# What is the oxidation numbers of (a) N in NH_4^+?

Mar 12, 2018

Well, as usual, the oxidation number of $H$ is $+ I$ as is typical....

#### Explanation:

And the sum of the individual oxidation numbers is equal to the charge on the ion...

And so ${N}_{\text{oxidation number}} + 4 \times {I}^{+} = + 1$

${N}_{\text{oxidation number}} = - I I I$

For a few more examples .... see this older answer.

Mar 12, 2018

$- 3$

#### Explanation:

We have the ammonium ion, $N {H}_{4}^{+}$.

As you can see from the formula, it has a $+ 1$ charge.

Since nitrogen is more electronegative than hydrogen, hydrogen will occupy a $+ 1$ charge. There are four hydrogen atoms in this ion, so the total charge of the hydrogens is $+ 1 \cdot 4 = + 4$.

Let $x$ be the oxidation number of $N$ in $N {H}_{4}^{+}$.

We got:

$x + \left(+ 4\right) = + 1$

Treating them as normal numbers, we get

$x + 4 = 1$

$x = 1 - 4$

$= - 3$

So, nitrogen would have a $- 3$ charge, which is also its usual charge.