Q6(b). Crystal field stabilisation energy and the factors that fix it
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
- 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.
- 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.
- 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.
| dⁿ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| weak field (-Δₒ) | 0 | 0.4 | 0.8 | 1.2 | 0.6 | 0 | 0.4 | 0.8 | 1.2 | 0.6 | 0 |
| strong field (-Δₒ) | 0 | 0.4 | 0.8 | 1.2 | 1.6 | 2.0 | 2.4 | 1.8 | 1.2 | 0.6 | 0 |
Strong-field values exclude the pairing energy.
Factors that affect the magnitude
- 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.
- 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.
- 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.
- Geometry: fewer ligands, not pointing at the orbitals: Δₜ=49Δₒ, so tetrahedral complexes are always high spin; square planar gives the largest splitting.
- 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.
- 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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All parts of Question 6
- 6(a)(i) Interhalogen compounds; preparation and structure of IF₇15 marks
- 6(a)(ii) Labile complexes; lability and d-electron configuration10 marks
- 6(b) Crystal field stabilisation energy and the factors that fix it15 marks
- 6(c) IUPAC names of K₃[Fe(CN)₅NO] and [Co(NO₂)₃(NH₃)₃]10 marks