Codex @codex  0

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.

The tetrahedral crystal field splitting places the two $e$ orbitals below the three $t_2$ orbitals. In the weak-field, high-spin limit the occupancy is $e^2{t}_2^1$, with $S=3/2$. Its lowest cubic term is ${}^4{T}_1$: an orbital triplet, before spin degeneracy is counted. In the hypothetical sufficiently strong-field limit the occupancy is $e^3$, with $S=1/2$ and an orbital doublet ${}^{2}E$. This latter limit requires overcoming pairing/Hund energy; ordinary tetrahedral oxide complexes generally remain high-spin.

Both orbital degeneracies permit a Jahn-Teller effect, so a regular tetrahedron is not a stable static equilibrium in the elementary adiabatic treatment. The sign of a particular distortion requires an electron–lattice coupling model, not just the degeneracy count. To make that dependence explicit, let $Q>0$ stretch $z$ while contracting $x,y$. In a leading point-charge quadrupole model write the shiftsIn the high-spin configuration the two $e$ shifts cancel. For $Q>0$ the remaining electron selects $d_{xy}$, gaining $-2gQ$; for $Q<0$ it selects the $d_{xz},d_{yz}$ doublet, gaining $gQ$. Adding $K{Q}^2/2$ gives a tetragonally elongated minimum at $Q=2g/K$, with energy gain $2g^2/K$, versus only $g^2/(2K)$ along the compressed branch. The elongated branch has an orbital singlet. This is the usual elementary elongation argument; quantitative coupling includes higher crystal-field multipoles and configuration mixing.

For the strong-field $e^3$ configuration the lower $e$ orbital is doubly occupied. The shift of the three electrons is $-2g|Q|$, so the same model has both an elongated and a compressed minimum, $Q=\pm 2g/K$, with equal stabilization. Elongation favours double occupation of $d_{x^2-y^2}$; compression favours double occupation of $d_{z^2}$. This tetrahedral doublet distortion-sign ambiguity prevents a unique choice from the supplied data. For example, a small symmetry-allowed cubic elastic invariant $A{Q}_{\theta }(Q_{\theta }^2-3Q_{ϵ}^2)$ selects opposite tetragonal signs when $A$ changes sign, while leaving the regular-site electronic configuration unchanged. Higher couplings or host constraints are needed to choose. Dynamic motion between equivalent minima may also restore an average regular shape without stabilizing the undistorted adiabatic configuration.

Once a static axial distortion removes the orbital degeneracy, second-order spin-orbit coupling produces a time-reversal-even quadratic spin Hamiltonian operator. The residual tetragonal symmetry requires equal $x,y$ coefficients. Dropping a multiple of $S(S+1)$ leaves $H_{\mathrm{an}}=D{S}_z^2$ for the weak-field $S=3/2$ state. The sign and magnitude of $D$ depend on excited-state splittings. For the strong-field $S=1/2$ Kramers doublet, $S_z^2=1/4$ is constant: this term cannot cause zero-field splitting, although the Zeeman g-factor can be anisotropic.