= Solution
The Co(III) centre in $A^1$ has $t_{2g}^6(e_g^*)^0$: every occupied d subset is orbitally balanced. \b[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 $t_{2g}^5(e_g^*)^2$. Its two $e_g^*$ 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 give
$$
\Delta E_{\rm elong}=4(-\delta)+1(2\delta)=-2\delta.
$$
For equal-magnitude compression the shifts reverse, and filling the lower $d_{xy}$ orbital first gives
$$
\Delta E_{\rm compress}=2(-2\delta)+3(\delta)=-\delta.
$$
The two singly occupied $e_g^*$ orbitals contribute zero net shift in either case. Thus \b[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.
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