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 areBecause 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 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.

Fix the positive lobes of the real atomic orbitals along the printed positive axes, and write ${\varphi }_1=p_x(1)$, ${\varphi }_4=p_x(4)$, ${\varphi }_5=p_y(5)$ and ${\varphi }_8=p_y(8)$. A coordinate model consistent with the figure is $r_1=(0,a,-b)$, $r_4=(0,-a,-b)$, $r_5=(a,0,b)$ and $r_8=(-a,0,b)$, with $a,b>0$. For the D2d molecular point group, take the two transverse twofold axes along $x+y$ and $x-y$, with vertical mirrors $xz$ and $yz$. The $A_2$ character table row on $(E,2S_4,C_2,2C_2^{\prime },2{\sigma }_d)$ is $(1,1,1,-1,-1)$. The character projection operator isThe eight signed images needed for this projection areso $P_{A_2}{\varphi }_1=({\varphi }_1-{\varphi }_4-{\varphi }_5+{\varphi }_8)/4$. Under the standard orthonormal atomic orbital approximation its squared norm is $4/16=1/4$. Therefore the normalized symmetry-adapted linear combination isAn overall minus sign describes the same quantum state; changing one local atomic orbital phase requires changing its coefficient too. If intersite orbital overlap were retained, normalization would instead use the overlap matrix.