In which of the following compounds, an element exhibits two different oxidation states.
A. NH₂OH

B. NH₄NO₃

C. N₂H₄

D. N₃H

We find out the oxidation number for Nitrogen for all options by using the unknown oxidation state method.

(i) NH₂OH

→ x= -1

(ii) NH₄NO₃

In this compound there are two nitrogen atoms bonded to two different types of groups. So we consider one molecular group at a time. We first consider NH₄⁺ as an ion and find out oxidation number for the other N atom. The oxidation number of NH₄ is +1.

∴ → x= +5

Now we know that for NO₃ the oxidation number is -1

∴ → x= -3

Thus N has two oxidation states -3 and +5. This was possible because there were two kids of molecules available for nitrogen to bond.

As in case of option (iii) & (iv) nitrogen is bonded to only hydrogen so by same calculations as above the oxidation number for N₂H₄ is -2 and that for N₃H is -1/3. The fraction -1/3 shows that one valence electron is being shared between three N atoms in N₃H.