JEE Chemistry Classification Of Elements and Periodicity In Properties MCQs Set 10

Answer: (c) \( \frac{1}{2} \)

Explanation:
1. Moseley's law states \( \sqrt{\nu} = a(Z-b) \).
2. So \( \nu^{1/2} \) is linear in Z, which gives n = 1/2.

Teacher's Note:
1. Moseley's law is the basis of the modern periodic law.
2. Remember: root of frequency is proportional to Z.

Answer: (d) 10

Explanation:
1. An ion with 55 protons is Cs, and Cs+ has 54 electrons, the configuration of Xe.
2. Xe has 1s, 2s, 3s, 4s and 5s filled, five s-subshells with 2 electrons each.
3. Number of s-electrons = 5 x 2 = 10.

Teacher's Note:
1. Count electrons, not protons, in an ion.
2. Each filled s-subshell holds 2 electrons.

Answer: (c) He < Xe < Kr < Ne

Explanation:
1. The electron gain enthalpy values of the noble gases are about He 48, Xe 77, Kr 96 and Ne 116 kJ/mol.
2. Arranged in increasing order this gives He < Xe < Kr < Ne.

Teacher's Note:
1. Ne has the most positive value because the added electron enters the compact n = 2 shell with strong repulsion.
2. Learn the order He, Xe, Kr, Ne.

Answer: (a) Si < Be < Mg < K

Explanation:
1. Si is a metalloid, so it has the least metallic character.
2. Be is a metal, and metallic character increases down group 2 from Be to Mg.
3. K lies to the left of Mg in the periodic table, so K is the most metallic.
4. Order: Si < Be < Mg < K.

Teacher's Note:
1. Metallic character increases down a group and decreases across a period.
2. Compare elements from different groups using their position.

Answer: (b) Both Statement I and Statement II are correct

Explanation:
1. The first ionization enthalpies are B 801, Al 577 and Ga 579 kJ/mol, so the drop from B to Al is large and from Al to Ga is almost nil. Statement I is correct.
2. Ga has the configuration [Ar]3d10 4s2 4p1, so its 3d orbitals are completely filled. Statement II is correct.
3. The poorly shielding 3d electrons raise the effective nuclear charge, which explains the small drop.

Teacher's Note:
1. Poor shielding by d electrons explains the anomaly of Ga.
2. Both statements are true here.

Answer: (b) 4

Explanation:
1. Acidic oxides: N2O3, NO2, Cl2O7 and SO2.
2. N2O, CO and NO are neutral oxides.
3. CaO and Na2O are basic oxides.
4. Total acidic oxides = 4.

Teacher's Note:
1. Oxides of non-metals in high oxidation states are acidic.
2. N2O, NO and CO are the common neutral oxides.

Answer: (d) A - IV, B - II, C - I, D - III

Explanation:
1. Z = 37 is Rb, an alkali metal, so s-block (IV).
2. Z = 78 is Pt, so d-block (II).
3. Z = 52 is Te, a chalcogen, so p-block (I).
4. Z = 65 is Tb, a lanthanoid, so f-block (III).

Teacher's Note:
1. Identify the element from its atomic number, then its block.
2. Z = 58 to 71 are the lanthanoids.

Answer: (a) V2O3 > V2O4 > V2O5

Explanation:
1. Oxides of a metal become more acidic as its oxidation state rises.
2. V2O3 (+3) is basic, V2O4 (+4) is amphoteric and V2O5 (+5) is mainly acidic.
3. So basicity decreases V2O3 > V2O4 > V2O5.

Teacher's Note:
1. Higher oxidation state means more acidic oxide.
2. Compare CrO, Cr2O3 and CrO3 in the same way.

Answer: (d) Ca2+ < K+ < Cl- < S2-

Explanation:
1. All four ions have 18 electrons, so they are isoelectronic.
2. For isoelectronic ions, radius decreases as the nuclear charge increases.
3. Nuclear charge: Ca2+ (20) > K+ (19) > Cl- (17) > S2- (16).
4. So the radii increase as Ca2+ < K+ < Cl- < S2-.

Teacher's Note:
1. Among isoelectronic ions, more protons mean a smaller radius.
2. Anions are bigger than cations of the same electron count.

Answer: (d) K+, Cl-, Ca2+, Sc3+

Explanation:
1. K+, Cl-, Ca2+ and Sc3+ all have 18 electrons.
2. Set (a) has 2, 10, 10 and 18 electrons, set (b) has 54, 36, 18 and 18, and set (c) has 10, 10, 10 and 18, so none of those is isoelectronic.

Teacher's Note:
1. Electrons = Z minus charge.
2. Count electrons for every ion in the set.

Answer: (b) \( \Delta_{eg}H(Se) < \Delta_{eg}H(S) \)

Explanation:
1. Values in kJ/mol: F -328, Cl -349, S -200, Se -195, Br -325, I -295, At -270, Te -190, Po -174.
2. Cl is more negative than F, I more negative than At, and Te more negative than Po, so (a), (c) and (d) are correct.
3. Se (-195) is less negative than S (-200), so \( \Delta_{eg}H(Se) > \Delta_{eg}H(S) \) and (b) is incorrect.

Teacher's Note:
1. Second-period elements O and F break the group trend because of small size.
2. Use sign carefully: more negative means a larger gain.