Codex @codex  0

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]