Chemistry · d- and f-block elements · chapter 4 of 10

Class 12 Chemistry Chapter 4: The d- and f-Block Elements case study question with answers

One case-based question of the kind the board asks: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks
A lighter flint is made of mischmetall, an alloy of about 95% lanthanoid metals with about 5% iron. The lanthanoids in it are so alike that separating them is difficult: all are reactive metals whose atoms share the 6s² configuration while the number of 4f electrons changes along the series. Their compounds are mostly of the tripositive ions, many of which give coloured solutions with sharp absorption bands. A few lanthanoids also form compounds in which they show an oxidation state other than +3, and such compounds are useful as oxidising or reducing agents in the laboratory.
Part 1 · 1 mark
Which of the following tripositive ions is colourless in aqueous solution?
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Answer: (d) Lu³⁺

Lu³⁺ is 4f¹⁴: with a filled f subshell no f–f transition is possible, so it is colourless. Pr³⁺, Nd³⁺ and Sm³⁺ have partly filled 4f sets and are coloured.

Part 2 · 1 mark
On going from La³⁺ to Lu³⁺, the ionic radius
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Answer: (b) decreases regularly

The lanthanoid contraction: the decrease is regular for the M³⁺ ions (not quite regular for the atoms), because 4f electrons shield the rising nuclear charge poorly.

Part 3 · 2 marks
Eu²⁺ is a strong reducing agent while Tb⁴⁺ is an oxidising agent. Explain both, using their electronic configurations.
Answer and marking scheme

Eu (Z = 63) is [Xe]4f⁷6s²; Eu²⁺ (4f⁷) forms by losing the two s electrons, and it readily loses one more electron to return to the common +3 state, so it reduces other species. Tb⁴⁺ has a half-filled 4f⁷ set, which favours its formation, but it readily gains an electron to return to +3, so it oxidises other species.

  • Eu²⁺ 4f⁷, gives up an electron to reach +3: reductant (1)
  • Tb⁴⁺ 4f⁷, takes an electron to reach +3: oxidant (1)
Or
E°(Ce⁴⁺/Ce³⁺) is +1.74 V, enough in principle to oxidise water. Explain why cerium shows the +4 state, and why Ce(IV) is nevertheless a useful analytical reagent in aqueous solution.
Answer and marking scheme

Ce (Z = 58) in the +4 state has a noble gas configuration ([Xe], 4f⁰), which favours Ce⁴⁺; it is a strong oxidant that reverts to the common +3 state. Its reaction with water is very slow, so Ce(IV) solutions are kinetically stable and serve as a good analytical oxidant.

  • Ce⁴⁺ has the noble gas configuration (4f⁰) (1)
  • oxidation of water is very slow, so the reagent is usable (1)
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