# Question #7cea3

Feb 9, 2016

$- 3$

#### Explanation:

The first thing to notice here is that you're dealing with an ion, which as you know is a chemical species that carries a net charge.

More specifically, the ammonium ion, ${\text{NH}}_{4}^{+}$, is known to carry a $1 +$ charge. This means that the oxidation numbers of all the atoms that make up the ion must add up to give $+ 1$.

Since one ammonium ion contains one nitrogen atom and four hydrogen atoms, you can say that

${\text{ON"_("N") + 4 xx "ON}}_{\textrm{H}} = 1$

Now, oxidation numbers are assigned by taking into account the difference in electronegativity between the atoms. More specifically, you assign oxidation numbers by distributing the bonding electrons to the more electronegative of the two atoms.

Nitrogen is much more electronegative than hydrogen, so it will take all the bonding electrons it shares with hydrogen.

Now, each hydrogen atom will have a $\textcolor{b l u e}{+ 1}$ oxidation state, since it just "lost" its electron to nitrogen. This means that you have

${\text{ON}}_{\textrm{N}} + 4 \times \left(\textcolor{b l u e}{+ 1}\right) = 1$

${\text{ON}}_{\textrm{N}} = 1 - 4 = - 3$

Now, does this make sense?

After nitrogen "takes" hydrogen's electrons, it will end up with four extra electrons, so its oxidation number should be equal to $\textcolor{b l u e}{- 4}$, right?

Here is where the net charge of the ion comes into play. The fact that the ammonium ion has a $1 +$ net charge means that nitrogen has to have a $\textcolor{b l u e}{- 3}$ oxidation number.