In which of the following compounds, an element exhibits two different oxidation states.
We find out the oxidation number for Nitrogen for all options by using the unknown oxidation state method.
→ x= -1
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.
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