Electronic configuration in [Ni(CN)4]2-

Question

$\ce{[Ni(CN)4]^2-}$ is a low spin , inner orbital complex having $\mathrm{dsp^2}$ hybridisation. According to Crystal Field Theory, $\ce{CN^-}$ is a strong field ligand that causes pairing of electrons i.e electrons occupy the orbitals of the lower $t_{2g}$ level completely before they occupy the higher $e_{g}$ level. If I am not wrong, Hund's rule of maximum multiplicity is followed while filling the orbitals of $t_{2g}$ or $e_{g}$.

$\ce{Ni^2+}$ has a configuration of $[Ar]3d^8$. If we write down the electronic configuration of $\ce{Ni^2+}$ in this complex, 6 of the d-electrons would occupy the $t_{2g}$ level and the remaining 2 would occupy then $e_{g}$ level.

If we search for diagrams of electronic configuration of $\ce{[Ni(CN)4]^2-}$, many of them show the last 2 occupying the same orbital in $e_{g}$ which seems to contradict Hund's rule in my view. But I am pretty sure that the the diagrams on the Internet are correct according to experimental data of magnetic moments and other features.

My question is whether strong field ligands can pair up electrons within orbitals of $t_{2g}$ and $e_{g}$ like in this case where the 2 $e_{g}$ electrons are paired up within one orbital instead of letting them occupy two separate ones. Or is there some other factor at play?

Answer 1

Your thinking is correct. Within a set of degenerate orbitals (i.e. equal in energy), the orbitals will fill one electron at a time, and only after all have one electron will they start to pair up.

The situation with $\ce{[Ni(CN)4]^{2-}}$ is that it has square planar geometry, so the two orbitals that are $e_g$ in an octahedral complex are separated in energy. This diagram from Chemistry LibreTexts shows it nicely:

Because of the separation in energy, the $\mathrm{d}_{xy}$ fills completely (two electrons) before any electrons fill the $\mathrm{}_{x^2-y^2}$.