# How do you find the oxidation number of N in NO (nitrogen oxide)?

$N$ in $N O$ has a formal oxidation number of $+ I I$
Since oxygen is slightly more electronegative than nitrogen, oxygen formally gets the electrons, to give ${N}^{2 +}$ and ${O}^{2 -}$.
$N O$ necessarily is a free radical (why necessarily?), and the single electron is placed on the nitrogen.