Question

Which of the following has been arranged in order of increasing oxidation number of nitrogen?

• A. $N{H}_3<N_2{O}_5<NO<N_2$
• B. $N{O}_2^{+}<N{O}_3^{-}<N{O}_2^{-}<N_3^{-}$
• C. $N{H}_4^{+}<N_2{H}_4<N{H}_{2}OH<N_{2}O$
• D. $N{O}_2<Na{N}_3<N{H}_4^{+}<N_{2}O$

Answer 1

The correct option is C $N{H}_4^{+}<N_2{H}_4<N{H}_{2}OH<N_{2}O$

Oxidation state of $H$ $=+1$

Oxidation state of $O$ $=-2$

For option A oxidation state of $N$ in $N{H}_3$ is -3,in $N_2{O}_5$ is +5,in $NO$ is +2. So this is not increasing order of oxidation state of nitrogen.

For option B oxidation state of $N$ in $N{O}_2^{+}$ is +5,in $N{O}_3^{-}$ is +5. So this is not increasing order of oxidation state of nitrogen.

For option C oxidation state of $N$ in $N{H}_4^{+}$ is -3,in $N_2{H}_4$ is -2,in $N{H}_{2}OH$ is -1 and in $N_{2}O$ is +1. So this is correct answer as compounds are arranged in increasing order of oxidation number of nitrogen.

For option A oxidation state of $N$ in $N{O}_2$ is -4,in $Na{N}_3$ is -1/3,in $N{H}_4^{+}$ is -3. So this is not increasing order of oxidation state of nitrogen.