Identify disproportionation reaction :

MCQ

A
$\text{CH}_4+2\text{O}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\text{CO}_2+2\text{H}_2\text{O}$
B
$\text{CH}_4+4\text{Cl}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\text{CCl}_4+4\text{HCl}$
C
$2\text{F}_2+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }2\text{F}^-+\text{OF}_2+\text{H}_2\text{O}$
✓
$2\text{NO}_2+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{NO}^-_2+\text{NO}^-_3+\text{H}_2\text{O}$

✓

Answer

Correct option: D.

$2\text{NO}_2+2\text{OH}^-\xrightarrow{ \ \ \ \ \ \ \ \ }\text{NO}^-_2+\text{NO}^-_3+\text{H}_2\text{O}$

Reactions in which the same substance is oxidized as well as reduced are called disproportionation reactions.
Writing the $O.N$. of each element above its symbol in the given reactions,

$\stackrel{-4+1}{\text{CH}}_4+\stackrel{0}{\text{2O}}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+4-2}{\text{CO}}_2+\stackrel{+1-2}{\text{2H}}_2\text{O}$
$\stackrel{-4+1}{\text{CH}}_4+\stackrel{0}{\text{4Cl}}_2\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+4-1}{\text{CCl}}_4+\stackrel{+1-1}{\text{4HCl}}\text{}$
$2\stackrel{0}{\text{F}}_2+\stackrel{-2+1}{\text{2OH}^-}\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{-1}{\text{2F}^-+}\stackrel{+2-1}{\text{OF}_2}+\stackrel{+1-2}{\text{H}_2\text{O}}$
$\stackrel{+4-2}{\text{2NO}_2}+\stackrel{-2+1}{\text{2OH}^-}\xrightarrow{ \ \ \ \ \ \ \ \ }\stackrel{+3-2}{\text{NO}^-_2}+\stackrel{+5-2}{\text{NO}^-_3}+\stackrel{+1-2}{\text{H}_2\text{O}}$

Thus, in reaction $(d), N$ is both oxidized as well as reduced since the $O . N $. of $N$ increases from $+4$ in $\mathrm{NO}_2$ to $+5$ in $ \mathrm{NO}^{-}3$ and decreases from $+4$ in $\mathrm{NO}_2$ to $+3$ in $\mathrm{NO}^{-} 2$.