Past exam of the chemistry B course of the University of Cambridge / 2023 / Paper 2 / B8 / c

A neutral, simple ${\mu }_2$-NO contributes three electrons to the two-metal unit overall. In neutral bookkeeping it is an LX-type bridge: a two-electron donor component plus the odd electron of neutral NO. For a symmetric bridge, this can be apportioned as $1.5$ electrons to each metal in a covalent count; it is not three electrons independently to each metal.

An equivalent ionic description represents the bridge as NO${}^{-}$ donating two electrons to each metal, four overall. The extra negative ligand charge must then be included when assigning metal oxidation states. Mixing the four-electron ionic donation with a neutral-NO charge assignment would double-count an electron.

The terminal nitrosyls are linear in both proposed electron-precise structures; the formal M–M bond orders are V–V double and Fe–Fe single. Use three-electron terminal linear NO donors, three-electron ${\mu }_2$-NO bridges, and five-electron neutral Cp donors in covalent counting.

For the vanadium dimer the two metal fragments together containA metal–metal bond of order $b$ adds $b$ to each local count, so $32+2b=36$ gives $b=2$ and 18 electrons per V. For an ionic cross-check, Cp is $-1$, each linear terminal NO is conventionally NO${}^{+}$, and each bridge is represented as NO${}^{-}$. ThusEach V then has $4+6_{\mathrm{Cp}}+2_{\mathrm{terminal}\mathrm{NO}}+4_{\mathrm{two}\mathrm{bridges}}+2_{\mathrm{V}=\mathrm{V}}=18$ electrons.

The iron species is monoanionic, and ${\mathrm{L}}^{-}$ is a pyrazolate bridge binding through its two nitrogens. This is a four-electron donor overall in ionic counting, or three overall in neutral counting. The carbonyl bridge supplies two electrons overall. The covalent fragment total is consequentlyTherefore $34+2b=36$ gives $b=1$. In the formal NO${}^{+}$ description,Each Fe receives $10+4_{\mathrm{two}\mathrm{NO}}+2_{\mathrm{pyrazolate}\mathrm{N}}+1_{\mu \mathrm{CO}}+1_{\mathrm{Fe}-\mathrm{Fe}}=18$ electrons. The negative formal oxidation state follows from the chosen nitrosyl assignment and the overall charge; omitting that charge would change the count.

Bent terminal NO ligands would supply fewer neutral-count electrons and demand much higher formal M–M bond orders to attain 18-electron counts. The linear assignments give the ordinary double and single bond solutions. Since NO is a non-innocent ligand, these oxidation states are formal bookkeeping limits, not measured metal charges; actual nitrosyl bending and M–M bond strengths require structural and spectroscopic evidence.^[ref]