Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

up vote 6 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.