 Solution

The Co(III) centre in $A^1$ has $t_{2g}^6(e_g^{\ast }{)}^0$: every occupied d subset is orbitally balanced. There is no first-order metal-centred Jahn–Teller driving force in $A^1$. This does not require all six experimental bond lengths to be equal, since pyridine and dioxolene are different donors.

The Co(II) centre in $A^2$ has $t_{2g}^5(e_g^{\ast }{)}^2$. Its two $e_g^{\ast }$ orbitals each contain one electron, so that subset gives no first-order preference. The $t_{2g}$ hole does give a weaker Jahn–Teller instability. To compare tetragonal distortions, take a positive small splitting parameter $\delta$ and retain each subset's orbital energy barycentre. On axial elongation let $d_{xz},d_{yz}$ shift by $-\delta$ and $d_{xy}$ by $+2\delta$. Five electrons then giveFor equal-magnitude compression the shifts reverse, and filling the lower $d_{xy}$ orbital first givesThe two singly occupied $e_g^{\ast }$ orbitals contribute zero net shift in either case. Thus the elementary tetragonal splitting model favours elongation for high-spin $d^7$, rather than equal electronic stabilization by the two signs. In a strictly sigma-only octahedral molecular orbital model $t_{2g}$ is nonbonding, so this weak effect is absent; it arises when the electrostatic and non-sigma interactions neglected in that diagram are restored. spin-orbit coupling, unequal donors and chelate constraints can modify the actual distortion, so the argument does not establish which experimental Co–donor bonds must lengthen.