# What is the nitrogen oxidation state in ammonia?

Jul 31, 2017

Ain't we got $\stackrel{- I I I}{N}$?

#### Explanation:

We have $N {H}_{3}$, and the electrons in each $N - H$ are assigned to the most electronegative element, i.e. to nitrogen....and so we gots $N \left(- I I I\right)$ and $3 \times H \left(+ I\right)$; the sum of the oxidation numbers is equal to the charge on the resultant species, i.e. $0$.

Note that this makes sense when we speak of dinitrogen reduction, i.e. the reduction of zerovalent dinitrogen gas.......

$\frac{1}{2} {\stackrel{0}{\text{N"_2(g) +3/2stackrel(0)"H"_2(g) rarr stackrel(-III)" NH}}}_{3} \left(g\right)$

....which is about the most important inorganic reaction on the planet. Here hydrogen is oxidized to $\stackrel{+ I}{H}$ and nitrogen reduced to $\stackrel{- I I I}{N}$. Capisce?