Question

Which of the following shows nitrogen with its increasing order of oxidation number?

• A. $NO < N _{2} O < NO _{2}< NO _{3}^{-}, NH _{4}^{+}$
• B. $NH _{4}^{+}< N _{2} O < NO _{2}< NO _{3}^{-}< NO$
• C. $NH _{4}^{+}< N _{2} O < NO < NO _{2}< NO _{3}^{-}$
• D. $NH _{4}^{+}< NO < N _{2} O < NO _{2}< NO _{3}^{-}$

Answer 1

Answer: (C)

$NH _{4}^{+}< N _{2} O < NO < NO _{2}< NO _{3}^{-}$

Answer 2

Oxidation state of $N$ in $NO$ is $+2$, in $N _{2} O$ is
$+1$, in $NO _{2}$ is $+4$, in $NO _{3}^{-}$is $+5$ and in $NH _{4}^{+}$is $-3$.
Nitrogen with increasing number of oxidation number is:
$NH _{4}^{+}< N _{2} O < NO < NO _{2}< NO _{3}^{-}$