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

Explanation:

Nitrogen has a higher electronegativity than Hydrogen. This means that the electron density of the electrons that form the bond between the Nitrogen and the Hydrogen will be pulled closer to the Nitrogen than the Hydrogen.

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.