# How do you find the oxidation number of nitrogen in these molecules?

## $H N {O}_{3}$ ${N}_{2} {H}_{4}$ $N {O}_{2}^{+}$ ${N}_{2}$ $N {O}_{2}^{-}$ $N {O}_{2}$ ${N}_{2} {O}_{4}$ ${N}_{2} O$ $N O$

Oct 27, 2016

$H N {O}_{3}$ $\text{Nitrogen +V}$

#### Explanation:

${N}_{2} {H}_{4} ,$ $\text{Nitrogen -II}$

$N {O}_{2}^{+} ,$ $\text{Nitrogen +V}$

${N}_{2} ,$ $\text{Nitrogen 0}$

$N {O}_{2}^{-} ,$ $\text{Nitrogen +III}$

$N {O}_{2} ,$ $\text{Nitrogen +IV}$

${N}_{2} {H}_{4} ,$ $\text{Nitrogen -II}$

${N}_{2} O ,$ $\text{Nitrogen +I}$, $N \equiv {N}^{+} - {O}^{-}$, the terminal nitrogen is zerovalent, and the central $N$ is $+ I I$, i.e. average $+ I$ state.

${N}_{2} {O}_{4} ,$ $\text{Nitrogen +IV}$

$N O ,$ $\text{Nitrogen +II}$

I think I have got these right. Oxygen is more electronegative than nitrogen so it gets the electric charge in determining oxidation state.