JEE Chemistry Classification of Elements and Periodicity in Properties MCQs Set 20

Answer: (c) 6

Explanation:
1. The effect grows with the poor shielding by 4f and 5d electrons, so it is greatest in period 6 (Tl, Pb, Bi).

Teacher's Note:
1. Inert pair effect increases down a group.
2. Heaviest p-block elements show it most.

Answer: (a) MnO

Explanation:
1. The oxidation state of Mn is +2 in MnO, +3 in K₃[Mn(CN)₆], +4 in MnO₂ and +7 in KMnO₄.
2. The lower the positive charge, the larger the ion, so Mn²⁺ in MnO is the largest.

Teacher's Note:
1. For the same element radius decreases as oxidation state increases.
2. Z_eff rises with charge.

Answer: (a) (1) and (3)

Explanation:
1. IE2 of Be (1757 kJ/mol) is greater than IE1 (899), so (1) is true.
2. C (1086) is greater than Be (899), so (3) is true; B (801) is less than Be.

Teacher's Note:
1. Successive ionisation energies increase.
2. Be has a filled 2s subshell.

Answer: (a) its small size and high electron density cause strong electron-electron repulsion on adding an electron

Explanation:
1. The 2p subshell of F is very compact, so the incoming electron is repelled strongly.
2. In Cl the 3p subshell is larger and repulsion is smaller.

Teacher's Note:
1. Second-period elements have anomalous electron gain enthalpy.
2. Vacant d orbitals are not the reason.

Answer: (a) O < S < F < Cl

Explanation:
1. Values in kJ/mol: O 141, S 200, F 328, Cl 349.

Teacher's Note:
1. Halogens have much larger values than chalcogens.
2. Second-period members are lower than third-period members.

Answer: (d) 2

Explanation:
1. Z = 119 is the first element of period 8: [Og] 8s¹, group 1.
2. Its valency is 1, so x + y = 1 + 1 = 2.

Teacher's Note:
1. The element after a noble gas starts a new alkali metal group.
2. Alkali metals have valency 1.

Answer: (c) 26

Explanation:
1. 5.92 B.M. corresponds to five unpaired electrons, \( \sqrt{5 \times 7} = 5.92 \).
2. M³⁺ is therefore d⁵, so the atom has 3d⁶ 4s², which is iron (Z = 26).

Teacher's Note:
1. Magnetic moment = \( \sqrt{n(n+2)} \) B.M.
2. d⁵ in +3 state means the neutral atom is d⁶s².

Answer: (d) 3

Explanation:
1. Cr³⁺ has a half-filled t_2g³ configuration which is especially stable in solution.

Teacher's Note:
1. Cr(III) is the most stable state in aqueous solution.
2. Cr(VI) is strongly oxidising.

Answer: (d) 3

Explanation:
1. The 4f electrons are buried deep, so all lanthanoids lose 6s² and one 5d or 4f electron to give Ln³⁺.

Teacher's Note:
1. The characteristic state of lanthanoids is +3.
2. Ce(+4) and Eu(+2) are exceptions.

Answer: (d) 8

Explanation:
1. Covalent radii: r_A = 5 and r_B = 3.
2. For negligible electronegativity difference, d(A-B) = r_A + r_B = 8.

Teacher's Note:
1. Bond length is the sum of the covalent radii when there is no polarity.
2. Average the two homonuclear distances.

Answer: (d) 30

Explanation:
1. Across the 3d series the radius first decreases and then increases due to electron repulsion once the 3d subshell is full.
2. Zn (134 pm) is larger than Cu (128 pm) and Ni (124 pm).

Teacher's Note:
1. In the later 3d series, increased pairing repulsion enlarges the atom.
2. Zn has Z = 30.

Answer: (c) 2

Explanation:
1. The huge jump between IE2 and IE3 shows two valence electrons.
2. An element with two valence electrons is in group 2.

Teacher's Note:
1. A sharp jump marks the end of the valence electrons.
2. Here it occurs after the second.

Answer: (b) 23

Explanation:
1. 3 eV per atom = 69 kcal/mol.
2. Moles = 10/30 = 1/3, so energy = 69/3 = 23 kcal.

Teacher's Note:
1. Convert eV per atom to kcal/mol first.
2. Then scale by moles.

Answer: (c) 5

Explanation:
1. B = 2.04. H (2.20), C (2.55), N (3.04), O (3.44) and F (3.98) are higher; Li (0.98) and Be (1.57) are lower.

Teacher's Note:
1. Compare Pauling values.
2. Hydrogen is slightly more electronegative than boron.

Answer: (d) Group 9 and group 18

Explanation:
1. Group 9: Co has 4s² but Rh has 5s¹, so the counts differ.
2. Group 18: He has 2 electrons, the others have 8.
3. Groups 13 and 11 have the same outer shell count throughout.

Teacher's Note:
1. Look for exceptions like He and Rh.
2. Check group 11: Cu, Ag, Au all have one s electron.

Answer: (b) Al and Sc

Explanation:
1. Al is always +3 and Sc (3d¹4s²) is +3 in its compounds.
2. Cu is +1 or +2 and Xe is +2, +4, +6 or +8.

Teacher's Note:
1. Three valence electrons gives +3.
2. Group 3 and 13 elements.

Answer: (d) S, O and H

Explanation:
1. S shows -2, +4, +6; O shows -2, -1, +2; H shows +1 and -1.
2. Zn shows only +2.

Teacher's Note:
1. Zn has a filled d subshell and loses only 4s².
2. Hydrogen shows both signs.

Answer: (b) Pb, Tl and Bi

Explanation:
1. Because of the inert pair effect, Pb(II), Tl(I) and Bi(III) are more stable than Pb(IV), Tl(III) and Bi(V).

Teacher's Note:
1. Heavy p-block elements prefer lower states.
2. Oxygen is not subject to inert pair effect.

Answer: (a) All three

Explanation:
1. Across period 2 radii decrease: Li 152, Be 111, B 88, N 74, O 66, F 64 pm.
2. All three orders are in agreement.

Teacher's Note:
1. Radius falls across a period.
2. Check the order against the table.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. IE2: O 3388 > F 3374; IE3: P 2915 > Al 2745; IE2: Cu 1958 > Zn 1733.
2. IE1 of Al (578) is slightly less than that of Ga (579), so (iii) is wrong.

Teacher's Note:
1. Poor shielding by 3d electrons raises Ga's IE.
2. O⁺ has a half-filled 2p³.

Answer: (b) S and Cl

Explanation:
1. S (-200) gains more readily than O (-141), and Cl (-349) more readily than F (-328).
2. N is a member whose egH is positive/zero.

Teacher's Note:
1. O and F are anomalous because of small size.
2. Third-period members have the higher affinity.

Answer: (d) (ii) and (iii) only

Explanation:
1. Electronegativity rises across a period: Li 0.98 < Be 1.57 < B 2.04 and C 2.55 < N 3.04 < O 3.44.
2. It falls down a group, so (i) and (iv) are wrong.

Teacher's Note:
1. Down a group electronegativity decreases.
2. F is greater than Cl.

Answer: (c) transition elements

Explanation:
1. In transition elements the differentiating electron enters the inner (n-1)d subshell, so properties change only slightly across a series (horizontal similarity) and also resemble those down a group.

Teacher's Note:
1. Horizontal similarity is a feature of d- and f-block elements.
2. Representative elements show mainly vertical similarity.

Answer: (a) The first member of the lanthanoid series is lanthanum

Explanation:
1. The lanthanoid series begins with cerium (Z = 58), where 4f filling starts.
2. The other statements are correct.

Teacher's Note:
1. Lanthanoids are the 14 elements Ce to Lu.
2. Actinoids are all radioactive.

Answer: (b) nitrogen family

Explanation:
1. 15 = [Ne]3s²3p³, 33 = [Ar]3d¹⁰4s²4p³, 51 = [Kr]4d¹⁰5s²5p³.
2. All have ns²np³, i.e. group 15, the nitrogen family.

Teacher's Note:
1. Same outer configuration means same group.
2. Group 15 has ns²np³.

Answer: (a) 1s², 2s², 2p⁶

Explanation:
1. Na = 1s²2s²2p⁶3s¹.
2. The ten core electrons screen the lone 3s electron.

Teacher's Note:
1. Only inner electrons shield.
2. The 3s electron has no other electrons of its shell.