# What are the nitrogen oxidation numbers in (i) "ammonia", and (ii) in "ammonium cation"?

Jul 12, 2017

The oxidation number of nitrogen in ammonia, $N {H}_{3}$, and ammonium cation, $N {H}_{4}^{+}$ are $- I I I$.......

#### Explanation:

See this old answer for some background; and here, and elsewhere.

Oxidation number is formally defined by the charge left on the metal centre (or the central atom), when all the bonding pairs of electrons are conceptually broken; with the charge, the electrons, assigned to the MOST electronegative atom.

If we gots $N {H}_{3}$, then if we $\text{break}$ the $N - H$ bonds we gets ${N}^{3 -} + 3 \times {H}^{+}$, i.e. clearly oxidation states of $N \left(- I I I\right)$ and $H \left(+ I\right)$ given that nitrogen is much more electronegative than hydrogen. And in fact to make the ammonia, we perform so-called $\text{dinitrogen reduction}$.......

${\stackrel{0}{N}}_{2} \left(g\right) + 3 {\stackrel{0}{H}}_{2} \left(g\right) \rightarrow 2 N {H}_{3} \left(g\right)$,

.......where the nitrogen atoms have CLEARLY undergone $\text{formal 3 electron reduction.}$

And in ammonium ion, the oxidation state of the nitrogen is necessarily the same. The sum of the oxidation numbers, $- I I I + 4 \times + I = + I$, the charge on the ion as required.........