Name the gas that can readily decolourise acidified ${\text{KMnO}}_4$ solution:

${\text{SO}}_2$ is a strong reducing agent. It reduces the purple ${\text{MnO}}_4^{-}$ (where Mn is $+7$) to the colorless ${\text{Mn}}^{2+}$ ion, thereby decolourising the solution.
$$2{\text{MnO}}_4^{-}+5{\text{SO}}_2+2{\text{H}}_2\text{O}\to 2{\text{Mn}}^{2+}+5{\text{SO}}_4^{2-}+4{\text{H}}^{+}$$