Past exam of the chemistry B course of the University of Cambridge / 2022 / Paper 2 / B6 / b

Neutral pyridine contributes no formal charge. If $n$ of the two dioxolene ligands are dianions, the combined ligand charge is $-2-n$, so the cobalt oxidation state is $2+n$. The distinct charge assignments are thereforeInterchanging the two ligand labels in the middle row does not make another charge combination. Co(IV) belongs in this unrestricted charge-balance list even though the subsequent magnetic analysis excludes it.

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.

water lies below pyridine and cyanide ligand in the spectrochemical series. At fixed metal and dioxolene charges, replacement by water decreases ${\Delta }_o$ and favours the high-spin limits: Co(II) remains $S=3/2$; Co(III) can become $S=2$ if the splitting falls below the pairing threshold. Replacement by cyanide increases ${\Delta }_o$ and favours Co(III), $S=0$, and Co(II), $S=1/2$. The two unchanged semiquinonate ligands would then give the Co(II) form a combined independent-spin moment of $3{\mu }_B$. The ranking predicts the direction of the change; two substituted donors alone do not guarantee a Co(III) spin crossover, since all six donors determine its splitting. A shift of the valence-tautomer equilibrium is also possible and is separate from this fixed-charge comparison.

Treat each semiquinonate ligand as a symmetric O,O chelate, with the free radical delocalized over the dioxolene framework. The ideal octahedral donor-connectivity structures are one trans form and two mirror-related cis forms, giving three stereoisomers. Trans refers to the two pyridine N donors, not to the two O atoms of a chelate, which must remain adjacent.

In the diagram, coloured curves join the two O donors belonging to one ligand. For an unambiguous coordinate specification, the trans form has Py at $+z,-z$, with chelate pairs $(+x,+y)$ and $(-x,-y)$. A cis pair has Py at $+z,+x$, with chelates $(-z,+y)$ and $(-x,-y)$; reflecting $y\mapsto -y$ gives the other cis form. Proper rotations cannot superpose those two cis arrangements. Ring conformations and pyridine rotations are not extra donor-connectivity isomers.

The Co(III) centre in $A^1$ has $t_{2g}^6(e_g^{\ast }{)}^0$: every occupied d subset is orbitally balanced. There is no first-order metal-centred Jahn–Teller driving force in $A^1$. This does not require all six experimental bond lengths to be equal, since pyridine and dioxolene are different donors.

The Co(II) centre in $A^2$ has $t_{2g}^5(e_g^{\ast }{)}^2$. Its two $e_g^{\ast }$ orbitals each contain one electron, so that subset gives no first-order preference. The $t_{2g}$ hole does give a weaker Jahn–Teller instability. To compare tetragonal distortions, take a positive small splitting parameter $\delta$ and retain each subset's orbital energy barycentre. On axial elongation let $d_{xz},d_{yz}$ shift by $-\delta$ and $d_{xy}$ by $+2\delta$. Five electrons then giveFor equal-magnitude compression the shifts reverse, and filling the lower $d_{xy}$ orbital first givesThe two singly occupied $e_g^{\ast }$ orbitals contribute zero net shift in either case. Thus the elementary tetragonal splitting model favours elongation for high-spin $d^7$, rather than equal electronic stabilization by the two signs. In a strictly sigma-only octahedral molecular orbital model $t_{2g}$ is nonbonding, so this weak effect is absent; it arises when the electrostatic and non-sigma interactions neglected in that diagram are restored. spin-orbit coupling, unequal donors and chelate constraints can modify the actual distortion, so the argument does not establish which experimental Co–donor bonds must lengthen.