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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

Q7(a)(ii). Lanthanide contraction and its effect on 4d and 5d metals

  • 10 marks
  • Answered
Question

What do you understand by “Lanthanide contraction”? How does it affect the chemistry of 4d and 5d block elements.

Answer in brief

The steady decrease in atomic and ionic radii from La to Lu (La³+ 116 pm to Lu³+ 98 pm, CN 8), because the 4f electrons shield the rising nuclear charge poorly. It cancels the normal increase in size down a group, so 5d metals are almost the same size as the 4d metals above them (Zr 160, Hf 159 pm) and the 4d/5d pairs are chemically very alike.

Key points

  • From Ce to Lu each added electron enters the inner 4f subshell while the nuclear charge rises by one. 4f orbitals are diffuse and shield poorly, so the effective nuclear charge felt by the outer 5s, 5p, 6s electrons rises step…
  • The total fall is about 18 % for Ln³+ (116 → 98 pm). Without it, Period 6 atoms after the lanthanides would be clearly larger than their Period 5 partners.
  • metallic radii (pm): Zr 160, Hf 159; Nb 147, Ta 147; Mo 140, W 141; Pd 137, Pt 139; Ag 144, Au 144. Normally size rises down a group (Ti 147 → Zr 160), but from 4d to 5d it does not.
  • same size and charge give almost identical properties: Zr and Hf occur together and are among the hardest pairs to separate (solvent extraction, ion exchange); Nb/Ta and Mo/W are similar pairs.

The full model answer

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

  1. From Ce to Lu each added electron enters the inner 4f subshell while the nuclear charge rises by one. 4f orbitals are diffuse and shield poorly, so the effective nuclear charge felt by the outer 5s, 5p, 6s electrons rises step by step and the atom or ion contracts.
  2. The total fall is about 18 % for Ln³+ (116 → 98 pm). Without it, Period 6 atoms after the lanthanides would be clearly larger than their Period 5 partners.

Effects on the 4d and 5d elements

  1. Equal sizes: metallic radii (pm): Zr 160, Hf 159; Nb 147, Ta 147; Mo 140, W 141; Pd 137, Pt 139; Ag 144, Au 144. Normally size rises down a group (Ti 147 → Zr 160), but from 4d to 5d it does not.
  2. Similar chemistry: same size and charge give almost identical properties: Zr and Hf occur together and are among the hardest pairs to separate (solvent extraction, ion exchange); Nb/Ta and Mo/W are similar pairs.
  3. High densities: the atomic mass nearly doubles from Zr to Hf at the same volume, so 5d metals are about twice as dense (Os, Ir, Pt are the densest elements).
  4. Higher ionisation energies, nobler metals: the 5d electrons feel a larger effective nuclear charge, so Pt and Au are more noble and less reactive than Pd and Ag.
  5. 3d differs from 4d/5d: 4d and 5d metals favour higher oxidation states, form low-spin complexes and metal–metal bonds more readily than 3d metals; the 3d member of a group stands apart, the 4d and 5d members go together.
  6. Yttrium: Y³+ is close in size to the heavy Ln³+ ions, so Y occurs with the heavy lanthanides; the basicity of Ln(OH)₃ also falls from La to Lu.

Application. the small, regular change in size within the series is what allows the lanthanides to be separated by ion exchange and solvent extraction; Zr with its Hf removed is needed for nuclear reactor cladding because Hf absorbs neutrons strongly.

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Model answer to Q7(a)(ii), 2016 Paper I: solution page 37
Solution page 37 of 46
Read the whole 2016 solution2016 question paper

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