What is the oxidation number of nitrogen in $N{H}_{2}OH$?

Let the oxidation number of $N$ be $x$.

The oxidation number of H and O is +1 and -2 respectively.

$\therefore$ $x+2(+1)+(-2)+1=0$

$\therefore$ $x+2-2+1=0$

$\therefore$ $x=-1$

Therefore, the oxidation number of nitrogen in ${NH}_{2}OH$ is $-1$.