 electrons from its valence shell (3rd shell).

The element C will readily gain one electron in its outermost 3rd shell to attain inert gas electronic configuration (Is²……3s²3p⁶). So, C will form an anion.

Valency of A = 0 (it has a complete octet of electrons in the outermost shell). Valency of B = 3 (by the loss of 3 electrons from the 3rd shell it will attain stable inert gas electronic configuration).

Valency of C = 1 (because by the gain of the electron, it can attain stable inert gas configuration).

Question 23. A, B, and C are three elements with atomic numbers group (8 + 2) = 10. 17, 18, and 20 respectively. Write their electronic configuration. Which one of them is a metal and which one is a non-metal? What will be the formula of the compound formed by the union of A and C? What may be the nature of valency involved in the formation of the above compound?
Answer: Electronic configuration of 1A: ls²2s²2p⁶3s²3p⁵

Electronic configuration of 10B: ls²2s²2p⁶3s²3p⁶

Electronic configuration of 2QC: ls²2s²2p⁶3s²3p⁶4s²

Element C is a metal as it can easily form a dipositive ion by the loss of two electrons from 4s -orbital. ElementA is an anon-metal as it can achieve inert gas configuration by accepting one electron in a 3p -subshell.

As already mentioned, the element C can easily, form a dipositive cation (C²⁺), while the element A readily forms a uninegative anion (A-).

So the elements A and C can combine to form the compound CA2.

The above-mentioned compound is electrovalent because it will be formed by the union of two A^– ions with one C²⁺ ion.

Question 24. Outer electronic configuration of 4 elements is as follows: 3d°4s1 3s²3p⁵ 4s²4p⁶ Electronic configuration of 10A: ls²2s²2p⁶ 3d⁸4s². Find their positions in the periodic table
Answer: This element (3d°4s1) is an s -block element. So it is an element of period group 1.

This element (3s²3p⁵) is a p -p-block element containing (2 + 5) or 7 electrons in the valence shell (n = 3). Soitis an element ofthe 3rd periodin group (10 + 2 + 5) = 17.

This element (4s²4p⁶) is a p -p-block element containing (2 + 6) or 8 electrons in the valence shell (n = 4). So it is an element of the 4thperiodin group (10 + 2 + 6) = 18.

This element (3d⁸4s²) is a d -d-block element containing 8 electrons in the d -orbital of the penultimate shell (n = 3) and 2 electrons in the s -orbital of the valence shell (n = 4). So, it is an element of the 4th period in group (8+2)=10

Question 25. Write the electronic configuration of the element with atomic number 35. What will be the stable oxidation states of the element?
Answer: Electronic configuration: ls²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁵. The most stable oxidation state is -1 because it can accept one electron to achieve inert gas configuration (Is²… 3d¹⁰4s²4p⁶).

Again in an excited state, it can also exhibit oxidation number +3 or +5 by forming a covalent bond by using its 3 or 5 odd electrons in its outermost shell.

Question 26. justify by mentioning two reasons, the inclusion of Ca and Mg in the same group of the periodic table.
Answer: Both Ca and Mg have similar electronic configurations. Both of them belong to s -s-block. Electronic configuration of Mg: ls²2s²2p⁶3s² Electronic configuration of Ca: ls²2s²2p⁶3s²3p⁶4s²

Both are typical divalent metals and form stable ionic compounds, for example; MgCl₂, CaCI₂; MgO, and CaO.

Question 27. Cu, Ag & Au are regarded as transition elements. Why?
Answer: Despite having filled d -orbitals (d10), copper, silver, and gold are considered transition elements because at least in one stable oxidation state of the atoms of these elements, d -orbital is partially filled with electrons, For example, the electronic configuration of Cu²⁺ is ls²2s²2p⁶3s²3p²3d94s².

Question 28. Write down the electronic configuration of the element having atomic number 21. Name two other elements of the same series to which this element belongs. Why do they belong to the same series?
Answer: Electronic configuration ofthe element (Atomicnumber= 21): ls²2s²2p⁶3s²3p⁶3d¹⁴s² From the electronic configuration, it can be said that the element belongs to the first transition series as its 3d -orbital is partially filled. Two other elements belonging to this series are Cr(24) and Mn(25). The 3d -orbitals of these elements are also partially filled. Their electronic configurations are—

Cr: ls²2s²2p⁶3s²3p⁶3d⁵4s¹

Mn: ls²2s²2p⁶3s²3p⁶3d⁵4s²

Question 29. Can Cu (At. no. 29) and Zn (At. no. 30) be called transition elements? Explain.
Answer: Although the Cu -atom in its ground state does not contain an incompletely filled d-orbital, Cu²⁺ has a partially filled d-orbital and so it is a transition element.

• Electronic config. of Cu: ls²2s²2p⁶3s²3p⁶3d¹⁰4s¹
• Electronic config. of Cu2+: ls²2s²2p⁶3s²3p⁶3d⁹4S²

In the ground state or any stable oxidation state of Zn, the d orbital is filled. So, it is not a transition element.

Question 30. The atomic numbers of three elements A, B, and C are 9, 13, and 17 respectively.

• Write their electronic configuration.
• Ascertain their positions in the periodic table.
• Which one is most electropositive and which one is most electronegative?

Answer: Electronic configurations of gA: ls²2s²2p⁵, 13B: ls²2s²2p⁶3s²3p¹, 17C: ls²2s²2p⁶3s²3p⁵

All three elements are p -p-block elements. Hence, their group and period numbers are as follows:

Question 27. Cu, Ag & Au are regarded as transition elements. Why?
Answer: Despite having filled d -d-orbitals (d10), copper, silver, and gold are considered transition elements because at least in one stable oxidation state of the atoms of
these elements, the -orbital is partially filled with electrons, For example, the electronic configuration of Cu²⁺ is ls²2s²2p⁶3s²3p⁶3d¹⁴s⁰

Question 28. Write down the electronic configuration of the element having atomic number 21. Name two other elements of the same series to which this element belongs. Why do they belong to the same series?
Answer: Electronic configuration ofthe element (Atomicnumber= 21) : ls²2s²2p⁶3s²3p⁶3d¹⁴s²

From the electronic configuration, it can be said that the element belongs to the first transition series as its 3d –orbital is partially filled.

Two other elements belonging to this series are Cr(24) and Mn(25). The 3d -orbitals of these elements are also partially filled. Their electronic configurations are—

• Cr: ls²2s²2p⁶3s²3p⁶3d⁵4s¹
• Mn: ls²2s²2p⁶3s²3p²3d5⁴s²

Question 29. Can Cu (At. no. 29) and Zn (At. no. 30) be called transition elements? Explain.
Answer: Although the Cu -atom in its ground state does not contain an incompletely filled d-orbital, cu 2+ has a partially filled d-orbital and so it is a transition element.

• Electronic config. of Zn: ls²2s²2p⁶3s²3p⁶3d¹⁰4s¹
• Electron Config. of Zn²⁺: ls²2s²2p⁶3s²3p⁶3d⁹4s⁰

Question 30. The atomic numbers of three elements A, B, and C are 9, 13, and 17 respectively.
Write their electronic configuration. Ascertain their positions in the periodic table. Which one is most electropositive and which one is most electronegative?
Answer: Electronic configurations of gA: ls²2s²2p5 13B: ls²2s²2p⁶3s²3p1, 17C: ls²2s²2p⁶3s²3p⁵

All three elements are p -block elements. Hence, their group and period numbers are as follows:

Element B can easily donate 3 electrons from its outermost shell to attain a stable inert gas configuration. So, it is the most electropositive element.

Elements A and C are electronegative because they can accept one electron to attain stable
inert gas electronic configuration.

These elements (A and C) have similar outer electronic configurations (ns2np5) but the size of A is smaller than that of C. So, the electronegativity of A is greater than that of C. Hence, A is the most electronegative element.

Question 31. Why is the atomic size of Ca²⁺ smaller than that of K⁺?
Answer: K+ and Ca²⁺ are isoelectronic (each contains 18 electrons). K⁺ contains 18 protons in its nucleus and Ca²⁺ contains 20 protons.

Several protons being in Ca²⁺  ion, the electrons in the outermost shell of Ca²⁺  ion will experience greater attractive force by the nucleus compared to K+. Consequently, Ca²⁺ has a smaller ionic size than K⁺ ions.

Question 32. The atomic radius of the chlorine atom is 0.99 A but the ionic radius of chloride (Cl-) is 1.81 A—explain.
Answer: In an anion, the total number of electrons being greater than that of the number of protons, attraction of the nucleus for the outermost electrons decreases.

Furthermore, due to mutual repulsion among electrons, the outermost orbit gets appreciably expanded.

As a result, the anionic radius becomes more than the radius of the parent atom. On account of this, the radius of Cl- (1.81 A) (18 electrons and 17 protons) is greater than the radius of the chlorine (0.99 A) atom (17 electrons and 17 protons).

Question 33. First ionization energy of elements increases with the atomic numbers of the elements in a period—why? Cite an exception.
Answer: In any period, with an increase in atomic number, the magnitude of the positive charge of the nucleus increases, but there is no addition of any new shell.

In consequence, the attractive force of the nucleus for the outermost electrons increases. So, the amount of energy required for the removal of an electron from the outermost shell of the atom (z.e., first ionization energy) in any period increases with the increase in atomic number.

An exception to this generalization is the nitrogen-oxygen pair. In the second period, oxygen belonging to group VIA has a lower value of first ionization energy than nitrogen of group. The reason can be ascribed to the stable electronic configuration of the nitrogen atom (ls²2s²2p³)

Question 34. The ionization potential of O is less than that of N—
explain.
Answer: The reason for such a difference may be explained based on their electronic configurations filled, its electronic configuration is highly stable.

So, a large amount of energy is required to form N+ ion by removal of 2pelectron.

On the other hand, the formation of 0+ ion by removal of one electron from a partially filled 2p -orbital requires less energy, since the 2p -orbital of 0+ion is half-filled, the electronic configuration assumes stability. Hence, oxygen has a lower ionization potential than nitrogen.

Question 35. Explain why the ionization potentials of inert gas are very high while that of alkali metals are very low. ses are
Answer: Outermost shells of inert gases contain octets of electrons. Besides this, each of the inner shells of inert gas elements is filled. Such configuration is exceptionally staMe conversion of a neutral inert gas atom into its ions try removal of an electron from the outermost shell requires large energy. As a result, they have high ionization potentials.

Configuration of the outermost and penultimate shell of alkali metals is {n-l)s2(n-1)p6nsl (except Li ). Thus loss of 1 electron from their outermost shell brings about a stable configuration of inert gases. Hence, the conversion of alkali metals to their ions requires comparatively less amount of energy. As a result, alkali metals have low values of ionization potential.

Question 36. Which member in each of the following pairs has
a lower value of ionization potential? F, Cl S, Cl Ar, K O Kr, Xe Na, Na+.
Answer: Cl has lower ionization enthalpy than F because electrons of 2p-orbital are more strongly attracted by the nucleus than the 3p-electrons in Cl.

(Note that effective nuclear charge on the outermost electrons is nearly the same for both and Cl).

S has lower ionization enthalpy than Cl because the size of S is greater than that of Cl and also the nuclear charge of is less than that of CL

K has a lower ionization potential than Ar as the outermost shell is filled with electrons in Ar. On the other hand, K can attain a stable configuration like the inert gas Arby the loss of only 1 electron.

Xe has lower ionization enthalpy than Kr because ionization enthalpy decreases on moving down a group in the periodic table.

Na (ls²2s²2p²⁶3s¹) has lower ionization enthalpy’ than Na+(ls²2s²2p⁶), because the former can attain inert gas-like electronic configuration by loss of 1 electron from its outermost shell, whereas the latter attains unstable electronic configuration (Is²2s²2p⁵) by loss of one electron from its outermost shell.

Question 37. Why is the value of electron-gain enthalpy negative?
Answer: When an electron is added to a neutral gaseous atom to form a negative gaseous ion, energy’ is usually liberated, i.e., the enthalpy change in the process is usually negative. So electron-gain enthalpies of most elements have negative values.

Question 38. Calculate the energy (in kj unit) required to convert all sodium atoms into sodium ions, present in 3.45 mg of its vapor. [I.P of sodium 490kJ mol-1 ]
Answer: Sodium atoms present \(=\frac{3.45}{23} \times 10^{-3}=1.5 \times 10^{-4} \mathrm{~mol}\)

Na(g) + = Na+(g) + = 490 kj. mol-1.

The energy required for the ionization of 1 mol sodium is 490 kj.

The energy required for ionization of 1.5 X 10-4 mol sodium is 490 X 1.5 X 10-4 kj = 73.5J

Question 39. A, B, C, and D are four elements of the same period, of which A and B belong to s -block. B and D react together to form B+D~. C and D unite together to produce a covalent compound, CD₂. What is the formula of the compound formed by A and D ? 0 What is the nature of that compound? What will be the formula and nature of the compound formed by the union of B and C
Answer: Since A and B are s -block elements of the same period, one of them is an alkali metal group-1A while the other is an alkaline earth metal of group-2A.

B and D react to form anionic compound B+D. Therefore, B is a monovalent alkali metal of group IA, and D is a monovalent electronegative element of group VIIA.

Hence, the other element A of the s -the block is a bivalent alkaline earth metal of group-2A. C and D combine to produce the covalent compound CD₂. Hence, C is a bivalent electronegative element belonging to group 4A.

The formula of the compound formed by the combination ofelectropositive bivalent element A with electronegative monovalent element D is AD₂

The compound is ionic or electrovalentinnature. A compound formed by reactions of electropositive monovalent element B with electronegative bivalent element C will have the formula B₂C. It is an electrovalent or ionic compound.

Question 40. What changes in the following properties are observed while moving from left to right along a period & from top to bottom in a group? Atomic volume, Valency, Electronegativity, Oxidising, and reducing powers.
Answer: On moving from left to right across a period, atomic volume first decreases and then increases. In a group, atomic volume increases with an increase in atomic number down a group.

Oxygen-based valency goes on increasing from left to right over a period but does not suffer any change down a group.

Electronegativity increases gradually from left to right across a period while it decreases down a group with an increase in atomic number.

Oxidizing power increases from left to right across a period and decreases from top to bottom in a group.

On the other hand, reducing power decreases from left to right in a period but it increases from top to bottom in a group.

Classification Of Elements And Periodicity In Properties Short Answer Type Questions

Question 1. Write the name of the element which is diagonally related to the element beryllium. Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively.
State the periodic table in which elements A and C belong. Write the formulas ofthe compounds formed between B and C and A and C. State the nature ofthe bonds.
Answer: Aluminium (Al)

• 11^A: ls²2s²2p⁶3s2³p²,
• 17^C: ls²2s²2p⁶3s²3p⁵

From the electronic configuration is evident that A is an s -a block element whereas C is a p -a block element.

The compound formed between B and C has the formula BC₄ and the nature of the bond is covalent. The compound formed by a combination of A and C is AC (electrovalent).

Question 2. Classify as basic, amphoteric, or acidic: BeO, A12O₃, CaO, SiO₂ State the modem periodic law.
Answer: BeO, A1₂O₃: amphoteric; CaO: basic; SiO₂: acidic

Question 3. Write the name of the element which is diagonally related to the element beryllium. Three elements A, B, and C have atomic numbers 11, 14, and 17 respectively. State the periodic table in which elements A and C belong. Write the formulas ofthe compounds formed between B and C and A and C. State the nature ofthe bonds.
Answer: 

1. Symbol of one transition element — Fe.
2. Symbol of one inner-transition element — Ce
3. Ionic radius is highest for O2- for the given case.
4. Electronegativity is lowest for Na for the given case.
5. Ionization energy is highest for Ar for the given case.

Question 4. State the group number in the modem periodic table where solid, liquid, and gaseous elements are present at room temperature. Identify solid, and liquid elements. Indicate the given elements as alkali metal, alkaline-earth metal, coinage metal, chalcogen: Li, Ca, S, Cu.
Answer: Group 17 of modem periodic table contains solid (iodine, I₂), liquid (bromine, Br₂), and gaseous (chlorine, Cl₂) elements at the same time

• Group 17 of the modern periodic table contains solid
• (iodine, I₂), liquid (bromine, Br₂) and gaseous
• (chlorine, Cl₂) elements at the same time.

Question 5. Among which of the following pairs of elements, the first has lesser ionization energy than the second—

1. Na, K
2. N, O
3. B, Be
4. Br, I

Answer: 3. B, Be

Question 6. Name a pair of elements that exhibit diagonal relationships.
Answer: Li and Mg.

Question 7. Name a chalcogen and an alkaline earth metal.
Answer: Oxygen (O) is an example of chalcogen and magnesium (Mg) is an alkaline earth metal

Question 10. Mention the names of the most electropositive and most
electronegative stable elements in the periodic table. 2 What is metalloid? Give one example.
Answer: The most electropositive stable element of the periodic table is cesium (Cs) and the most electronegative stable element of the periodic table is fluorine (F).

Question 11. Which has greaterionisation energy and why—S or P?
Answer: Phas greaterionisation energy than S. According to Hund’s
rule the half-filled electronic configuration is highly stable
and the outermost 3p -orbital of the P atom is half-filled.

To produce P+ ions by removing an electron from the 3p orbital ofPrequireshuge energyresulting a very high value of ionization energy. On the other hand, S atom has a partially filled 3p -orbital which attains a stable half-filled electronic configuration in the S+ ion.

Therefore, to produce S to S+ comparatively lower energy is required resulting in a lower-value of ionization energy.

Question 13. Arrange as directed: SiO₂, NO₂, A1₂O₃, C1O₂ (Increasing acidic property) Na+, Al³⁺, F-, Cl- (Increasing ionic radius) Arrange the following ions in the increasing order of their ionic radius. F-, Mg²⁺ Al³⁺, O^2-
Answer: AI₂O₃ < SiO₂ < NO₂ < C1O₂ Al³⁺ < Na+ < F- < Cl-

Question 14. Arrange the following ions in the increasing order of their ionic radius. F-, Mg²⁺, Al³⁺, O^2-
Answer: Al³⁺ < Mg²⁺ < F^– < O^2-

Question 15. Which ofthe given elements has the lowest electron affinity

1. C
2. P
3. O
4. S

Answer: 2. P

Question 16. Arrange the following elements according to their decreasing oxidising ability: I, Br, F, Cl.
Answer: F > Cl > Br >I

Question 17. Arrange as per instruction: Na₂O, B₂O₃, AI₂O₃, MgO (increasing order of basicity) Be, 0, Cl, I (increasing order of electro-negativity)
Answer: B₂O₃ < A1₂O₃ < MgO < Na₂O ; Be <I < Cl < O

Question 19. Two atoms X and Y have electronic configurations [He] 2sz₂p₃ and [Ne] 3s₂3p₂ respectively. Which period and group of the periodic table do they belong to? The second electron affinity of an element is always endothermic. Explain.
Answer: X — Period 2, Group-15. Y — Period 3, Group-14.

Question 20. Electron affinity ofCl is greater than that of. Explain. Indicate the basic/acidic behavior of the hydrides of the following elements: F, C, O, and N Metallic properties of elements in the periodic table.
Answer: HF: acidic, CH₄: neutral, H₂O: neutral, NH₃: basic (From left to right along a period the relative acidity of the hydrides ofthe elements increase).

Question 21. Metallic property of elements down the group in the periodic table
Answer: HF: acidic, CH₄: neutral, H₂O: neutral, NH₃: basic (From left to right along a period the relative acidity of the hydrides ofthe elements increase).

Question 22. Arrange the following in increasing order of ionic radius: Na⁺, F^–, O^2-, Al³⁺, N^3-.
Answer: Al³⁺ < Na⁺ < F^– < O^2- < N³

Question 23. Why does nitrogen have a higher ionization enthalpy than that of Oxygen? Arrange the following in increasing order of acidity: NO₂, A1₂O₃, SiO₂, CIO₂
Answer: Na < A1 < Mg < Si

Question What will be the order of, Mg, A1, and SI in terms of the first ionization enthalpy?
Answer: Mg²⁺ < Na⁺ < F^– < O^2-

Question 25. Arrange the following ions in order of increasing ionic radii: Na+, F-, O2-, Mg2+.
Answer: Mg2+ < Na+ < F” < O2-

Question 26. Why is the electron-gain enthalpy of oxygen less than
that of sulfur? Arrange the following metal oxides in terms of ascending order of basicity: ZnO, MgO, CaO, CuO.
Answer: Electron-gain enthalpy of O is less than that of S as the 2p -orbital of the outermost shell of O-atom is much smaller in size than the 3p -orbital of the outermost shell of the S-atom.

So, the additional electron-electron repulsive force produced due to the addition of one electron to the 2p -orbital of the oxygen atom from outside is more than the additional electron-electron repulsive force developed by the addition of one electron to the 3p -orbital of the S-atom. So the electron-gain enthalpy of the O-atom becomes less than that S-atom.

⇒ \(\mathrm{CuO}<\mathrm{ZnO}<\mathrm{MgO}<\mathrm{CaO}\)

Question 27. Why is the first ionization enthalpy of helium maximum among all the elements? Arrange the given compounds in terms of ascending order of oxidizing property: HC1, HBr, HI, HF
Answer: Configuration of He is Is₂, i.e., all electrons of He are present in the Is -orbital. These electrons are attracted very strongly by the nucleus. Further, there are no inner electronic orbitals to shield these electrons from the pull of the nucleus.

So removal of an electron from the Is -orbital requires a large amount of energy. Therefore, the first ionization enthalpy of He is maximum. Hydrogen halides do not show oxidising properties. However, the reducing power of the hydrogen halides follows the sequence: HF <HCl < HBr < H

Question 28. Determine the position of an element in the long form of the periodic table if its electronic configuration is [18Ar]3d104s2.
Answer: It is a d -block element. Hence period no. = principle quantum no. of the outermost shell = 4 and its group no. = total electronin 3d and 4s -orbital = 10 + 2 = 12.

Question 29. Mention the name and the position of two elements, one which is most electronegative and the other most positive in the periodic table.
Answer: Most electronegative element =F (group 17, period-2)

Most electropositive element = Cs (group 1, period-6)

Question 30. Which of the following two elements has a diagonal relationship? Li, Be, A1, and Si Between 2gCu and igK which one has higher ionization enthalpy and why
Be and Al

• Electronic configuration of 19K: ls²2s²2p⁶3s²3p⁶4s¹
• Electronic configuration of 2gCu: ls²2s²2p⁶3s²3p⁶3d¹⁰4s¹

The nucleus of the Cu-atom contains 10 more protons than that of the K-atom.

Additional nuclear pull on the outermost electron (4s1) in Cu-atom is not counter-balanced by the shielding effect of ten 3d-electrons because orbitals have poor screening effects. Thus the effective nuclear charge acting on the 4s-electron of copper is greater than that acting on the 4s-electron of

Question 31. Arrange the following ions in ascending order of radius: Na+, F-, O2-, Mg2+
Answer: Mg²⁺ < Na+ <F^– < O^2-

Question 32. Is the electronegativity of Sn²⁺ and Sn⁴⁺ equal or different?
Answer: Elements in a higher oxidation state have higher electronegativity compared to the elements in a lower oxidation state. Thus Sn⁴⁺ is more electronegative than Sn₂

Question 33. What is the oxidation state of TI in the compound t₃? Which is the stronger oxidizing agent between CO2 and PbO2 and why?
Answer: +1

PbO2. Due to the inert pair effect, Pb is very unstable in the +4 oxidation state. Thus Pb⁴⁺ can be easily reduced to Pb2.

Question 34. Which iw is more stable between BC1₃ and T1C1₃ and why? What is the oxidation state of Zn in Zn-Hg Arrange the following elements in the increasing order of their first ionisation enthalpy.IX Be. Na, Mg 56 Arrange the following elements in the decreasing order of their electro-negativity. Si N. F. Cl
Answer: Because of the poor shielding effect by the inner d – and/- electrons, the inert pair effect is maximum for Tl.

Thus the most stable oxidation state of TL is +1 and not +3. Therefore TIClg is unstable. B does not exhibit an inert pair effect and thus BC1₃ is stable.

Question 37. The outermost electronic configuration of the atom of an element is 3s-3p3. Mention the position of the element in the long periodic table.

Why is electron gain enthalpy of oxygen less than that of sulfur?
Answer: Because of the poor shielding effect by the inner d – and/- electrons, the inert pair effect is maximum for Tl. Thus the most stable oxidation state of TL is +1 and not +3. Therefore TIClg is unstable. B does not exhibit an inert pair effect and thus BC13 is stable.

Classification Of Elements And Periodicity In Properties Solved NXERT Exercise

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 th<? 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 the 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 sixth period 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 in the case of6th period continues till a 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 a 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) of the 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 of the third period and group-17 Is 3ia³p1² 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 & 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 and 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 (N^3-) 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 (S^2-) ion [16 + 2]. chloride (Cl ion [17 + 1], potassium (K⁺) ion [19 -1], calcium (Ca2+) ion [2O^-2].

There are (12-2) = 10 electrons in Mg2t. Isoelectronic species of Mg²⁺ are nitride (N^3-) ton [7 + 3], oxide (O^2-) 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 (Sr^2-) Ion [30^2-].

Question 12. Consider the given species: N^2-, O^2-, I^1-, 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 of 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 N^3- < O^2- <F^– < Na⁺ < Mg²⁺ < Al³⁺. Therefore, the order of increasing ionic radii is: Al³⁺ < Mg²⁺ < Na+ < F- < O^2- < N^3-.

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 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:

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 the 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 when 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 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

Question 21. Would you expect the second electron-gain enthalpy of 0 as positive, more negative, or less negative than the first? Justify your answer.
Answer: There are sue valence electrons in oxygen and it requires two more electrons to complete its octet. So, the O-atom accepts one electron to convert into O~ ion and in the process liberates energy. Thus, the first electron-gain enthalpy of oxygen is negative.

⇒ \(\mathrm{O}(g)+e \rightarrow \mathrm{O}^{-}(g)+141 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\left(\Delta_i H_1=-v e\right)\)

However, when another electron is added to O^– to form an O2-ion, energy is absorbed to overcome the strong electrostatic repulsion between the negatively charged O ion and the second incoming electron. Thus, the second electron-gain enthalpy of oxygen is positive.

⇒ \(\mathrm{O}^{-}(\mathrm{g})+e \rightarrow \mathrm{O}^{2-}(\mathrm{g})-\left(780 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\right)\left(\Delta_i H_2=+v e\right)\)

Question 23. How would you react to the statement that the electronegativity of N on the Pauling scale is 3.0 in all the nitrogen compounds?
Answer: The electronegativity of any element depends on the hybridization state and oxidation state of that element in a particular compound, i.e., the electronegativity of an element varies from compound to compound. For example, the electronegativity of Natom varies as sp³ —N < sp²—N < sp—N. So, the, given statement is not correct.

Question 24. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Answer: Isotopes of an element have the same number of electrons and similar electronic configurations. So their nuclear charge and atomic radii are identical. Consequendy, two isotopes of the same element are expected to have the same ionization enthalpy.

Question 25. What are the major differences between metals and non-metals?
Answer: Metals have a strong tendency to lose electrons to form cations they are strong reducing agents, have low ionization enthalpies, have less negative electron gain enthalpies, low electronegativity, and form basic oxides and ionic compounds.

Non-metals have a strong tendency to accept electrons to form anions. They are strong oxidizing agents, have high ionization enthalpies, have high negative electron-gain enthalpies, and high electronegativity, and form acidic oxides and covalent compounds.

Question 26. Use the periodic table to answer the given questions. Identify an element with 5 electrons in the outer subshell. Identify an element that would tend to lose 2 electrons. Identify an element that would tend to gain 2 electrons. Identify the group having metal, non-metal, liquid, and gas at room temperature
Answer: Fluorine. Its configuration is ls²2s²2p⁵

Magnesium. Its configuration is ls2²s2²p⁶3s². So, Mg loses 2 electrons from its outermost shell to form Mg²⁺ and attains a stable configuration.

Oxygen. Its configuration is ls²2s²2p⁴ So, O gains 2 electrons to form O^2- and attains stable configuration. Group-17. The metallic character of astatine (At) is much greater than its non-metallic character and its melting point is very high (302°C).

So, astatine is considered as a metal. So in group-17 there is a metal (At), non-metals (F₂, Cl₂, Br₂, I₂), liquid (Br₂) and gas (F₂, Cl₂).

Question 28. The order of reactivity of group-1 LI < Na < K < Rb < Cs whereas that of group-17 elements Is F > Cl > Br >I. Explain.
Answer: There is only one electron in the valence shell of the elements of group 1. Thus, they have a strong tendency to lose this single electron. The tendency to lose electrons depends on the ionization enthalpy. As ionization enthalpy decreases down the group, the correct order of increasing reactivity of group 1 elements is Li < Na < K < Rb < Cs. On the other hand, there are 7 electrons in the valence shell of the elements of group-17.

Thus, they have a strong tendency to gain a single electron. The tendency to gain electrons depends on the electrode potentials of the elements. As the electrode potential of elements decreases down the group, the correct order of activity is F > Cl > Br >I.

Alternate explanation: In the case of halogens, their reactivity increases with the increase in electron-gain enthalpy.

Order of electron-gain enthalpy: F < Cl > Br >I. As electron gain enthalpy decreases from Cl to, the order of reactivity also follows this sequence. However, fluorine is the most reactive halogen as its bond dissociation energy is verylow.

Question 29. Assign the position of the element having outer electronic configuration: ns₂np₄ for n = 3, 0 (n-l)d₂ns₂ for n = 4 & (n-2)/7(n-l)d1ns₂ for tt = 6, in the periodic table.
Answer: (T) As n= 3, the element belongs to the period. Since the last electron enters the p-orbital, the given element is a p-block element. For p-block elements, group no. of the element = 10+no. of electrons in the valence shell. The element is in the (10+6) = 16th period. As n = 4, the element belongs to the fourth period.

Since is present in the element, it is a block element. For d-block elements, group no. of the element = no. of ns electrons + no. of(n-l) f electrons = 2+2 = 4. Therefore, the element is in the 4th period.

As n – 6, the element belongs to the sixth period. Since the last electron enters the f-orbital, the element is a f-block element. All f-block elements are situated in the third group of the periodic table.

The first (A1H1) and second (A1₂) ionization enthalpies (klmol-1) and the (AegH) electron gain enthalpy (in kj.mol-1 ) of a few elements are given below: Which of the above elements is likely to be: the P least reactive clement. the most reactive metal.

the most reactive non-metal. the least reactive non-metal. the metal which can form a stable binary halide of the formula MX₂(X = halogen).

Question 30. The metal that can form a predominantly stable covalent halide of the formula MX (X = halogen)? Element V is the least reactive metal as it has highest first ionization enthalpy & positive electron-gain enthalpy.
Answer: Element is the most reactive metal as it has the lowest first ionization enthalpy & low negative electron-gain enthalpy.

Element 3 is the most reactive non-metal because it has a very high first ionization enthalpy and very high negative electron-gain enthalpy.

Element 4 is the least reactive non-metal because it has a high negative electron-gain enthalpy but not so high first ionization enthalpy.

Element 4 has low first and second ionization enthalpy. Again, the first ionization enthalpy of this element is higher than those ofthe alkali metals. Thus, the given element is an alkaline earth metal and can form a stable binary halide ofthe formula MX₂.

The first ionization enthalpy of elements is low but its second ionization enthalpy is high. So, it is an alkali metal and can form stable covalent halide (MX)

Predict the formulas of the stable binary compounds that would be formed by given pairs of elements: OLi and O,0Mg and N, A1 and, Si and O, P and F, Element with atomic numbers 71 and F.

Question 31. In the modern periodic table, period indicates the value of atomic number atomic mass principal quantum number azimuthal quantum number
Answer: Each period in the modern periodic table begins with the filling of a new shell. So, the period indicates the value of a principal quantum number.

Question 32. Which of the following statements related to the modern periodic table is incorrect?

• p -block has 6 columns because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
• d -block has 8 columns, as a maximum of 8 electrons can occupy all the orbitals in a d -subshell.
• Each block contains many columns equal to the number of electrons that can occupy that subshell.
• Block indicates the value of azimuthal quantum number (l) for the last subshell that received electrons in building up electronic configuration

Answer: The statement is incorrect because d -the block has 10 columns because a maximum of ten electrons can occupy all the orbitals in a d -subshell.

Question 33. Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?

1. Valence principal quantum number (n)
2. Nuclear charge (Z)
3. Nuclear mass
4. Number of core electrons.
5. Nuclear mass does not affect the valence shell electrons (such as, and H have similar chemical properties.
6. The size of isoelectronic species F-, Ne, Na+ is affected by: nuclear charge (Z)
7. valence principal quantum number (n)

Higher Order Thinking Skills

Question 1. What will be the name (IUPAC) and symbol if the element with atomic number 119 is discovered? Write its electronic configuration. Also, write the formulas of the stable chloride and oxide of this element.
Answer: IUPAC name : Ununennium, Symbol: Uue

Atomic number ofthe element =119 = 87 + 32

It is known that the element with atomic number 87 is francium (Fr). Fr belongs to group 1 in the 7th period of the periodic table.

So, the element with atomic number 119 will take its position in group 1 and 8th period just below francium(Fr).

The electronic configuration of this element will be [UuojBs1, (where Uuo = Ununoctium, Z = 118). It will be an alkali metal with valency=1

If the symbol ofthe element is ‘M’ then the formulas of its stable chloride and oxide will be MCI and M20 respectively.

Question 2. Formulas of oxide and chloride of an element A are A205 and AClg respectively. Which group of Mendeleev’s periodic table will the element belong to? State whether it is a metal or a non-metal.
Answer: Hiking oxygen as standard, the valency ofthe element A is 5, and taking chlorine as standard, the valency of A is 3. Since oxygen-based valency (8-chlorine-based valency) is the same as that ofthe group number of the element, the element will be of group Mendeleev’s periodic table. Thus, itis a non-metal.

Question 3. A and B are two elements with atomic numbers 9 and 17 respectively. Explain why the element A is a more powerful oxidizing agent than the element B.
Answer: Electronic configuration of gA: ls²2s²2p⁵. Electronic configuration of 1? B: ls²2s²2p⁶3s²3p⁵. Both A and B can accept electrons to form the anions of A- and B_ having inert gas electronic configuration. However, the anions A- are more stable than B- because of the smaller atomic size of A. So A is a stronger oxidizing agent than B.

Question 4. Elements A, B, and C have atomic numbers (Z- 2), Z, and (Z +1) respectively. Of these, B is an inert gas. Which one of these has the highest electronegativity? Which one of these has the highest ionization potential? What is the formula of the compound formed by the combination of A and C? What is the nature of the bond in this compound?
Answer: Since element B (atomic no =Z) is an inert gas, the element ‘A’ with atomic number (Z- 2) is included in group 6A. On the other hand, the element C, having an atomic number (Z + 1) must belong to groups (alkali metal). Hence, the electronegativity of the element A is maximum.

The element B, being an inert gas, has the highest value of ionization potential.

The valency of the element A, belonging to group (8- 6) = 2, and that ofthe element C, being an element of group IA, is 1. Therefore, the formula of the compound formed by A and C will be C₂A.

Being a strongly electronegative element and C being a strongly electropositive element complete their octet through gain and loss of electrons respectively. So, the nature of the bond formed between C and A in C₂A isionic or electrovalent bond.

Question 5. Atomic radius of 1(2)Ne is more than that of gF —why?
Answer: Fluorine forms diatomic molecules, thus the atomic radius of fluorine is a measure of half of the internuclear distance in its molecule (i.e., half of the covalent bond length of F₂ molecule) but neon being an inert gas, its atoms are incapable of forming covalent bonds by mutual combination amongst themselves.

The only force that comes into play between the atoms is the weak van der Waals force. So a measure of the atomic radius of Ne is equal to its van der Waals radius but the van der Waals radius is always greater than the covalent radius.

Thus, the atomic radius of neon is larger than that of fluorine. Furthermore, due to an increase in the number of electrons in the outermost 2p -orbital of Ne, there occurs an increase in electron-electron repulsion. So 2p-orbital of Ne suffers expansion leading to its increased atomic radius.

Question 6. The first electron affinity of oxygen is negative but the second electron affinity is positive—explain.
Answer: When an electron is added to the valence shell of an isolated gaseous O-atom in its ground state to form a negative ion, energy is released.

Because a neutral oxygen atom tends to complete its octet with electrons. So, the electron affinity of oxygen is an exothermic process and its value is negative.

When an extra electron is added to an O- ion, that second electron experiences a force of repulsion exerted by the negative charge ofthe anion. So, first, this process requires a supply of energy from an external source.

This accounts for the endothermic nature of second electron affinity and has a positive value.

Question 7. The electron affinity of sodium is negative but magnesium has a positive value—why?
Answer: Electronic configuration of nNa: ls²2s²2p⁶3s¹ Electronic configuration of 12Mg: ls22s22p63s2 Addition of one electron to the 3s -orbital of Na leads to a comparatively stable electronic configuration with fulfilled orbital.

So, this process of the addition of electrons to Na is an exothermic, process. So, the electron affinity of Na is negative. On the other hand, Mg has fulfilled 3s orbital and that has a stable electronic configuration.

So the addition of an electron to the 3p -orbital destabilizes the electronic configuration of Mg.

Additional energy is required for the addition of electrons to the outermost shell of Mg i.e., this process is endothermic and thus the value of electron affinity ofMg is positive.

Question 8. If the electron affinity of chlorine is 350 kJ*moI-1, then what is the amount of energy liberated to convert 1.775 g of chlorine (existing at atomic state) to chloride ions completely (in gaseous state)
Answer: Atomic mass of chlorine = 35.5 g .mol 1
The energy liberated in the conversion of 35.5 g of Cl to Clion =350 kj
Energyliberatedin the conversion of1.775 g ofCl to Cl-

\(\text { ion }=\frac{350}{35.5} \times 1.775=17.5 \mathrm{~kJ}\)

Question 9. The second ionization enthalpy of Mg is sufficiently high and the second electron-gain enthalpy of O has a positive value. How do you explain the existence of Mg²⁺ O^2- rather than Mg+O-?
Answer: The lattice energy of an ionic crystal depends on the force of
attraction between the cations and anions \(\left(F \propto \frac{q_1 q_2}{r^2}\right)\)

So, the magnitude of lattice energy increases as the charges on the cation and anion increase. Consequently, the lattice energy of Mg²⁺ O^2- is very much greater than that of Mg⁺ O^–.

The lattice energy of Mg²⁺ O^2- is so high that it exceeds the unfavorable effects of the second ionization enthalpy of Mg and the second electron-gain enthalpy of 0.

So, Mg²⁺ O^2- is a stable ionic compound, and its formation is favored over Mg⁺ O^–.

Question 10. The atomic numbers of some elements are given below. Classify them into three groups so that the two elements in each group exhibit identical chemical behavior: 9, 12, 16, 34, 53, 56.
Answer:

\(\begin{array}{|c|c|}
\hline \begin{array}{c}
\text { Atomic } \\
\text { no. }
\end{array} & \text { Electronic configuration } \\
\hline 9 & 1 s^2 2 s^2 2 p^5 \\
\hline 12 & 1 s^2 2 s^2 2 p^6 3 s^2 \\
\hline 16 & 1 s^2 2 s^2 2 p^6 3 s^2 3 p^4 \\
\hline 34 & 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^4 \\
\hline 53 & 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^5 \\
\hline 56 & 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 d^{10} 5 s^2 5 p^6 6 s^2 \\
\hline
\end{array}\)

Elements with atomic numbers 9 and 53 belong to the -block and they have similar outer electronic configurations (ns2np5). So they will exhibit similar chemical properties. Their group number = (10 + 2 + 5) = 17. Elements with atomic numbers 12 and 56 belong to s -block and they have similar outer electronic configurations (ns2).

So they will exhibit similar chemical properties. Their group number = 2. Elements with atomic numbers 16 and 34 belong to the -block and they have similar outer electronic configurations (ns²np²).

So, they will exhibit similar chemical properties. Their group number = (10 + 2 + 4) = 16. So, based on similarity in chemical properties, the given elements are divided into three groups :

⇒ \(\begin{tabular}{|c|c|c|c|}
\hline & Group-2 & Group-16 & Group-17 \\
\hline \multirow{2}{*}{\begin{tabular}{c}
Atomic \\
number
\end{tabular}} & 12 & 16 & 9 \\
\cline { 2 – 4 } & 56 & 34 & 53 \\
\hline
\end{tabular}\)

Question 11. Which one is more basic and why—MgO & A1₂O₃?
Answer: MgO is more basic than A1₂O₃. Mg & A1 are elements of group-2A and 3A of period-3 respectively. Since in a periodic table, metallic property decreases along a period from left to right, the metallic property of Mg is greater than that of A1. Again the oxide of an element with more metallic character is more basic than that with less metallic character. Thus, the basic character of MgO will be more than that of A1₂O₃.

Question 12. Though the nuclear charge of sulfur is more than that of phosphorus, yet the ionization potential of phosphorus is relatively high”—why?
Answer: \(\begin{aligned}
& { }_{15} \mathrm{P}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p_x^1 3 p_y^1 3 p_z^1 \text { and } \\
& { }_{16} \mathrm{~S}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p_x^2 3 p_y^1 3 p_z^1
\end{aligned}\)

3p -orbital of the outermost shell of the P -atom being half filled, this electronic configuration is very stable. So, the removal of one 3p -electron to produce a P+ ion requires a sufficiently high amount of energy.

On the other hand, the amount of energy required for removing one electron from a partially filled 3p -orbital of the S -atom to yield an S+ ions relatively less, since the half-filled 3p -orbital of S+ assumes the extra stability due to the loss of this electron. This accounts for the higher value of ionization potential of phosphorus, relative to sulfur.

Question 13. Mg has relatively higher ionization enthalpy than A1 although the atomic number of the latter is more than the former—explain why.
Answer: . Electronic configuration of 12Mg \(: 1 s^2 2 s^2 2 p^6 3 s^2\)

Electronic Configuration of \({ }_{13} \mathrm{Al}: 1 s^2 2 s^2 2 p^6 3 s^2 3 p^1\)

The penetration effect of the s-electron is greater than that of the electron. So it is easier to remove the 3p-electron from the outermost shell of Al.

Furthermore, the removal of this electron gives Al+, which has filled 3s-orbital (stable electronic configuration) in the outermost shell. On the other hand, Mgatom has filled 3s-orbital (stable electronic configuration) in its ground state.

Removal of one electron from the 3s-orbital of Mg-atom will require a large amount of energy because the resulting Mg+ ion will have a less stable electronic configuration (ls²2s²2p⁶3s¹). Furthermore, it is rather difficult to remove an electron from the s-orbital having a greater penetration effect. So ionization enthalpy ofMg is greater than that of A.

Question 15. Compare the atomic radii of fluorine and neon.
Answer: Fluorine and neon are the members of the second period having atomic numbers 9 and 10 respectively. The covalent radius of fluorine (halogen) is expressed in terms of its covalent radius, while that of neon (noble gas) is expressed in terms of its van der Waals radius. Since by definition, van der Waals radii are always greater than covalent radii, therefore, the atomic radius of neon is greater than that of fluorine.

Question 16. Why are electron-gain enthalpy of Be and N positive?
Answer: The fact that Be and N have positive electron-gain enthalpy values can be explained by considering the given electron-gain processes.

⇒ \(\mathrm{Be}\left(1 s^2 2 s^2\right) \stackrel{+e}{\longrightarrow} \mathrm{Be}^{-}\left(1 s^2 2 s^2 2 p_x^1\right)
(fully filled 2 s-subshell)\)

⇒ \(
\mathrm{N}\left(1 s^2 2 s^2 2 p_x^1 2 p_y^1 2 p_z^1\right) \stackrel{+e}{\longrightarrow} \mathrm{N}^{-1}\left(1 s^2 2 s^2 2 p_x^2 2 p_y^1 2 p_z^1\right)
(half-filled 2 p-subshell)\)

It is observed that the stable electronic configuration of both Be and N -atoms are disturbed by the addition of an electron to each of them.

Consequently, such electron addition processes involve the absorption of energy and hence, both Be and N have positive electron-gain enthalpy values.

Question 17. The electron affinity of lithium is negative but the electron affinity of beryllium is positive”—why?
Answer: Electronic configuration of 3Li: ls²2s¹ Electronic configuration of 4Be: ls²2s² On addition of an electron to Li-atom, 2s -orbital of Li becomes filled with electrons and consequently that electronic configuration attains stability.

This process is accompanied by the liberation of energy. On the other hand, Be has a stable electronic configuration with a frilly-filled 2s subshell.

When an electron is added to Be-atomic occupies 2p -subshell causing destabilization ofthe stable electronic configuration.

This process is accompanied by the absorption of heat. Naturally electron affinity of Li is negative but the electron affinity ofBe is positive.

Question 18. LiCO₂ in spite of being an alkali metal carbonate, is sparingly soluble in water like MgCO₃ — explain.
Answer: The electronegativities of Li and Mg are very close to each other. Furthermore, both Li+ and Mg2+ have similar ionic potential) values and both of them have high polarising power.

So Li and Mg show some similarities in their chemical properties. Due to their small ionic size and high polarising power, their carbonates are somewhat covalent. So, both LiCO₃ (alkali metal carbonate) and MgCO₃ are sparingly soluble in water.

Classification Of Elements And Periodicity In Properties Multiple Choice Questions

Question 1. An element belongs to Group 15 and the third period of the periodic table. Its electronic configuration will be—

1. ls²2s²2p²
2. ls²2s²2p⁴
3. ls²2s²2p⁶3s²3p³
4. ls²2s²2p⁶3s²3p²

Answer: 3. ls²2s²2p⁶3s²3p³

The electronic configuration of the valence shell of group 15 elements in the periodic table is ns²np³ where n is the tire period number.

Therefore, the element located in the third period has the electronic configuration ls²2s²2p⁶3s²3p³

Questions 2. Which one ofthe following has the lowest ionization energy—

1. ls²2s²2p⁶
2. ls²2s²2p⁵
3. ls²2s²2p⁶3s¹
4. ls²2s²2p³

Answer: 2. ls²2s²2p⁵

The electronic configuration ls²2s²2p⁶3s¹ is that of an alkali metal. In a certain period of the periodic table, the ionization potential of alkali metals is the lowest.

Question 3. If 2nd the ionization 1st ionization energy of the atom-atom is— is 13.6 eV, then the

1. 27.2 eV
2. 40.8 eV
3. 54.4 eV
4. 108.8eV

Answer: 3. 54.4 eV

The first ionization energy of the -atom is calculated as:

⇒ \(\frac{2 \pi^2 m Z^2 e^4}{h^2}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=13.6 \mathrm{eV}\)

The second ionization energy of the He atom is = 13.6 X Z2 eV = 13.6 X 22 eV = 54.4 eV.

Questions 4. The stable bivalency of Pb and trivalency of Bi is—

1. Due to d contraction in pb and bi
2. Due to relativistic contraction of the 6s -orbitals of pb
3. And bi, leading to an inert pair effect
4. Due to the screening effect
5. Due to the attainment of noble gas configuration.

Answer: 2. Due to relativistic contraction of the 6s -orbitals of pb

Due to the relativistic contraction of 6s -orbital, an inert pair effect is observed as a result ofwhich, the lower oxidation states of elements attain stability on moving down a group in the periodic table. Thus lead shows stable bivalency while bismuth shows stable trivalency

Question 5. Which of the following is correct?

1. Radius of ca²⁺ < cl- < s²
2. The radius of cl- <s^2-ca²⁺
3. Radius of s^2- = cl- = ca²⁺
4. Radius of s^2- cl- < ca²⁺

Answer: 1. Radius of ca²⁺ < cl- < s²⁺

The number of electrons in Ca²⁺, S^2- and CI^–= 18, i.e., all three ions are isoelectronic.

For any ion, ionic radius oc(e/Z) [where, e = electron number and Z = atomic number]. Therefore, the correct order of the radii ofthe given three ions is—Ca²⁺ < Cl^– < S^2-

Question 6. For BC1₃, A1C1₃, and GaCl₃, the increasing order of ionic character is

1. BC1₃ < A1C1₃ < GaCl₃
2. GaCl₃ < A1C1₃ < BC1₃
3. BCl₃ <GaCl₃ <AlCl₃
4. A1C1₃ < BC1₃ < GaCl₃

Answer: 3. BCl₃ <GaCl₃ <AlCl₃

The ionic character of a compound depends on the polarising power of the cation present in the compound. The greater the magnitude of polarising power, the lesser will be the ionic character of the cation.

Polarisingpower of B³⁺, Al³⁺, and Ga³⁺ follows the order: Al³⁺ < Ga³⁺ < B³⁺, Thus, the order of increasing ionic character is— BCl₃ < GaCl₃ < AICI₃.

Question 7. The hydrides of the first elements in groups 15-17, namely NH₃, H₃O, and HF respectively show abnormally high values for ting and boiling points. This is due to—

1. Small sizes of n, 0 and
2. The ability to form extensive intermolecular-bonding
3. The ability to form extensive intramolecular h-bonding
4. Effective van der waals interaction

Answer: 2. The ability to form extensive intramolecular-bonding

The ionization potential values of, O, and F are quite high and thus, they form intermolecular H-bonding in the compounds, NH₃, H₃O, and HF. As a result, these compounds show abnormally high values for melting and boiling points.

Question 8. Decreasing the basic character of K₂O, BaO, CaO, and MgO is

1. K₂O > BaO > CaO > MgO
2. K₂O > CaO > BaO > MgO
3. MgO > BaO > CaO > K₂O
4. MgO > CaO > BaO > K₂O

Answer: 1. K₂O > BaO > CaO > MgO Alkali metal oxides are highly basic in nature. Down the group in the periodic table, the ionization potential of the alkaline earth metals decreases.

Question 9. Amongst Be, B, Mg, and A₁ the second ionization potential is maximum for—

1. B
2. Be
3. Mg
4. Al

Answer: 1. B

The electronic configuration of B (5) is ls²2s²2p¹. The second ionization potential of B is the maximum because an electron has to be removed from the filled 2s orbital which will require a high amount of energy.

Question 10. An element X belongs to the period and fifteenth group of the periodic table. Which of the following statements is true—

1. It has a filled s -s-orbital and a partially filled d -d-orbital
2. It has filled s -and p -orbital and a partly filled d -orbital
3. It has filled s -and p orbitals and a half-filled d -orbital.
4. It has a half-filled p -p-orbital and filled s and d – d-orbital

Answer: 4. It has a half-filled p -orbital and filled s and d – orbital The element X is positioned at the 4th period and 15th group. Hence the element is As. (Atomic mass = 33) Electronic configuration ofthe element: ls²2s²2p⁶3s²3p⁶3d¹⁰4s²43³ Thus the element has a half-filled p -p-orbital and a filled s – and a d –orbital

Question 11. Which of the following atoms should have the highest 1st electron affinity—

1. F
2. O
3. N
4. C

Answer: 1. F

Decreasing order of electron affinity: F > O > N > C.

Question 12. Which of the set of oxides are arranged in the proper order of basic, amphoteric acidic

1. SO₂,P₂O₅,CO
2. BaO, A1₂O₃, SO₂
3. CaO, SiO₂> A1₂O₃
4. CO₂, A1₂O₃, CO

Answer: 2. BaO→basic, A1₂O₃ amphoteric, SO₂ → acidic

Question 13. Which ofthe following orders presents a correct sequence of the increasing basic nature ofthe given oxides—

1. A1₂O₃ < MgO < Na₂O < K₂O
2. MgO < K₂O < A1₂O₃ < Na₂O
3. Na₂O < K₂O < MgO < A1₂O₃
4. K₂O < Na₂O < A1₂O₃ < MgO

Answer: 1. A1₂O₃ < MgO < Na₂O < K₂O

With the increase in the value of electropositivity of metals, the basic nature of their oxides also increases.

Question 14. The increasing order of the ionic radii of the given isoelectronic species is —

1. S^2- , Cl^2- Ca²⁺ , K
2. Ca²⁺ , K⁺ , Cl^– S^2-
3. K⁺ , S^2- , Ca²⁺ , Cl^–
4. Cl-, Ca²⁺, K²⁺, S²⁺

Answer: 2. Ca²⁺ , K⁺ , Cl^– S^2-

For any atom Orion, ionic radius oc(e/Z) [e=number of electrons and Z= atomic number]. Since the given ions are isoelectronic their electron number is also the same.

It means the higher the atomic number lower the ionic radius of the ion. Therefore, the correct order of the ionic radii ofthe ions is Ca²⁺, K⁺, Cl^– S^2-

Question 15. The first ionization potential of Na is 5.1 eV. The value of electron-gain enthalpy of Na+ will be—

1. +2.55ev
2. -2.55ev
3. -5.1ev
4. -10.ev

Answer: 3. -5.1ev

For any element, value offirstionisation potential and that of the electron-gain enthalpy of its unipositive ion. However, for electron gain enthalpy energy is liberated while for ionization potential energy is absorbed. Hence, they have the same magnitude but opposite sign.

Question 16. Which of the given represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se, and Ar—

1. Ca<Ba<S<Se<Ar
2. Ca < S < Ba < Se < Ar
3. S < Se < Ca < Ba < Ar
4. Ba < Ca < Se < S < Ar

Answer: 4. Ba < Ca < Se < S < Ar

On moving from left to right across a period in the periodic table, the values of ionization enthalpy of elements increase, while moving down a group it decreases.

Question 17. Ionic radii (A) of N^3- , O^2- and F^–

1. 1.71, 1.40 and 1.36
2. 1.36, 1.40 and 1.71
3. 1.71, 1.36 and 1.40
4. 1.36, 1.71 and 1.40

Answer: 2. 1.36, 1.40 and 1.71

The boiling point of Xe is the highest because the boiling point increases down the group from He to Rn due to an increase in van der Waals force of attraction as the size of the atom increases.

Question 19. Which of the following atoms has the highest first ionization energy

1. Rb
2. Na
3. K
4. Sc

Answer: 4. Sc

Question 20. The group having isoelectronic species is-

1. O^2- , F^– , Na⁺ Mg²⁺
2. O^– , F^– ,Na⁺   Mg²⁺
3. O^2- F^– , Na, Mg²⁺
4. O^– F^–, Na⁺, Mg²⁺

Answer: 1. O^2-, F^–, Na⁺ Mg²⁺

Question 21. Both lithium and magnesium display several similar properties due to the diagonal relationship, however, the incorrect one, is—

1. Both form nitrides
2. Nitrates of both li and mg yield nO₂ and O₂ on heating
3. Both form soluble carbonates
4. Both form soluble bicarbonates

Answer: 3. Both form soluble carbonates

Despite having a diagonal relationship, Li and Mg differ in carbonate salts. Magnesium forms basic carbonate [3MgCO₃, Mg(OH)₂, 2H₂O] but lithium like other alkali metals forms normal carbonate [Li2CO₃] salts. Hence, O-2, F+, and Mg²⁺ are isoelectronic species.

Question 22. What is the value of electron-gain enthalpy of Na+ if IEX

1. -5.1 eV
2. -10.2 eV
3. +2.55 eV
4. +10.2 eV

Answer: 1. -5.1 eV

Question 23. In which of the following arrangements, the given sequence is not according to the indicated against it—

1. HO < H₂S < H₂Se < H₂Te: increasing pKa values
2. NH₃ < PH₃ < ASH₃ < SbH₃: increasing acidity
3. CO₂<SiO₂<SnO₂<PbO₂: increasing oxidising power
4. HF < HC1 < HBr <HI: increasing acidic strength

Answer: 1. HO < H₂S < H₂Se < H₂Te: increasing pKa values

On moving down a group in the periodic table, the covalent character of the hydrides of the elements increases and so their acidity gradually increases. Therefore, the acidic character of the hydrides of the elements of group VA and VILA follows the order— NH₃< PH₃ < AsH₃ < SbH3 and HF < HCI < HBr < HI On the other hand, the oxidizing power of the oxides in option follows the trend:

CO₂ < SiO₂ < SnO₂ < PbO₂

Question 24. Identify the wrong statement among the following —

1. The atomic radius of the elements increases as one moves down the first group ofthe periodic table
2. The atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table
3. Amongst isoelectronic species, the smaller the positive charge on the cation, the smaller the ionic radius
4. Amongst isoelectronic species, the greater the negative charge on the anion, the larger is the the ionic radius

Answer: 3. Amongst isoelectronic species, the smaller the positive charge on the cation, the smaller the ionic radius

On moving from left to right across a period. The proton number and magnitude of the nuclear charge of elements increase gradually leading to a corresponding decrease in the atomic radii of the elements.

Again, moving down a group, with an increase in the atomic number of elements, their atomic radii gradually decrease due to the addition of new shells. For isoelectronic ions, with a decrease in the positive charge of cations and with an increase in the negative charge of anions, ionic radii increase.

Question 25. The reason for lanthanoid contraction is—

1. Negligible screening effect of f-orbitals
2. Increasing nuclear charge
3. Decreasingnuclear charge
4. Decreasing screening effect

Answer: 1. Negligible screening effect of f-orbitals.

Question 26. Be2+ is isoelectronic with which ofthe following ions?

1. H⁺
2. Li⁺
3. Na⁺
4. Mg2⁺

Answer: 2. Li⁺

⇒ \(\begin{array}{|c|c|c|c|c|c|}
\hline \text { Ion } & \mathrm{Be}^{2+} & \mathrm{H}^{+} & \mathrm{Li}^{+} & \mathrm{Na}^{+} & \mathrm{Mg}^{2+} \\
\hline \text { No. of electron } & 2 & 0 & 2 & 10 & 10 \\
\hline
\end{array}\)

Question 27. Which of the following orders of ionic radii is correctly represented?

1. H→ H+→H
2. F→ O₂→Na+
3. Na⁺ → F^–→ O^2-
4. Al³⁺ > Mg²⁺ > N₃

Answer: 2. F→ O₂→Na+

None of the options is correct; An atom loses an electron (s) to form a cation. Thus, the radius of the formed cation is less than the parent atom.

Again an atom gains electron (s) to form an anion. So, the radius of the formed anion is greater than the parent atom. Therefore, H- > H > H+. Now, for isoelectronic species, with an increase in their atomic number, their ionic radii decrease. Therefore, the correct orders of ionic radii are O^2-→F¯ > Na+ and N^3- > Mg²⁺ > Al³⁺

Question 28. The species Ar, K+, and Ca2+ contain the same number of electrons. In which order do their radii increase—

1. Ca²⁺ <K+<Ar
2. K+ < Ar < Ca²⁺
3. Ar < K+ < Ca²⁺
4. Ca²⁺ < Ar < K⁺

Answer: 1. Ca²⁺ <K+<Ar

For isoelectronic species atomic/ionic radii decrease with the increase of nuclear charge. Hence increasing order atomic/ionic radii 20 Ca²⁺ < igK+ < 18Ar.

Question 29. Because of lanthanide contraction, which of the following pairs of elements have nearly the same atomic radii (Numbers in the parenthesis are atomic numbers)—

1. Zr(40) and Hf(72)
2. Zr(40) and Ta(73)
3. Ti(22) and Zr(40)
4. Zr(40) and Nb(41)

Answer: 1. Zr(40) and Hf(72)

Zr and Hf have the same atomic radii due to the lanthanide contraction.

Question 30. In which ofthe following options the order of arrangement does not agree with the variation of property indicated against it

1. I < Br < Cl < F (increasing electron-gain enthalpy)
2. Li<Na<K<Rb (increasingmetallic radius)
3. Al3+ < Mg2+ < Na+ < F_ (increasing ionic size)
4. B<C<N<0 (increasing first ionization enthalpy)

Answer: 1. I < Br < Cl < F (increasing electron-gain enthalpy) The increasing order of negative electron gain enthalpy; I < Br < F < Cl and that of first ionization energy: B < C < O < N.

Question 31. The element Z =114 has been discovered recently. It will belong to which of the following family/groups and electronic configuration—

1. Carbon family, [rn]5/I4⁶d¹⁰7s²7p²
2. Oxygen family, [rn]5/14⁶d¹⁰7s²7p⁴
3. Nitrogen family, [rn]5/14⁶d¹⁰7s²7p⁶
4. Halogen family, [rn]5/I4⁶d¹⁰7s²7p⁵

Answer: 1. Carbon family, [rn]5/i4⁶d¹⁰7s²7p²

The electronic configuration of the element having atomic mass 114 is [rn]5/i4⁶d¹⁰7s²7p² The outer electronic configuration of the element is the same as that of carbon. So the element should belong to the carbon family.

Question 32. Among CaH₂, BeH₂, BaH₂ the order of ionic character

1. BeH₂<CaH₂<BaH₂
2. CaH₂ < BeH₂ < BaH₂
3. BeH₂ < HaH₂ < CaH₂
4. BaH₂ < BeH₂ < CaH₂

Answer: BeH₂<CaH₂<BaH₂

Down a group metallic character increases. Thus ionic character of the metal hydride increases down the group. Hence least ionic compound is BeH₂.

Question 33. Which ofthe following oxides is most acidic—

1. MgO
2. BeO
3. BaO
4. CaO

Answer: 2. BeO

Down a group, the metallic character of the elements increases. Hence down the group basic character of the metallic oxide increases. Thus BeO has the most acidic character. Actually, it is an amphoteric oxide whereas others are basic oxides.

Question 34. The first ionization enthalpy of Na, Mg, and Si are 496, 737, and 776 kj.mol-1 respectively. What will be the first ionization enthalpy potential ofA1inkj mol-1 —

1. >766 kj – mol-1
2. >496 and < 737 k).mol-1
3. >737 and < 766 kj.mol-1
4. >496 kj- mol-1

Answer: 2. >496 and < 737 k).mol-1

The ionization enthalpy of A1 is lower than that of Mg as the 3p1 electron of A1 is easier to remove than to remove an electron from the fully-filled 3s -orbital of Mg.

Question 35. Which is correct regarding the size of the atom—

1. B < Ne
2. Na > K
3. N < O
4. V > Ti

Answer: 2. The atomic radii of noble gases are the largest in their respective periods. This is due to the reason that noble gases have only van der Waals radii.

Question 36. An element (X) belongs to the fourth period and fifteenth group of the periodic table. Which one of the following is true regarding the outer electronic configuration of (X)? It has—

1. Partially filled d -d-orbitals and filled orbital,
2. Filled s -orbital and filled p orbitals.
3. Filled s -orbital and half-filled p -orbitals.
4. Half-filled d -orbitals and filled s -orbitals.

Answer: 3. filled s -orbital and half-filled p -orbitals.

The electronic configuration of (X) can be written as X: ls²2s²2p⁶3s³3p⁶4s²3d¹⁰4p³ So, element (X) has filled s -and d -orbitals and half-filled p -orbitals.

Question 36. Which of the following transformations produces the maximum amount of energy—

1. M-(g)→M(g)
2. M(g)→M-(g)
3. M⁺(g)→M³⁺(g)

Answer: 4. M²⁺+(g)→M³⁺(g)

Question 37. The amount of energy released when 106 atoms of iodine in a vapor state are converted into- ions is 4.8 x 10-13 J. What is the electron affinity of iodine in kJ mol.1 —

1. 489
2. 289
3. 259
4. 389

Answer: 2. 289

Question 38. The elements that occupy the peaks of ionization energy potential curve are—

1. Na, K, Rb, Cs
2. Cl, F, Br, I
3. Na, Mg, Cl, I
4. He, Ne, Ar, Kr

Answer: 4. He, Ne, Ar, Kr

Question 39. The electronic configuration ofthe atom having maximum difference second and third ionization enthalpies is—

1. ls²2s²2p⁶3s²
2. ls²2s²2p⁶3s²3p¹
3. ls²2s²2p⁶3s²3p²
4. ls²2s²2p⁶3s¹

Answer: 1. ls²2s²2p⁶3s²

Question 40. Identify the least stable ion amongst the following—

1. Li-
2. Be-
3. B-
4. C-

Answer: 2. Be-

Question 41. If each orbital can accommodate a maximum of four electrons, the number of elements in the third period of the periodic table will be—

1. 10
2. 12
3. 14
4. 16

Answer: 4. 16

Question 42. Three elements X, Y, and Z are present in the third short period and their oxides are ionic, amphoteric, and giant molecules respectively. The correct order of atomic numbers of X, Y, and Z is—

1. Z< Y<X
2. X<Z<Y
3. Y<Z<X
4. X<Y<Z

Answer: 4. X<Y<Z

Question 43. A gaseous mixture of He, Ne, Ar, and Kr is irradiated with photons of frequency appropriate to ionize Ar. The ion(s) present in the mixture will be—

1. Only
2. Ar+ and He+
3. Ar+ and Ne
4. Ar+ and Kr+

Answer: 4. Ar+ and Kr+

Question 44. Boiling point of Kr & Rn are -152°C & -62°C respestively. Then the boiling point of Xe is expected to be—

1. -92C
2. -87C
3. -107C
4. 77C

Answer: 3. -107C

Question 45. Which ofthe following is smallest in size—

1. Li+ (aq)
2. Na+ (aq)
3. K+ (aq)
4. Rb+ (aq)

Answer: 4. Rb+ (aq)

Question 46. The atomic radius of Li is 1.23 A and ionic radius of Li+ is 0.76 A. Percentage of the volume occupied by a single valence electron in Li is—

1. 35
2. 52.5
3. 76.4
4. 83.72

Answer: 3. 76.4

Question 47. The number of valence electrons in element A is 3 and that in element B is 6. The most probable compound from A and B is—

1. A₂B
2. AB₂
3. A₆B₃
4. A₂B₃

Answer: 4. A₂B₃

Question 48. The ionic radius of ‘Cr’ is the minimum in which of the following compounds—

1. K₂CrO₄
2. CrF₃
3. CrO₂
4. CrCl₃

Answer: 1. K₂CrO₄

Question 49. The correct order of radii is—

1. N < Be < B
2. F^– <O^2- <N^3-
3. Na < Li < K
4. Fe³⁺ <Fe²⁺<Fe⁴⁺

Answer: 2. F^– <O^2- <N^3-

Question 50. The atomic number of the inert gas in which the total number of d -electrons is equal to the difference in the numbers of total p -and s -electrons is-

1. 18
2. 36
3. 54
4. 86

Answer: 2. 36

Question 51. The element with atomic number 118 (Uuo) has been discovered recently. Which ofthe following is not expected for this element—

1. It is a radioactive and unstable element
2. It is a solid at room temperature
3. Its ionization enthalpy is minimal in the group
4. It has a 7s²7p⁶ outer shell configuration

Answer: 2. It is a solid at room temperature

Question 52. Which occupy the peak ofthe atomic volume curve—

1. Transition Elements
2. Halogens
3. Alkali Metals
4. Alkaline earth metals

Answer: 3. Alkali Metals

Question 53. The ionization energy will be the maximum for the process

1. Ba BA²⁺
2. Be Be²⁺
3. Cs C+^3-
4. Li li²⁺

Answer: 2. Be Be²⁺

Question 54. The correct order of second ionization potential of carbon, nitrogen, oxygen, and fluorine is

1. C>N>O>F
2. O>N>F>C
3. O>F>N>C-
4. F>O>N>C

Answer: 3. O>F>N>C-

Question 55. The most reactive metal is

1. Na
2. Fe
3. Hg
4. Cs

Answer: 4. Cs

Question 56. You are given Avogadro’s number of atoms of a gas X. If the atoms are converted into X+(g) by energy AH, then the ionization enthalpy of X is

1. \(\frac{2 \Delta H}{N_A}\)
2. \(\frac{2 N_A}{\Delta H}\)
3. \(\frac{\Delta H}{2 N_A}\)
4. \(\frac{N_A}{\Delta H}\)

Answer: 1. \(\frac{2 \Delta H}{N_A}\)

Question 57. The second electron affinity of oxygen is +744 kj.mol-1 then the second electron affinity of sulphur is—

1. -200kj mol-1
2. +450 kj – mol-1
3. +800 kj – mol-1
4. 1200 kl – mol-1

Answer: 2. +450 kj – mol-1

Question 58. What would be the atomic number of the next halogen if discovered in the future—

1. 116
2. 117
3. 118
4. 119

Answer: 2. 117

Question 59. An element belongs to group 17 and the fourth period of the periodic table. Which of the following properties will be shown by the element—

1. Gaseous, non-metallic
2. Liquid, non-metallic
3. Solid, non-metallic
4. Solid, metallic

Answer: 2. Liquid, non-metallic

Question 60. Five successive ionization enthalpies of an element are 800, 2427, 3638, 25024 & 32824kj. mol-1 respectively.

The number of valence electrons ofthe element is—

1. 3
2. 5
3. 4
4. 2

Answer: 1. 3

Question 61. Choose the correct set of ionic radii among the following—

1. Ti³⁺<Mn⁷⁺
2. 35ci-<37ci-
3. p³⁺→p⁵⁺
4. K+→C1-

Answer: 3. p³⁺>p⁵⁺

Question 62. Electron affinity(kj.mol-1 ) of three halogens X, Y, and Z are -349, -333, and -325. X, Y, Z are respectively—

1. F₂, Cl₂ and Br2
2. Cl₂, F₂ and Br₂
3. Br₂, Cl₂ and F₂
4. Br₂, F₂ and Cl₂

Answer: 1. F₂, Cl₂ and Br₂

Question 63. The correct order of ionic radii is—

1. N^3- → O^2- →F^– → Na⁺
2. N^3- → Na+ → O^2- →F^–
3. Na³⁺ → O^2- → N^3- →F^–
4. O^2- → F^–→ Na²⁺ → N^3-

Answer: 1. N^3- → O^2- →F^– → Na⁺

Question 64. The ionization enthalpy of lithium is 520 kj. mol-1. The amount required to convert 210 mg of -atoms in a gaseous state into Li+ ions is—

1. -520kj
2. -173.3 kj
3. 2.47 kj
4. 15.60 kj

Answer: 4. 15.60 kj

Question 365. Electronic configuration ls²2s²2p⁶3s²3p⁶3d⁹ indicates—

1. Mentalization
2. Non-metal atom
3. Non-mental anion
4. Mental atom

Answer: 1. Mentalcation

Question 66. In the periodic table metals used as catalysts belong to

1. F-block
2. d- block
3. p-block
4. s-block

Answer: 2. d- block

Question 67. The oxide of an element having configuration [Ne]3s1 is

1. Amphoteric
2. Basic
3. Acidic
4. Neutral

Answer: 2. Basic

Question 68. Which one of the following orders is not by the property stated against it—

1. F → Cl → Br →I : oxidisingpower
2. F →Cl → Br →l : electronegativity
3. F→F → Cl→Cl → Br→Br → I→I: bond dissociation energy
4. HI → HBr → HC1 → HF: acidic property in water

Answer: 3. F→F → Cl→Cl → Br→Br → I→I: bond dissociation energy

Question 69. In which ofthe following process energy is absorbed

1. F^–F^–
2. H^–h^–
3. Cl^–cl^–
4. O^–O^2-

Answer: 4. O^–O^2-

Question 70. In which of the following pairs do both the metals have identical values of metallic radii

1. Ag and Au
2. Cu And Ag
3. Cu and Zn
4. Zn and Hg

Answer: 1. Ag and Au

Question 71. In which of the following pairs the difference between the covalent radii ofthe two metals is maximum—

1. K, Ca
2. Mn, Fe
3. Co, Ni
4. Cr, Mn

Answer: 1. K, Ca

Question 72. The increasing order of the first ionization enthalpies of the elements B, P, S, and F is—

1. F<S<P<B
2. P<S<B<F
3. B<P<S<F
4. B<S<P<F

Answer: 4. B<S<P<F

Question 73. Which group of atoms have nearly the same atomic radius—

1. Na, K, Rb, Cs
2. Li, Be, B, C
3. Fe, Co, Ni, Cu
4. F, Cl, Br,I

Answer: 3. Fe, Co, Ni, Cu

Question 74. The lattice energy is highest for

1. CsF
2. RbF
3. NaF
4. KF

Answer: 3. NaF

Question 75. The correct order of lattice energy for lithium halides is-

1. Lil > LiBr > LiCl > LiF
2. LiCl > LiF > LiBr > Lil
3. LiBr > LiCl > Lil > LiF
4. LiF > LiCl > LiBr > Lil

Answer: 4. LiF > LiCl > LiBr > Lil

Question 76. An element with atomic number 20 is most likely to combine with the element whose atomic number is

1. 11
2. 16
3. 18
4. 10

Answer: 2. 16

Question 77. K+, Cl-, Ca2+, and S2- ions are isoelectronic. The decreasing order of their size is—

1. S^2- → Cl- → K⁺→ Ca²⁺
2. Ca²⁺ → K⁺ →> Cl- → S^2-
3. K⁺ → Cl^–→ Ca2+ → S^2-
4. Cl^– → S^2-→ Ca2+ → K⁺

Answer: 1. S^2-→ Cl^– → K+ → Ca²⁺

Question 78. Which ofthe following oxides of Cr is amphoteric

1. CrO
2. CrO₃
3. Cr₂O₃
4. None of the theses

Answer: 3. Cr₂O₃

Question 79. The element with atomic number 118 has been discovered recently. In the periodic table, the element will occupy its position in the family of

1. Halogens
2. Insert gases
3. Chalcogens
4. Alkaline earth metals

Answer: 2. Insert gases

Question 80. In the relation, Electronegativity \(=\frac{0.359 Z_{\text {eff }}}{r^2}+0.744, r \text { is }-\)

1. Ionic radius
2. Metallic radius
3. Van der Waals radius
4. Covalent radius

Answer: 4. Covalent radius

Question 81. Select the endothermic processes—

1. H→H^–
2. O→ O^2-
3. Cr³⁺
4. Ar→Ar+

Answer: 2. O→O^2-

Question 82. Which of the following sequences contain the atomic number of only representative elements—

1. 3,33,53,87
2. 7,17,25,37,48
3. 2,10,22,86
4. 9,35,51,88

Answer: 1. 3,33,53,87

Question 83. Which of the given elements will gain one electron more readily in comparison to other elements of their group—

1. S
2. Na
3. O
4. Cl

Answer: 1. S

Question 84. MgSO₄ is soluble in water but BaS04 is not because—

1. The hydration enthalpy of MgSO₄ is more than its lattice enthalpy
2. The lattice enthalpy of MgSO₄ is greater than its hydration enthalpy
3. The lattice enthalpy of BaSO₄ is greater than its hydration enthalpy
4. The hydration enthalpy of BaSO₄ is greater., than its lattice enthalpy

Answer: 1. The hydration enthalpy of MgSO₄ is more than its lattice enthalpy

Question 85. Consider the following ionization steps—

• M(g)→M+(g) + e-; AH = 100 eV
• M(g)→M2+(g) + 2e-; AH = 250 eV

Select the correct statements—

1. Af H1 ofM(g) =100eV
2. H1 of M+(g) =150 eV
3. AlH₂ of M(g)=250eV
4. AlH₂ of M(g)=150 eV

Answer: 1. M(g)→M+(g) + e-; AH = 100 eV

Question 86. In which of the following sets of atomic numbers, all elements are in the same group—

1. 8,16,24
2. 3,11,37
3. 12,38,56
4. 10,18,54

Answer: 2. 3,11,37

Question 87. Which of the following elements are artificially made and do not exist in nature—

1. Bi
2. Ge
3. Tc
4. At

Answer: 3. Tc

Question 88. Choose the pairs in which the IEX of the first element is greater than the IEl of second element but in the case of IE₂ order is reversed—

1. P, S
2. F, O
3. Mg, Al
4. N, O

Answer: 1. P, S

Question 89. Which ofthe following ions can form complexes—

1. Al³⁺
2. Cu²⁺
3. Cr³⁺
4. Fe²⁺

Answer: 1. Al³⁺

Question 90. Which ofthe following pairs contain metalloid

1. In, Tl
2. Ge, Ga
3. As, Te
4. I, Bi

Answer: 2. Ge, Ga

Question 91. Which of the following elements exist in a liquid state at room temperature—

1. Selenium
2. Germanium
3. Cesium
4. Gallium

Answer: 2. Germanium

Question 92. Which of the following elements can form two or more chlorides—

1. Cu
2. Vs
3. Hg
4. Cs

Answer: 1. Cu

Question 93. Which of the following hydrides rarely show basic properties—

1. PH₂
2. SbH₃
3. AsH₃
4. BiH₃

Answer: 2. SbH₃

Question 94. The correct sequences of increasing electropositive character are

1. Fe<Cu<Mg
2. Cu < Fe < Mg
3. Cs < Ca < Sr
4. Ca<Sr<Cs

Answer: 2. Cu < Fe < Mg

Question 95. Select the correct sequence of electron-gain enthalpy—

1. S > Se > O
2. Cl > F >I
3. F > Cl >I
4. S > O > Se

Answer: 3. F > Cl >I

Question 96. The correct orders of electronegativity are

1. C < Si < P < N
2. Si < P < C < N
3. P < Se < S < N
4. S < P < Se < N

Answer: 3. P < Se < S < N

Classification Of Elements And Periodicity In Properties Very Short Type Questions

Question 1. Who proposed the ‘law of octaves’?
Answer: Newland

Question 2. For which of the elements, the ‘Law of octave’ is not applicable?
Answer: Heavy mental

Question 3. Identify the element predicted by Mendeleev as ekaaluminium.
Answer: Gallium

Question 4. Mention the name of the group in the periodic table, which contains solid, liquid, and gaseous elements.
Answer: Group 72 or 17

Question 5. Mention the position of the alkali metals in the periodic table.
Answer: Group 1 A or 1

Question 6. Give an example of an element whose atomic mass has been corrected by Mendeleev’s periodic table.
Answer: Be

Question 7. Who proved that atomic number but not atomic
mass is the more proper determining factor of the periodicity of elements?
Answer: Moseley

Question 8. Give an example of d -a block element which is not regarded as a transition element.
Answer: Zn

Question 9. How many rare earth elements are present in the periodic table? Give an example-
Answer: 14

Question 10. Give the common electronic configuration of d-block elements.
Answer: (n-1) s1-10 ns 1-2

Question 11. Name the transition element having the lowest atomic mass.
Answer: Se

Question 12. Which element gives brick red coloration to the flame? In which block, the element is in the periodic table?
Answer: Ca, s-block

Question 13. Mention the respective block of the elements having atomic numbers 7, 11 and 22.
Answer: p,s,d

Question 14. What do you mean by representative elements?
Answer: Elements Of S and P- block’s

Question 15. Give two terminal elements of rare earth element series.
Answer: Ce, lu

Question 16. How will you detect the starting and ending of a period in the periodic table?
Answer: By electronic configuration.

Question 17. Which block contains inner transition elements?
Answer: F

Question 18. Give an example of a transition element, which has filled d-subshellinits ground state.
Answer: Cu

Question 19. Why the ions or atoms of the transition elements
are paramagnetic?
Answer: Due to the presence of unpaired d -electrons, 20. Cations of group-1A and 2A,

Question 20. Give an example of a diamagnetic.
Answer: Cations of group-1A and 2A,

Question 21. Why and Co called ferromagnetic?
Answer: They can easily be converted into magnets.

Question 22. Which element gives golden-yellow coloration to the flame?
Answer: Na

Question 23. Why Mg cannot be identified by flame test?
Answer: They can easily be converted into magnets.

Question 24. Which are called ‘noble metals’?
Answer: Elements of 5d -series (especially Pt, Au, and Hg ),

Question 25. Give the electronic configuration of the outermost shell of lanthanides.
Answer: 4f1-14 H5d0-116s2

Question 26. Which of the groups in the periodic table contain all the metallic elements?
Answer: Group-2A

Question 27. Give the names of the ‘noble gas’ elements present in the second and fifth periods.
Answer: Ne, Xe

Question 28. Mention the name and atomic number of the element present in group 13 of the third period.
Answer: A1.13

Question 29. Give the electronic configuration of the fifth element ofthe first transition series.
Answer: Electronic configuration of Mn

Question 30. Identify the transition element(s): K, Mn, Ca, Cs, Fe, Cu, pb.
Answer: Mn, Fe, Cu, Pb,

Question 31. Name two elements that do not give a flame test.
Answer: Be, Mg

Question 32. Which lanthanide elements have only 1 electron in a 5dsubshell?
Answer: Lu

Question 33. Write the outermost electronic configuration of chalcogens.
Answer: ns2 np4

Question 34. What will be the position in the periodic table of the element having electronic configuration ls22s22p4?
Answer: Second period, Grop -16

Question 35. Mention the position of the pnictogens in the long form of the periodic table.
Answer: 15

Question 37. Which element has the highest oxidising property?
Answer: Fluorine

Question 38. Give one example of each of metal, nonmetal, or metalloid presenting the p -block of the periodic table.
Answer: Pb.N.As

Question 39. Give the names of two non-metals present in s -block of the periodic table.
Answer: H2.He

Question 40. What is the unit of electron affinity?
Answer: kj-mol-1

Question 41. Is the value of the electron affinity of an element zero?
Answer: Yes

Question 42. Between Fe2+ and Fe3+, which is smaller in size, and 6. Why?
Answer: Fe3+, Z/e ratio is higher,

Question 43. Which element of each pair has higher electron affinity?

• Br. Cl
• F.Cl
• O.S

Answer:

• Cl
• Cl
• S

Question 44. The first ionization potential of carbon is 11.2 eV. State whether the value of first ionisation potential of silicon is same or greater or less than that of carbon
Answer: Lower

Question 45. Arrange s,p,d & /-subshells according to the screening power.
Answer: S>p>d> f

Question 46. Which element has the lowest ionization potential?
Answer: Cs

Question 47. Which element has the highest ionization potential?
Answer: He

Question 48. What is the unit of ionization potential?
Answer: eV atom-1

Question 49. What is the change observed in the covalent
character of the oxides of the elements starting from Na to Cl in the third period?
Answer: Increases

Question 50. Arrange Mg, Al, Si, and Na in the increasing order of their ionization potentials.
Answer: Na < Al < Mg < Si

Classification Of Elements And Periodicity In Properties Fill In The Blanks

Question 1. Mendeleev’s periodic law was similar to the law proposed by the scientist______________.
Answer: Lothar Meyer

Question 2. In Mendeleev’s periodic table, the______________incomplete period.
Answer: Seventh

Question 3. The starting elements of even series are K, Rb, and ______________period is an
Answer: Cs

Question 4. The starting elements of odd series are______________Au.
Answer: CU

Question 5. Meneleev’s triad elements are Ag and are the ______________ fundamental property of the element.
Answer: Transition elements

Question 7. ‘The elements from 58Ce to 71Lu are called ______________
Answer: lanthanides

Question 8. Be, Mg, Ca are called______________1L
Answer: Alkaline Earth

Question 9. Cu, Ag, Au are called______________metals.
Answer: Coinage

Question 10. S, Se, Te are called ______________
Answer: Chalcogens,

Question 11. The potential of s -block elements is
Except Be and______________ the s -block elements response to the flame test.
Answer: Low

Question 12. Except Be and______________the s -block elements response to the flame test.
Answer: Mg

Question 13. The s -block elements of the fourth, fifth, and sixth periods can form complex compounds by vacant d -orbital. in as they have ______________.
Answer: Coordinate Covalency

Question 14. Noble metals are chemically______________.
Answer: Insert

Question 15. F-block elements are _ the presence of odd electrons. block elements generally form colored in nature due to the omplex compounds.
Answer: Paramagnetic

Question 16. Block elements Generally from colored complex compounds ______________.
Answer: d

Question 17. Zn, Cd, and are not they are d -block elements.
Answer: Hg

Question 18. The element with electronic configuration ls22s22p4 is presentin group______________.
Answer: 16

Question 19. The general electronic configuration of transition
elements is______________
Answer: (n-1)d1-10 ns 1-2

Question 20. Effective nuclear charge = total nuclear change ______________.
Answer: Screening Constant,

Question 21. The IUPAC name of the element having an atomic number
150 is______________
Answer: Unpentrilum

Question 22. For homonuclear diatomic molecule, covalent radius = ______________x intemuclear distance.
Answer: \(=\frac{1}{2}\)

Question 23. The internuclear distance of the HC1 molecule is 1.36 A and the covalent radius of the chlorine atom is 0.99 A. Thus, the covalent radius ofhydrogen atom will be______________.
Answer: 0.37

Question 24. The covalent radius of an element is ______________ der Waals radius.
Answer: Shorter

Question 25. Anionic radius is ______________ radius.
Answer: Greater

Question 26. On moving from left to right across a period, the acidic property of oxide of element______________.
Answer: Increases

Question 27. Hydrides of most of the non-metals are ______________nature.
Answer: Colvent and nonpolar

Question 28. The first ionization potential of carbon is. the second ionization potential is ______________.
Answer: Increases

Question 29. Among the halogens,______________ nature.
Answer: Idonine

Question 30. Electron affinity of Be and are ______________ almost same.
Answer: Mg

Question 31. In a particular energy Level(orbit), the Followers the orders s>p>f.
Answer: Screening effect,

Question 32. In the case of elements belonging to the same group, ionic radii with increases in atomic number ______________.
Answer: Increases

Question 33. F-,Ar,Mg²⁺,Rb⁺ are inons ______________
Answer: Isoelectronic

Question 34. The ionization enthalpy of Cu and K can be explained based on______________.
Answer: Screening effect

Question 35. of the atom of any element and the first ionization enthalpy of its anion (unit -ve charge) are the same.
Answer: Electron affinity

Question 36. Atomic mass = Atomic volume x______________.
Answer: Density

Question 37. Low solubility of Li2CO³ and MgCO³ in water can be
explained by______________.
Answer: Diagonal relationship.