When a colourless gas is passed through bromine water only decolourisation take place. The gas is

Bromine water is an oxidising agent. It oxidises $\mathrm{SO}_2$ to $\mathrm{SO}_3$ and itself get reduced to colourless HBr .
$\begin{aligned}<br/>& \underset{\text { Orange red }}{\mathrm{Br}_2}+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Colourless }}{2 \mathrm{HBr}}+[\mathrm{O}] \\<br/>& \underset{\text { Colourless }}{\mathrm{SO}_2}+[\mathrm{O}] \longrightarrow \mathrm{SO}_3 \\<br/>& \mathrm{H}_2 \mathrm{O}+\mathrm{SO}_3 \longrightarrow \mathrm{H}_2 \mathrm{SO}_4<br/>\end{aligned}$
$\mathrm{H}_2 \mathrm{~S}$ is also oxidised by bromine water but in this case, decolourisation is also accompanied by formation of sulphur.