 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), $t_{2g}^6(e_g^{\ast }{)}^0$, 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.