Which of the following is a disproportion reaction?

Options

A. `CH_(4) + 2O_(2) to CO_(2) + 2H_(2)O`
B. `CH_(4) + 4Cl_(2) to C Cl_(4) +4HCl`
C. `2F_(2) + 2OH^(-) to 2F^(-) + OF_(2) + H_(2)O`
D. `2NO_(2) + 2OH^(-) to NO_(2)^(-) + NO_(3)^(-) + H_(2)O`

Answer

D

Explanation

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