Octahedral crystal field splitting

In an octahedral complex, six ligands sit along the $\pm x$, $\pm y$ and $\pm z$ axes. The five d orbitals no longer have equal energy. The ligands' electron pairs repel electrons in some orbitals more than in others.

Two sets of orbitals

• $e_g$ ($d_{z^2}$, $d_{x^2-y^2}$) point straight at the ligands. They are pushed up.
• $t_{2g}$ ($d_{xy}$, $d_{xz}$, $d_{yz}$) point between the ligands. They are pushed down.

The gap is $\Delta_\mathrm{o}$. The splitting keeps the average energy unchanged (the barycentre), so the two sets sit at

The crystal field stabilisation energy for $x$ electrons in $t_{2g}$ and $y$ in $e_g$ is then

plus a pairing energy $P$ for each extra pair compared with the free ion.

High spin or low spin?

For d¹–d³ and d⁸–d¹⁰ there is only one way to fill the orbitals. For d⁴–d⁷ the fourth electron has a choice:

• if $\Delta_\mathrm{o} < P$, it goes up into $e_g$, giving high spin with the most unpaired electrons
• if $\Delta_\mathrm{o} > P$, it pairs in $t_{2g}$, giving low spin

For example, Fe²⁺ (d⁶) gives:

The spin-only magnetic moment $\mu = \sqrt{n(n+2)}\,\mu_\mathrm{B}$ separates them: 4.90 $\mu_\mathrm{B}$ against zero.

What makes $\Delta_\mathrm{o}$ large

1. The ligand. The spectrochemical series runs from weak to strong field: I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO. π-donors sit at the weak end. π-acceptors sit at the strong end.
2. Higher oxidation state of the metal. M³⁺ splits more than M²⁺.
3. Going down a group. 4d and 5d orbitals are larger and overlap better. Their complexes are almost always low spin.