= Solution
Two contributions must be separated: the underlying contraction of the metal ion across the series and the configuration-dependent <crystal field stabilization energy>. Increasing <effective nuclear charge> across the relevant divalent ions generally contracts the metal orbitals and favours shorter metal–chlorine bonds. Superimposed on this, a weak-field tetrahedral chloride environment places the $e$ set below the $t_2$ set, at $-0.6\Delta_t$ and $+0.4\Delta_t$ relative to the barycentre. The high-spin occupancies and corresponding <crystal field stabilization energies> are
$$
\begin{array}{c|c|r}
d\text{ count}&\text{occupancy}&\mathrm{CFSE}/\Delta_t\\\hline
6&e^3t_2^3&-0.6\\
7&e^4t_2^3&-1.2\\
8&e^4t_2^4&-0.8\\
10&e^4t_2^6&0
\end{array}
$$
Because the <tetrahedral crystal field splitting> increases on bringing the ligands inward, the stabilization is especially effective at shortening the $d^7$ bond. Adding the next <electron> to the upper $t_2$ set reduces that stabilization; in a simple sigma-bonding description this set also has more metal–ligand antibonding character. That change explains the upward deviation from $d^7$ to $d^8$ despite the general contraction trend. By $d^{10}$ the <crystal field stabilization energy> is zero, but the background contraction can still produce the shortest bond in the plotted set. Thus \b[the $d^7$ shortening and $d^8$ rebound reflect orbital filling, while the overall decrease reflects contraction]. A zero <crystal field stabilization energy> does not mean no chemical bonding. Absolute distances also involve <covalent bonds>, repulsion and ligand interactions; this qualitative model does not determine them numerically. No inference about the missing $d^9$ distance is required.
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