D & F Block Elements MCQs for NEET — Chemistry Questions with Answers

FeSO₄ will require the least amount of acidified KMnO₄ for complete oxidation.

(a) Lanthanoid contraction is the regular decrease in atomic and ionic radii of lanthanides. This is due to the imperfect shielding or poor screening effect of orbitals due to their diffused shape, which unable to counterbalance the effect of the increased nuclear charge. Hence, the net result is a contraction in size of lanthanoids.

Since OH^- are generated from weak acid (H₂O), a weak acid (like CO₂) should be used to remove it because of strong acid (HCl) reverse the reaction. KOH increases the concentration of OH^-, thus again shifts the reaction in backward side. CO₂ combines with OH^- to give carbonate which is easily removed. SO₂ reacts with water to give strong acid, so it cannot be used.

(b) Interstitial compounds are obtained when small atoms like H, B, C. N etc., fit into the lattice of other elements. These retain metallic conductivity. These resemble the parent metal in chemical properties (reactivity) but differ in physical properties like hardness, melting point etc.But they are chemically inactive

This can be understood on the basis of E° values for M²⁺/M.

E°/V       Cr      Mn    Fe     Co
M²⁺/M -0.90 -1.18 -0.44 -0.28

E° value for Mn is more negative than expected from general trend due to extra stability of half-filled Mn²⁺ ion.

Thus,the correct order should be:-
Mn>Cr>Fe>Co

An examination of E° values for redox couple M³⁺/ M²⁺ shows that Cr²⁺ is strong reducing
agent (E° M³⁺/M²⁺=0.41V) and liberates H₂ from dilute acids.

2Cr²⁺ (aq) + 2H⁺ (aq)→ 2Cr³⁺(aq) + H₂↑(g)

K₂Cr₂O₇ + 3Na₂SO₃ + 4H₂SO₄ $\to$3Na₂SO₄ + K₂SO₄ + Cr₂(SO₄)₃ + 4H₂O

Key Idea Colour is obtained as a consequence of d-d (or f-f) d-d (or M transition. and for d-d (or f-f) transition, presence of unpaired electrons is the necessary condition.

La³⁺(Z=57)=[Xe] 4f⁰5d⁰6s⁰ (no unpaired electron)

Ti³⁺(Z=22)=[Ar] 3d¹4s⁰ (one unpaired electron)

Lu³⁺(Z=71)=[Xe]4f¹⁴5d⁰6s⁰ (no unpaired electron)

Sc³⁺(Z=21)=[Ar] 3d⁰ 4s⁰ (no unpaired electron)

Hence, only Ti³⁺ will exhibit colour in aqueous solution.

(c) The regular decrease in the radii of lanthanide ions from La3+ to Lu3+ is known as lanthanide contraction. It is due to the greater effect of the increased nuclear charge than that of the screening effect (shield effect). As a result of lanthanide contraction, the atomic radii of element of 4d and 5d one close just above them in their respective group, so the properties of 4d and 5d transition element shows the similarities.

(b) More number of oxidation states are exhibited by the acinoids than by the lanthanoids due to lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals.