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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2016 · Paper I

Q5(e). Oxidation states of the lanthanides other than +3

  • 10 marks
  • Answered
Question

In general, the principal oxidation state of lanthanide elements is +3, but some elements also exhibit other oxidation states. Explain.

Answer in brief

+3 is general because the sum of the first three ionisation energies is repaid by large lattice or hydration energies, while the 4f electrons are too buried to be removed further. Other states appear when they give an empty (f⁰), half-filled (f⁷) or filled (f¹⁴) shell or come close to it: Ce⁴+, Tb⁴+, Pr⁴+; Eu²+, Yb²+, Sm²+, Tm²+.

Key points

  • The lanthanides are [Xe]4fⁿ6s² (or 4fⁿ⁻¹5d¹6s²). The two 6s electrons and one 4f or 5d electron are lost easily; the ions Ln³+ are [Xe]4fⁿ (n=0 for La, 14 for Lu).
  • The 4f orbitals are contracted inside the 5s²5p⁶ shell. Once three electrons have gone, the remaining 4f electrons are held tightly: the fourth ionisation energy is too large to be repaid by the extra lattice or hydration energy.
  • For Ln²+ the extra lattice/hydration energy of the more highly charged Ln³+ more than pays for IE₃. So +3 is the most stable state with a uniformity found nowhere else in the periodic table; the standard potentials…
  • The extra stability of an empty, half-filled or filled 4f subshell (exchange energy) lowers the energy of a state that reaches it:

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Why +3 is the principal state

  1. The lanthanides are [Xe]4fⁿ6s² (or 4fⁿ⁻¹5d¹6s²). The two 6s electrons and one 4f or 5d electron are lost easily; the ions Ln³+ are [Xe]4fⁿ (n=0 for La, 14 for Lu).
  2. The 4f orbitals are contracted inside the 5s²5p⁶ shell. Once three electrons have gone, the remaining 4f electrons are held tightly: the fourth ionisation energy is too large to be repaid by the extra lattice or hydration energy.
  3. For Ln²+ the extra lattice/hydration energy of the more highly charged Ln³+ more than pays for IE₃. So +3 is the most stable state with a uniformity found nowhere else in the periodic table; the standard potentials E°(Ln³+/Ln) are all near -2.3 V.

Why some show +2 or +4

  1. The extra stability of an empty, half-filled or filled 4f subshell (exchange energy) lowers the energy of a state that reaches it:
  2. +4: Ce⁴+ (4f⁰) is the most important; it is a strong, useful oxidising agent. Tb⁴+ (4f⁷) and Pr⁴+ (4f¹) occur in the oxides formed in air, Tb₄O₇ and Pr₆O₁₁, which contain Ln(III) and Ln(IV).
  3. +2: Eu²+ (4f⁷) is the most stable; EuSO₄, EuCO₃, EuI₂ are known and its solutions are stable. Yb²+ (4f¹⁴), Sm²+ (4f⁶) and Tm²+ (4f¹³) also form, but Sm²+ and Yb²+ reduce water to hydrogen.
  4. The rule is not the whole story: Pr⁴+ (f¹), Sm²+ (f⁶) and Tm²+ (f¹³) are not special configurations. The final balance of ionisation energy against lattice and hydration energy decides; configurations next to f⁰, f⁷, f¹⁴ simply gain most from it.

Application. the odd states are used to separate lanthanides: Ce is removed from mixtures by oxidation to Ce(IV), Eu by reduction to Eu(II) (precipitated as EuSO₄); ceric ammonium nitrate, (NH₄)₂[Ce(NO₃)₆], is a standard oxidimetric reagent.

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Model answer to Q5(e), 2016 Paper I: solution page 29
Solution page 29 of 46
Read the whole 2016 solution2016 question paper

All parts of Question 5