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

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.