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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 · 2020 · Paper I

Q8(d). Four short and two long Cu–F bonds in CuF₂: Jahn–Teller distortion

  • 10 marks
  • Answered
Question

In the crystal structure of CuF₂, the Cu²⁺ is six-coordinated with four F⁻ at a distance of 1.93 Å and two F⁻ at 2.27 Å. Explain the reason(s) for this.

Answer in brief

Cu²⁺ is d⁹ (t_(2g)⁶e_g³): the odd electron can sit in either e_g orbital, a degenerate ground state. By the Jahn–Teller theorem the octahedron distorts; elongation along z lowers the energy, giving four short (1.93 Å) and two long (2.27 Å) Cu–F bonds.

Key points

  • In a regular octahedron the three e_g electrons can be arranged as d_(z²)²d_(x²-y²)¹ or d_(z²)¹d_(x²-y²)²: equal energies, so the ground state is orbitally degenerate.
  • a non-linear molecule in a degenerate electronic state distorts to remove the degeneracy and lower its energy.
  • the two axial F⁻ move out. d_(z²) falls by δ₁/2 and d_(x²-y²) rises by δ₁/2. With two electrons in the lower and one in the upper orbital the net gain is 2(-(δ₁)/(2))+1(+(δ₁)/(2))=-(δ₁)/(2).
  • d_(z²) is full and repels the axial ligands (long bonds, 2.27 Å); d_(x²-y²) is half full, so the four equatorial F⁻ are less screened and come closer (1.93 Å). The t_(2g) set also splits (d_(xz),d_(yz) below d_(xy)), but it is…

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. In a regular octahedron the three e_g electrons can be arranged as d_(z²)²d_(x²-y²)¹ or d_(z²)¹d_(x²-y²)²: equal energies, so the ground state is orbitally degenerate.
  2. Jahn–Teller theorem: a non-linear molecule in a degenerate electronic state distorts to remove the degeneracy and lower its energy.
  3. Elongation along z: the two axial F⁻ move out. d_(z²) falls by δ₁/2 and d_(x²-y²) rises by δ₁/2. With two electrons in the lower and one in the upper orbital the net gain is 2(-(δ₁)/(2))+1(+(δ₁)/(2))=-(δ₁)/(2).
  4. Bond lengths: d_(z²) is full and repels the axial ligands (long bonds, 2.27 Å); d_(x²-y²) is half full, so the four equatorial F⁻ are less screened and come closer (1.93 Å). The t_(2g) set also splits (d_(xz),d_(yz) below d_(xy)), but it is full, so it gives no net gain.
  5. Strong effect: the uneven occupation is in the e_g set, which points at the ligands, so the distortion is large. d⁴ high-spin (Cr²⁺, Mn³⁺) and low-spin d⁷ ions show it too; d³, d⁸ and d¹⁰ ions do not.

Application. the same distortion makes [Cu(H₂O)₆]²+ exchange its two axial waters very fast, makes log K₅ and log K₆ for ammine complexes of Cu(II) small, and broadens the single visible band that gives Cu(II) salts their blue colour.

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Model answer to Q8(d), 2020 Paper I: solution page 43
Solution page 43 of 45
Read the whole 2020 solution2020 question paper

All parts of Question 8