# Oxidation number for nitrogen in ammonium sulfate [duplicate]

I'm trying to find the oxidation number of $\ce N$ in $\ce{(NH4)2SO4}$.

The answer is supposed to be $-3$ but I can't figure out the logic behind this.

Ammonium sulfate is an ionic compound, composed of two ammonium ions ($\ce{NH_{4}^{+}}$) and one sulfate ion ($\ce{SO_{4}^{2-}}$). As such, you can neglect the sulfate entirely from your consideration of the oxidation state of the nitrogen, as well as treating each ammonium ion as a separate entity. Given the total charge on the ammonium cation, it should be easy to see why the nitrogen has the specified oxidation state.

To easily determine the oxidation number on an individual atom in an ion it is necessary to write an equation.

First determine what the oxidation number of your known atoms are.

In this case, we know the oxidation number for H is +1.

Then set this value equal to the overall net charge of the ion.

In this case, it is +1.

Our equation now looks like this: 1(4) = 1, You use the multiplier of 4 to indicate that the ammonium ion has 4 hydrogen.

Next substitute a variable in the equation for the missing oxidation number:

+1(4) + N = +1

Solve for N.

+1(4) + N = +1

N + 4 = +1

N = +1 - 4

N = -3

Thus, the oxidation number for Nitrogen is -3.