What is the oxidation state of N in NH3?

Oct 26, 2015

The oxidation state of $N$ in the ammonia molecule is $- I I I$.

Explanation:

Oxidation number is the charge left on the central atom when each of the bonding pairs are broken, with the charge going to the most electronegative atom. If I break the $N - H$ bonds I am left with ${N}^{3 -}$ and $3 \times {H}^{+}$.

What are the oxidation states of nitrogen in $N O$ and $N {O}_{2}$?

We could represent the oxidation of ammonia $N \left(- I I I\right)$, to nitrate, $N \left(V +\right)$, as an exercise...

$N {H}_{3} + 3 {H}_{2} O \rightarrow N {O}_{3}^{-} + 9 {H}^{+} + 8 {e}^{-}$

Note that we often speak of dinitrogen reduction to ammonia, the which represents a formal six electron reduction with respect to dinitrogen...

$\frac{1}{2} {N}_{2} \left(g\right) + 3 {H}^{+} + 3 {e}^{-} \rightarrow N {H}_{3} \left(g\right)$

Aug 11, 2018

The oxidation state of Nitrogen in$N {H}_{3}$ ammonia is -3
There are three sets of bonds between Nitrogen and Hydrogen in $N {H}_{3}$ The negative electron density in each of these three bonds will be pulled closer to the Nitrogen resulting in a negative three charge on the Nitrogen.