Question

How to count the oxidation number of NH₄⁺ N₂H₄ NH₂OH and N₂O?

Answer 1

1. The oxidation of nitrogen in NH4+ is -3. The hydrogen atoms have +1 and there are four of them. And net charge on the molecule is +1. Let nitrogen's oxidation state be x. Therefore, (4×(+1)) + x = +1. Hence, x is -3.
2. Hydrazine is N2H4. H has its usual oxidation state of +1 (since it is in a compound and not attached to a metal atom). The sum of oxidation states is the charge of the species, which is zero. So to calculate the oxidation state of N (call it n) we set 2n + 4 = 0, which gives n = -2.

3.We have: NH2OH

Usually, the oxidation state of O and H are −2 and +1, respectively.

This is not one of the exceptional cases, so these values can be used. We will let the oxidation state of N be x:

⇒x+2(+1)+(−2)+(+1)=0

⇒x+2−2+1=0

⇒x=−1

Therefore, the oxidation state of N inNH2OH is −1.

4.The oxidation number for oxygen (With the exception of peroxide compounds) is -2, and the sum of the oxidation numbers has to be equal 0 (In neutral compounds), so if nitrogen * 2 = 2, then nitrogen = 1