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The more radially extended Pt $5d$ orbitals give strong metal–ligand overlap and covalent bonding. This helps compensate the energetic cost of reaching Pt(IV), and the large octahedral ligand field splitting strongly stabilizes its low-spin $d^6$ configuration, with a crystal field stabilization energy contribution $-2.4{\Delta }_o$ before separately counting pairing and the full bond energies. Ni $3d$ bonding is less effective in this respect; Pd $4d$ compounds also generally favour the very stable $d^8$ M(II) arrangement more strongly than Pt does. The stronger bonding and large ligand-field stabilization available to Pt(IV) make oxidation to $+4$ more favourable relative to its congeners. This is a comparative trend, not a claim that Ni(IV) or Pd(IV) complexes cannot exist.

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.

In the ideal tetrahedral point-charge model, the five metal d orbitals split into a lower twofold $e$ set and an upper threefold $t_2$ set. Relative to their weighted barycentre their energies are $-3{\Delta }_t/5$ and $+2{\Delta }_t/5$, where ${\Delta }_t>0$ is the separation. For occupancies $e^{n_e}{t}_2^{n_t}$, the crystal field stabilization energy is $(-0.6n_e+0.4n_t){\Delta }_t$. Under weak-field high-spin filling it is $-0.6{\Delta }_t$, $-1.2{\Delta }_t$, $-0.8{\Delta }_t$ and zero for $d^6,d^7,d^8,d^{10}$ respectively. Shorter metal–ligand distances generally increase the splitting, so orbital filling can superimpose nonmonotonic bond-length changes on an ionic-size trend. The model alone does not determine an absolute bond length.