Crystal-field theory: splitting the d orbitals

In 1936 Pauling and Coryell weighed haemoglobin solutions in a magnetic field. Haemoglobin without oxygen was pulled into the field, as if each haem carried 4 unpaired electrons; with CO bound it became diamagnetic, with no unpaired electrons at all. The same contrast turns up on the bench: a strong magnet tugs at pale-green $\ce{FeSO4.7H2O}$ crystals but not at yellow $\ce{K4[Fe(CN)6].3H2O}$. Both contain $\ce{Fe^2+}$; both are d⁶. Six d electrons — why are four of them unpaired in one case and none in the other?

In the free ion the five d orbitals have the same energy (they are degenerate). Once ligands approach, the degeneracy breaks: the two orbitals pointing at the ligands rise and form the eg set; the three lying between the ligands fall and form the t₂g set. The gap between them is the octahedral splitting Δo.

The splitting obeys a “barycentre rule”: the total rise equals the total fall, so eg goes up by $\tfrac{3}{5}\Delta_\text{o}$ and t₂g down by $\tfrac{2}{5}\Delta_\text{o}$. The extra stability gained by putting electrons into the lower t₂g set is the crystal-field stabilisation energy (CFSE).

For d¹–d³ each electron takes its own t₂g orbital; there is no choice. The fourth electron has two options:

• enter eg with its spin parallel to the others — paying an extra Δo;
• pair up in an occupied t₂g orbital — paying the pairing energy P.

It takes the cheaper route. When Δo < P electrons would rather climb, giving the high-spin arrangement with the most unpaired electrons; when Δo > P they would rather pair, giving low spin. Only d⁴, d⁵, d⁶ and d⁷ have both options; d⁸–d¹⁰ again have just one.

Δo depends on the ligand (the spectrochemical series of the next lesson), on the metal’s oxidation state ($\ce{[Fe(H2O)6]^2+}$ 10 400 cm⁻¹, $\ce{[Fe(H2O)6]^3+}$ 14 300 cm⁻¹) and on the period (4d and 5d metals split far more and are almost always low spin).

In a tetrahedron no d orbital points straight at a ligand, but $d_{xy}$, $d_{xz}$ and $d_{yz}$ come closer to the ligands than the other two, so the pattern inverts: three t₂ orbitals above, two e below. With only four ligands, none aimed at an orbital, the splitting is also much smaller, $\Delta_\text{t} \approx \tfrac{4}{9}\Delta_\text{o}$ — well below the pairing energy — so tetrahedral complexes are almost always high spin.

A square plane is an octahedron with the two ligands on the z axis removed: $d_{x^2-y^2}$ is left alone at the top. A d⁸ ion puts all 8 electrons into the four lower orbitals, and the empty $d_{x^2-y^2}$ brings a large stabilisation, so $\ce{[Ni(CN)4]^2-}$ is square planar and diamagnetic, whereas weak-field $\ce{[NiCl4]^2-}$ is tetrahedral with 2 unpaired electrons.