# What is the oxidation number of nitrogen in the nitrate ion?

The oxidation number of nitrogen in nitrate anion is $+ V$.
Oxygen is more electronegative than nitrogen, hence such an operation will give ${O}^{2 -} \times 3 + O {N}_{N} = - 1$, where $O {N}_{N}$ is the oxidation number of nitrogen (and it is understood that the oxidation number of $O$ $=$ $- I I$). Clearly $O {N}_{N} = + V$. Of course, as you probably already know, this is a formalism; nitrogen really hasn't lost 5 electrons. We persist teaching these formalisms in order to be to able to balance oxidation/reduction reactions, where such formalisms are quite useful.
What are the oxidation states of nitrogen in ammonia, $N {H}_{3}$, $N {O}_{2}$, $N O$, and dinitrogen, $N \equiv N$? Please report your results in this thread.