Asserton :- `KClO_(3)rarrKClO_(4)+KCl`
This is a disporportionation type reaction.
Reason :- The reaction in which one substance oxidise or reduce is known as disproportionation reaction.

Explanation

Reaction in which the same substance is oixidised as well as reduced are called disproportionation reaction writing the O.N of each element above its symbols in the following reaction we have
(a) `C H_(4)+2O_(2)rarrCO_(2)+2H_(2)O (b) CH_(4)+4CI_(4)rarrC CI_(4) + 4HCI (c ) 2F_(2)+2OH^(-)rarr2F^(-)+OF_(2)+H_(2)O (d) 2 N O_(2)+2OH^(-)rarr NO_(2)^(-)+ANO_(3)^(-)+H_(2)O^(-2)`
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)^(-)` Thus option (d) is correct