Assertion: Conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ reacts with $\mathrm{KCl}$ to give ${\mathrm{Cl}}_2$ gas.

Reason : $\mathrm{HCl}$ cannot be oxidised by conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$

The reaction involved is:

$2\mathrm{HX}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{SO}}_2+{\mathrm{X}}_2+2{\mathrm{H}}_2\mathrm{O}\text{ for }\mathrm{X}=\mathrm{I}/\mathrm{Br}$

here I and Br are very strong oxidizing agents whereas $Cl$ is a milder reducing one hence it cannot be converted to free chlorine ions so it cannot be oxidized by sulphuric acid.

A similar goes with KCl and KI i.e KI can give iodine but KCl cannot.