Identify the disproportionation reaction.

Options

A. `CH_(4)+2O_(2)toCO_(2)+2H_(2)O`
B. `CH_(4)+4Cl_(2)toC Cl_(4)+4HCl`
C. `2F_(2)+2OH^(-)to2F^(-)+OF_(2)+H_(2)O`
D. `2NO_(2)+2OH^(-)toNO_(2)^(-)+NO_(3)^(-)+H_(2)O`

Answer

D

Explanation

Reactions in which the same substance is oxidised as well as reduce are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions.
(a) `overset(-4)(C)overset(+1)(H_(4))+2overset(@)(O_(2))tooverset(+4)(C)overset(-2)(O_(2))+2overset(+1)(H_(2))overset(-2)(O)`
(b) `overset(-4)(C)overset(+1)(H_(4))+4Coverset(0)(l_(2))to overset(+4)(C)overset(-1)(C)l_(2)+4overset(+1)(H_(2))overset(-1)(Cl)`
( c) `2overset(0)(F_(2))+2overset(-2)(O)overset(+1)(H)to2overset(-1)(F^(-))+overset(+2)(O)overset(-1)(F_(2)^(-))+overset(+1)(H_(2))overset(-2)(O)`
(d) `2overset(+4)(N)overset(-2)(O_(2))+2overset(-2)(O)overset(+1)(H^(-))tooverset(+3)(N)overset(-2)(O_(2)^(-))+overset(+5)(N)overset(-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)^(-)`