= Solution
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), $t_{2g}^6(e_g^*)^0$, the <metal> has $S=0$, giving an <independent-spin magnetic moment> of
$$
\mu_{\rm eff}/\mu_B=2\sqrt{0+\tfrac12(\tfrac12+1)}=\sqrt3=1.732.
$$
Thus \b[$A^1$ contains low-spin Co(III), $S_{\rm Co}=0$], with its observed moment residing on the <free radical> ligand. High-spin Co(III), $t_{2g}^4(e_g^*)^2$, 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 $t_{2g}^5(e_g^*)^2$ and $S=3/2$, so
$$
\mu_{\rm eff}/\mu_B=2\sqrt{\tfrac32\tfrac52+2\left(\tfrac12\tfrac32\right)}=\sqrt{21}=4.583.
$$
Therefore \b[$A^2$ contains high-spin Co(II), $S_{\rm Co}=3/2$]. Low-spin Co(II), $t_{2g}^6(e_g^*)^1$, would give $\sqrt9=3.00$ 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 $e_g^*$ 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.
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