Crystal Field Theory — Octahedral Splitting and CFSE
Crystal field theory begins with one deliberately crude assumption: treat the ligands as point negative charges and ask what they do to the energies of the five d orbitals on the metal. The assumption is wrong in detail — real metal–ligand bonds are substantially covalent — yet the model predicts colours, magnetic moments, ionic radii and lattice energies across the whole first transition series. This article develops the octahedral case properly and works out crystal field stabilisation energy for every dn configuration.
Why the five d orbitals split into three plus two
Place six ligands along the ±x, ±y and ±z axes. Two of the d orbitals point directly at the ligands: dz² and dx²−y². The other three point between the axes: dxy, dyz and dxz.
An electron in an orbital pointing at a negative ligand is repelled more strongly, so those two rise in energy while the three pointing between the ligands fall. The result is two sets, labelled by their symmetry in the Oh point group.
| Set | Orbitals | Orientation | Energy relative to barycentre |
|---|---|---|---|
| eg (upper, 2 orbitals) | dz², dx²−y² | Point at the ligands | +0.6 Δo = +6 Dq |
| t2g (lower, 3 orbitals) | dxy, dyz, dxz | Point between the ligands | −0.4 Δo = −4 Dq |
The gap between them is Δo, the octahedral splitting parameter (also written 10 Dq). The two levels straddle a weighted average, the barycentre: three orbitals drop by 0.4Δo and two rise by 0.6Δo, so 3 × 0.4 = 2 × 0.6 = 1.2 and the centre of gravity is preserved. That requirement is where the odd-looking 0.4 and 0.6 come from.
Crystal field stabilisation energy
Here nt2g and neg are the electron counts in each set, and P is the pairing energy — the cost of forcing two electrons into the same orbital, counted only for pairs that would not exist in the free ion. A negative CFSE means the complex is stabilised relative to a hypothetical spherical field.
High spin or low spin: one comparison decides it
For d⁴ to d⁷ there is a genuine choice. The fourth electron can either go into the upper eg set (costing Δo) or pair up in the lower t2g set (costing P). Whichever is cheaper wins.
Configurations d¹, d², d³, d⁸, d⁹ and d¹⁰ have no choice at all — high spin and low spin descriptions are identical for them. This is why questions about spin state are always about the middle of the series.
The complete CFSE table
Values in units of Δo. The pairing term is written separately because it depends on the metal, not on the geometry.
| dn | High-spin config. | HS CFSE | HS unpaired | Low-spin config. | LS CFSE | LS unpaired |
|---|---|---|---|---|---|---|
| d¹ | t2g¹ | −0.4 | 1 | same | −0.4 | 1 |
| d² | t2g² | −0.8 | 2 | same | −0.8 | 2 |
| d³ | t2g³ | −1.2 | 3 | same | −1.2 | 3 |
| d⁴ | t2g³eg¹ | −0.6 | 4 | t2g⁴ | −1.6 + P | 2 |
| d⁵ | t2g³eg² | 0 | 5 | t2g⁵ | −2.0 + 2P | 1 |
| d⁶ | t2g⁴eg² | −0.4 | 4 | t2g⁶ | −2.4 + 2P | 0 |
| d⁷ | t2g⁵eg² | −0.8 | 3 | t2g⁶eg¹ | −1.8 + P | 1 |
| d⁸ | t2g⁶eg² | −1.2 | 2 | same | −1.2 | 2 |
| d⁹ | t2g⁶eg³ | −0.6 | 1 | same | −0.6 | 1 |
| d¹⁰ | t2g⁶eg⁴ | 0 | 0 | same | 0 | 0 |
Check two rows by hand so the pattern sticks. Low-spin d⁶ is t2g⁶: CFSE = 6 × (−0.4) = −2.4 Δo, the largest possible, which is why [Co(NH₃)₆]³⁺ and [Fe(CN)₆]⁴⁻ are famously inert. High-spin d⁵ is t2g³eg²: CFSE = 3(−0.4) + 2(+0.6) = exactly zero, which is why Mn²⁺ and Fe³⁺ show no crystal field preference for any geometry.
Worked CFSE calculations in real units
Worked example 1 — a d¹ complex. [Ti(H₂O)₆]³⁺ has Δo ≈ 20 300 cm⁻¹.
Configuration t2g¹, so CFSE = −0.4 × 20 300 = −8120 cm⁻¹.
In kJ/mol: 8120 × 0.011963 = 97.1, so CFSE = −97.1 kJ/mol.
No pairing term is needed: a single electron cannot pair with anything.
Worked example 2 — the same metal ion, two ligands, two spin states. Compare [Fe(H₂O)₆]²⁺ and [Fe(CN)₆]⁴⁻. Both are Fe(II), d⁶, with pairing energy P ≈ 17 600 cm⁻¹. Take Δo ≈ 10 400 cm⁻¹ for the aqua complex and ≈ 33 000 cm⁻¹ for the cyanide.
[Fe(H₂O)₆]²⁺: Δo = 10 400 < P = 17 600, so high spin, t2g⁴eg².
CFSE = [4(−0.4) + 2(+0.6)] × 10 400 = (−1.6 + 1.2) × 10 400 = −0.4 × 10 400 = −4160 cm⁻¹ = 4160 × 0.011963 = −49.8 kJ/mol.
Unpaired electrons = 4, so μspin-only = √(4 × 6) = 4.90 BM. Pale green, paramagnetic, kinetically labile.
[Fe(CN)₆]⁴⁻: Δo = 33 000 > P = 17 600, so low spin, t2g⁶.
Orbital term = 6(−0.4) × 33 000 = −2.4 × 33 000 = −79 200 cm⁻¹ = −947 kJ/mol.
Pairing correction: the free d⁶ ion has one pair; t2g⁶ has three; so two extra pairs cost 2P = +35 200 cm⁻¹.
Net CFSE = −79 200 + 35 200 = −44 000 cm⁻¹ = 44 000 × 0.011963 = −526 kJ/mol.
Unpaired electrons = 0, diamagnetic. The difference between −49.8 and −526 kJ/mol is the quantitative reason low-spin cyanido complexes are so much more stable and so much more inert than their aqua analogues.
Worked example 3 — deciding a spin state quickly. For Co(III), d⁶, P ≈ 21 000 cm⁻¹. [CoF₆]³⁻ has Δo ≈ 13 000 cm⁻¹ (< P) → high spin, 4 unpaired, μ = 4.90 BM. [Co(NH₃)₆]³⁺ has Δo ≈ 23 000 cm⁻¹ (> P) → low spin, diamagnetic. Both values sit close to P on opposite sides, so this pair is near the spin-crossover boundary, where temperature or pressure can switch the spin state.
What controls the size of Δo
- The ligand. The spectrochemical series runs I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < bipy < phen < NO₂⁻ < CN⁻ < CO. π-donor ligands sit at the weak end, π-acceptor ligands at the strong end.
- Oxidation state. Higher charge pulls the ligands closer and raises Δo by roughly 50% going from M²⁺ to M³⁺.
- Position in the group. Δo rises by about 30–50% from 3d to 4d and again from 4d to 5d, which is why virtually every second- and third-row complex is low spin.
Consequences you should be able to quote
Jahn–Teller distortion. An unevenly occupied degenerate set is unstable, so the complex distorts to remove the degeneracy — strongly when the imbalance is in the eg set, which points at the ligands: high-spin d⁴ (eg¹), low-spin d⁷ (eg¹) and d⁹ (eg³). Hence Cu(II) complexes are almost always tetragonally elongated, with four short and two long bonds. A t2g imbalance distorts only weakly.
The double-humped hydration enthalpy curve. Plot hydration enthalpies of the M²⁺ ions from Ca²⁺ to Zn²⁺ and you get two humps, with Ca²⁺ (d⁰), Mn²⁺ (d⁵) and Zn²⁺ (d¹⁰) lying on a smooth line through them — precisely the three configurations with zero CFSE. Subtract the CFSE from each measured value and the curve straightens.
Other geometries. In a tetrahedral field the labels invert: the e set (dz², dx²−y²) lies lower at −0.6Δt and t₂ higher at +0.4Δt. With only four ligands, none on the axes, Δt ≈ (4/9)Δo for the same metal and ligands — always less than P in the first row, so first-row tetrahedral complexes are effectively always high spin. Square planar is the opposite extreme, arising from a large tetragonal distortion and favoured by d⁸ ions with strong-field ligands: diamagnetic [Ni(CN)₄]²⁻ against paramagnetic tetrahedral [NiCl₄]²⁻.
Errors that recur in CFSE questions
- Reversing the sign convention. t2g is −0.4Δo and eg is +0.6Δo, not the other way round; a stabilised complex has a negative CFSE.
- Counting all pairs instead of extra pairs. For low-spin d⁶ the correction is 2P, not 3P, because the free ion already has one pair.
- Asking whether d³ or d⁸ is high or low spin. Those configurations have only one possible arrangement.
- Claiming a first-row tetrahedral complex is low spin. Δt is about four-ninths of Δo and never beats P here.
- Using CFSE alone to compare two different metals. CFSE is a small correction on top of much larger electrostatic terms; it explains trends within a series, not absolute stabilities.
- Forgetting that high-spin d⁵ and d¹⁰ have zero CFSE. This is the fact that anchors the hydration-enthalpy argument.
Handle the numbers without breaking your concentration. The ABC Chemistry Calculator Suite covers the arithmetic these questions sit on — unit conversion between nm, cm⁻¹ and kJ/mol, electron configurations, quantum numbers, an interactive periodic table and a scientific constants reference.
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