Question

The oxidation state of $N$ in $N{H}_{4}N{O}_3$ is

• A. +2
• B. +1
• C. -3
• D. +5

Answer 1

Answer: (C), (D)

+5

Oxidation number of N in $N{H}_{4}N{O}_3$

If we go by the usual method $N_2{H}_4{O}_3;2X+4\times (+1)+3\times (-2)=0$

$2X+4-3=0orX=+1$ (wrong)

No doubt $N_2{H}_4{O}_3$ has two Nitrogen atoms. One $N$ has negative oxidation number (attached to H) and the other has positive oxidation number (attached to $O$). Hence the evolution should be made separately for $N{H}_4^{+}$ and $N{O}_3^{-}$.

For the oxidation number of $N$ in ammonium ion, $N{H}_4^{+}$: $x+4\times (+1)=+1orx=-3$

For $N{O}_3^{-}$ $x+3\times (-2)=-1orx=+5$