For tetrahedral CFSE, Nickel's last 6 electrons will go in the t2g set of orbitals. Since it is already the higher energy orbital, pairing should NOT happen right? However, all internet sources I found for NiCN4 show that pairing happens in the t2g. Shouldn't electrons not pair and follow hunds rule? Isn't this wrong? Electrons should not be able to pair and would follow general hund's rule


1 Answer 1


You do not have $t_{2g}$ or for that matter even $t_2$ (there's a "g", meaning centrally symmetric, only if the coordination geometry has that symmetry). You don't have a tetrahedral coordination at all.

When you have a $d^8$ central core with a strong field ligand, both crystal field theory and ligand field theory predict that a strong extra stabilization occurs by going to a square planar configuration, and so it generally does. $\ce{Ni(CN)4^{2-}}$ is actually a classic example of such a square planar coordination.

This square planar coordination no longer has the right symmetry to give a $t_2$ set and an $e$ set; instead the orbitals split in an entirely different way with the net result that one of the orbitals is much more destabilized than all the others. From https://www.jove.com/science-education/11462/crystal-field-theory-tetrahedral-and-square-planar-complexes:

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Then the eight electrons go into the four lower-energy orbitals where perforce they are all paired.


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