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A high-spin complex retains more unpaired electrons when the energy required for pairing exceeds the relevant orbital splitting. An octahedral high-spin $d^6$ center has $t_{2g}^4{e}_g^2$ and $S=2$.

For $n$ unpaired electrons, the spin-only magnetic moment iswhere ${\mu }_{\mathrm{B}}$ is the Bohr magneton and the second expression uses $g\simeq 2$.

For the square, a large separation of the highest $B_{1g}$ level favours pairing in the lower four molecular orbitals: $S=0$, $n=0$, ${\mu }_{\mathrm{eff}}=0$. If the cost of occupying $B_{1g}$ were smaller than the relevant electron pairing energy, a high-spin complex with two unpaired electrons and $S=1$ could instead occur, giving $\sqrt{8}{\mu }_{\mathrm{B}}\simeq 2.83{\mu }_{\mathrm{B}}$.

For ideal tetrahedral $d^8$, the occupation is $e^4{t}_2^4$. Hund's rules distribute four electrons over the three degenerate $t_2$ molecular orbitals as one pair and two single occupations. Thus $S=1$ and ${\mu }_{\mathrm{eff}}\simeq 2.83{\mu }_{\mathrm{B}}$. Merely increasing the $e$–$t_2$ splitting does not create a conventional paired tetrahedral $d^8$ alternative: four electrons must still occupy $t_2$.

Charge balance fixes each metal as ${\mathrm{M}}^{\mathrm{II}}$. In an octahedral crystal field the lower $t_{2g}$ set is $(d_{xy},d_{xz},d_{yz})$ and the upper $e_g$ set is $(d_{z^2},d_{x^2-y^2})$. The measured spin-only magnetic moments select the following occupations:The result follows from $S=n_{\mathrm{unpaired}}/2$ and $2\sqrt{S(S+1)}=\sqrt{n_{\mathrm{unpaired}}(n_{\mathrm{unpaired}}+2)}$. Manganese(II) and iron(II) are high-spin complexes: their octahedral splitting is too small to compensate the additional pairing cost of their low-spin alternatives. Copper(II) is $d^9$, with one unpaired electron regardless of which member of the upper pair is singly occupied. spin-orbit coupling and exchange interactions can produce deviations from the spin-only estimate in real measurements; no such correction is required to explain the supplied values.

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.

The metal is Cr(II), $d^4$. A chloride ligand produces a relatively weak ligand field, so the octahedral center is a high-spin complex with $t_{2g}^3{e}_g^1$, $S=2$. The singly occupied $e_g$ pair is orbitally degenerate: $d_{z^2}$ and $d_{x^2-y^2}$ cannot retain identical occupancies once the distortion selects one. This is a strong setting for the Jahn-Teller effect.

On tetragonal elongation, the two axial Cr–Cl distances increase. Repulsion is reduced most for $d_{z^2}$, so it falls relative to $d_{x^2-y^2}$; $d_{xz},d_{yz}$ also fall relative to $d_{xy}$. The single $e_g$ electron occupies the lower $d_{z^2}$-derived level. The electronic stabilization offsets an elastic cost until a distorted equilibrium geometry is reached. The high-spin $d^4$ occupancy drives an elongated, rather than regular, octahedron. The three lower-set electrons remain one per orbital in this simple diagram; it is primarily the unequal $e_g$ occupancy that produces the strong distortion.

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.