Question

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

Answer 1

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_"oxidation number"+4xxI^+=+1`

`N_"oxidation number"=-III`

For a few more examples .... see [this older answer.](https://socratic.org/questions/assign-oxidation-numbers#570916)

Answer 2

`-3`

Explanation:

We have the ammonium ion, `NH_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*4=+4`.

Let `x` be the oxidation number of `N` in `NH_4^+`.

We got:

`x+(+4)=+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.