Crystal Field Theory: Splitting in a Tetrahedral Field

When only four ligands surround a metal ion in a tetrahedral arrangement, rather than six in an octahedral one, the geometry of approach relative to the d-orbitals changes completely, with two important consequences.

First, in a tetrahedral complex, none of the four ligands approaches directly along any of the Cartesian $x$, $y$, or $z$ axes — instead, they approach along directions that point between the axes, toward the corners of a tetrahedron inscribed within a cube. This exactly reverses which set of d-orbitals experiences the stronger repulsion. The $e$ set ($d_{z^2}$ and $d_{x^2-y^2}$, which point directly along the axes) now finds itself pointing away from the incoming ligands and experiences comparatively weak repulsion, so it is lowered in energy, by $0.6\,\Delta_t$ below the barycentre. The $t_2$ set ($d_{xy}$, $d_{yz}$, $d_{xz}$, which point between the axes) now finds itself pointing more nearly toward the ligand directions and experiences stronger repulsion, so it is raised, by $0.4\,\Delta_t$. This is the exact opposite ordering to the octahedral case, where the analogous "between-the-axes" set ($t_{2g}$) was the lower-energy set.

Second, the overall size of the splitting, $\boldsymbol{\Delta_t}$, is intrinsically much smaller than $\Delta_o$ for the same metal ion and the same ligands at a comparable distance — as a useful rule of thumb, $\Delta_t \approx \tfrac{4}{9}\Delta_o$. This arises from two combined geometric factors: there are only four ligands instead of six (fewer point charges doing the repelling), and none of them lies directly on a d-orbital's axis of maximum electron density (weaker repulsion per ligand, even for the more-affected $t_2$ set, than the head-on repulsion the $e_g$ set experiences in an octahedral field). …