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A low-spin complex has the smaller-spin occupancy favored when ligand-field splitting outweighs the cost of pairing electrons in the lower orbitals. An octahedral low-spin $d^6$ center has $t_{2g}^6{e}_g^0$ and total spin zero.

A Co(III) centre has $d^6$, and charge balance requires one closed-shell catecholate ligand and one $S=1/2$ semiquinonate ligand. For low-spin Co(III), $t_{2g}^6(e_g^{\ast }{)}^0$, the metal has $S=0$, giving an independent-spin magnetic moment ofThus $A^1$ contains low-spin Co(III), $S_{\mathrm{Co}}=0$, with its observed moment residing on the free radical ligand. High-spin Co(III), $t_{2g}^4(e_g^{\ast }{)}^2$, would have $S=2$ and give $\sqrt{27}=5.20$ under the stated independent-spin approximation, so it is inconsistent with this measurement.

For Co(II), $d^7$, charge balance instead requires two $S=1/2$ semiquinonate ligands. High-spin Co(II) has $t_{2g}^5(e_g^{\ast }{)}^2$ and $S=3/2$, soTherefore $A^2$ contains high-spin Co(II), $S_{\mathrm{Co}}=3/2$. Low-spin Co(II), $t_{2g}^6(e_g^{\ast }{)}^1$, would give $\sqrt{9}=3.00$ when the two ligand spins are included. These four checks distinguish both oxidation state and metal spin.

The larger octahedral splitting at Co(III) exceeds the effective pairing energy and favours a low-spin complex; the smaller octahedral splitting at Co(II) favours a high-spin complex with occupation of both $e_g^{\ast }$ orbitals before further pairing. Interconversion also transfers an electron between cobalt and a dioxolene ligand, so this is valence tautomerism, rather than a metal-only spin change. Squared moments add here because the supplied model treats the centres as independent; adding the three spins first would be a different, exchange-coupled model.

cisplatin has square planar molecular geometry and a $5d^8$ metal centre. Choose $x$ and $y$ along the four Pt–donor directions. The strongly antibonding $d_{x^2-y^2}$ orbital is highest and remains empty; the other four d orbitals are doubly occupied. The square-planar splitting is represented by the conventional qualitative orderThe ordering among the lower occupied levels can depend on the ligand set, and the cis arrangement removes exact fourfold symmetry. The robust result is the filled lower four levels and empty highest level.

satraplatin has approximately octahedral molecular geometry around Pt(IV), $5d^6$. Its large octahedral splitting favours a low-spin complex:All three $t_{2g}$ orbitals are paired. Mixed ligands remove exact degeneracies, but do not change this qualitative occupancy.

iron remains Fe(II), $d^6$, in both states; the change is a spin crossover, rather than a change in oxidation state. At the higher temperature the high-spin complex has $t_{2g}^4{e}_g^2$: four unpaired electrons, $S=2$, andThis matches the observed moment. At low temperature a low-spin complex with $t_{2g}^6{e}_g^0$ has $S=0$ and shows diamagnetism. Cooling switches high-spin $d^6$ to low-spin $d^6$. Emptying the metal–ligand antibonding $e_g$ orbitals strengthens and shortens the Fe–N bonds, accounting for the contraction by roughly $0.2\text{A˚}$.

The octahedral splitting and electron pairing energy must be sufficiently close for both states to be accessible. The low-spin state gains electronic energy, while the high-spin state has greater spin degeneracy and usually greater vibrational entropy because of its softer bonds. The resulting Gibbs free energy balance favors high spin at higher temperature. The two endpoint observations do not by themselves specify hysteresis or transition cooperativity; those require a temperature series. Coupled structural and magnetic measurements distinguish this explanation from a mere bond-length anomaly.

A coordination complex with four carbon-bound cyanide ligands and a formal nickel(II), $d^8$, centre. Square planar molecular geometry permits a low-spin complex with no unpaired electrons.