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

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Answer 1 · 2016-06-27T20:07:20

$N$ in $NO$ has a formal oxidation number of $+II$

Since oxygen is slightly more electronegative than nitrogen, oxygen formally gets the electrons, to give $N^(2+)$ and $O^(2-)$ .

$NO$ necessarily is a free radical (why necessarily?), and the single electron is placed on the nitrogen.

How do you find the oxidation number of NN in NONO (nitrogen oxide)?