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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

Q6(b). Crystal field stabilisation energy and the factors that fix it

  • 15 marks
  • Answered
Question

Explain crystal-field stabilization-energy. What are the factors that affect its magnitude?

Answer in brief

CFSE is the lowering of the energy of the d electrons when a ligand field splits the d orbitals, measured from the mean (barycentre) energy. In an octahedron each t_(2g) electron counts -0.4 Δₒ and each e_g electron +0.6 Δₒ. Its size depends on the metal ion (charge, period, number of d electrons and spin), the ligands, and the geometry.

Key points

  • In a gaseous ion the five d orbitals are degenerate. Six ligands at the corners of an octahedron repel electrons in d_(x²-y²) and d_(z²) (pointing at them) more than in d_(xy), d_(xz), d_(yz): the set splits into e_g (+0.6 Δₒ)…
  • CFSE=(-0.4x+0.6y) Δₒ (+ pP) where p is the number of extra pairs formed relative to the gaseous ion and P the pairing energy.
  • d³ [Cr(NH₃)₆]³+: -1.2 Δₒ; Δₒ=21 500 cm⁻¹, so CFSE =-25 800 cm⁻¹=-308.6 kJ mol⁻¹. d⁶ low spin: -2.4 Δₒ+2P; d⁵ high spin: 0.
  • a higher charge pulls the ligands closer: Δₒ(aqua) for Co²+ 8400, for Co³+ 16 750 cm⁻¹ (about double); Fe²+ 9350, Fe³+ 14 000 cm⁻¹.

The full model answer

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

What CFSE is

  1. In a gaseous ion the five d orbitals are degenerate. Six ligands at the corners of an octahedron repel electrons in d_(x²-y²) and d_(z²) (pointing at them) more than in d_(xy), d_(xz), d_(yz): the set splits into e_g (+0.6 Δₒ) and t_(2g) (-0.4 Δₒ), keeping the barycentre.
  2. For t_(2g)^xe_gʸ: CFSE=(-0.4x+0.6y) Δₒ (+ pP) where p is the number of extra pairs formed relative to the gaseous ion and P the pairing energy.
  3. Examples: d³ [Cr(NH₃)₆]³+: -1.2 Δₒ; Δₒ=21 500 cm⁻¹, so CFSE =-25 800 cm⁻¹=-308.6 kJ mol⁻¹. d⁶ low spin: -2.4 Δₒ+2P; d⁵ high spin: 0.
dⁿ012345678910
weak field (-Δₒ)00.40.81.20.600.40.81.20.60
strong field (-Δₒ)00.40.81.21.62.02.41.81.20.60

Strong-field values exclude the pairing energy.

Factors that affect the magnitude

  1. Oxidation state of the metal: a higher charge pulls the ligands closer: Δₒ(aqua) for Co²+ 8400, for Co³+ 16 750 cm⁻¹ (about double); Fe²+ 9350, Fe³+ 14 000 cm⁻¹.
  2. Period (4d, 5d metals): larger d orbitals overlap better with the ligands, so Δₒ rises down a group: [Co(NH₃)₆]³+ 22 900, [Rh(NH₃)₆]³+ 34 000 cm⁻¹ (×1.48). Pairing energy also falls, so 4d and 5d complexes are almost always low spin.
  3. Nature of the ligand (spectrochemical series): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < bpy < NO₂⁻ < CN⁻ < CO. For Cr³+: Cl⁻ 13 700, H₂O 17 400, NH₃ 21 500, en 21 900, CN⁻ 26 600 cm⁻¹. π-donors lower Δ, π-acceptors raise it.
  4. Geometry: fewer ligands, not pointing at the orbitals: Δₜ=49Δₒ, so tetrahedral complexes are always high spin; square planar gives the largest splitting.
  5. Number of d electrons and spin state: CFSE follows the double-humped pattern of the panel; whether the strong-field values are reached depends on Δₒ against P.
  6. Metal–ligand distance: Δ falls steeply as the M–L distance grows, so shorter bonds (higher charge, compression) raise it.

Application. CFSE explains the double-humped curves of hydration and lattice enthalpies of the M²⁺ ions from Ca²+ to Zn²+ (d⁰, d⁵, d¹⁰ on the base line), the preference of Ni²+ and Cr³+ for octahedral sites in spinels, and the normal/inverse spinel structures (Fe₃O₄ inverse).

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Model answer to Q6(b), 2016 Paper I: solution page 33
Solution page 33 of 46