Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 2 / B7 / c / Solution

Start from an octahedral molecular geometry with ligands on $\pm x,\pm y,\pm z$, with $e_g=(d_{z^2},d_{x^2-y^2})$ above $t_{2g}=(d_{xy},d_{xz},d_{yz})$. Remove the $-z$ donor while keeping the other five positions. Loss of axial repulsion lowers $d_{z^2}$ substantially; $d_{x^2-y^2}$ still points directly at four basal donors and remains highest. Loss of an axial donor also lowers the orbitals with $z$ dependence relative to $d_{xy}$. The remaining fourfold axis keeps $d_{xz}$ and $d_{yz}$ degenerate.

The usual qualitative square-pyramidal molecular geometry diagram, shown in the last panel above, is thereforeThe corresponding $C_{4v}$ labels are $b_1,a_1,b_2,e$. Energies can be re-zeroed to the new orbital barycenter after ligand removal. Symmetry fixes the degeneracy and labels, but does not universally fix the separation of $d_{z^2}$ and $d_{xy}$: a weak or distant axial donor can reverse this middle pair on approach to the square-planar limit. The displayed ordering assumes an appreciable fifth axial interaction.