1.

Identify the disproportionation reaction.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» Correct Answer - d
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)^(-)`


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