Identify the disproportionation reaction.

Options

A. `CH_4 + 2O_2 rarr CO_2 + 2H_2O`
B. `CH_4 +4Cl_2 rarr "CCl"_4 + 4HCl`
C. `2F_2 + 2OH^(-) rarr 2F^(-) +OF_(2) +H_2O`
D. `2NO_2 + 2OH^(-) rarr NO_(2)^(-) + NO_(3)^(-) +H_2O`

Answer

D

Explanation

Reactions in which the same substance is oxidised as well as reduced are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions.
(a) `overset(-4)Coverset(+1)H_4+2overset(0)O_2 rarroverset(+4)Coverset(-2)O_2+2overset(+1)H_2overset(-2)O`
(b) `overset(-4)Coverset(-1)H_4+4Coverset(0)l_2 rarroverset(+4-1)"CCl"_4+4overset(+1)Hoverset(-1)(Cl)`
(c) `2overset(0)F_2+2overset(-2)Ooverset(+1)Hrarr2overset(-1)F^(-)+overset(+2)Ooverset(-1)(F_(2)^(-))+overset(+1)H_(2)overset(-2)O`
(d) `2overset(+4)Noverset(-2)O_(2)+2overset(-2)Ooverset(+1)(H^(-) ) rarroverset(+3)Noverset(-2)(O_(2)^(-))+overset(+5)Noverset(-2)(O_(3)^(-))+overset(+1)(H_(2))overset(-2)O`
Thus, in reaction (d), N is both oxidised as well as reduced since the O.N. of N increases from +4 in `NO_2` to +5 in `NO_(3)^(-)` and decreases from +4 in `NO_2` to +3 in `NO_(2)^(-)`