= Solution
\b[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 \b[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>.
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