# What is the "oxidation number" of nitrogen in "ammonia, ammonium ion, and nitrate ion"?

Aug 2, 2017

We gots $\stackrel{- I I I}{N}$..........

#### Explanation:

While oxidation number is a formalism, it is a convenient formalism. For ions and molecules, the SUM of the oxidation numbers ALWAYS equals the charge on the ion.

TO illustrate, let's take $N {H}_{3}$, $N {H}_{4}^{+}$, and $N {O}_{3}^{-}$; now hydrogen and oxygen in their compounds typically take oxidation numbers of $+ I$ and $- I I$ respectively, and they do here.......

ANd thus for ammonia, $3 \times \left(+ I\right) + \stackrel{\text{oxidation number}}{N} = 0$. Clearly we gots $\stackrel{\text{-III}}{N}$.

And for ammonium, $4 \times \left(+ I\right) + \stackrel{\text{oxidation number}}{N} = + 1$. Clearly we gots $\stackrel{\text{-III}}{N}$ again.

But for nitrate ion, oxygen is MORE electronegative than nitrogen, and it gets the electrons, $3 \times \left(- I I\right) + \stackrel{\text{oxidation number}}{N} = - 1$. Clearly we gots $\stackrel{\text{+V}}{N}$.

Capisce?

What about $N O$, $N {O}_{2}$, ${N}_{2} {O}_{5}$, ${N}_{2} {O}_{4}$?