Answer: Option A
:
A
First - let us write down the reaction:
$S{O}_2+2H_{2}S⟶3S+2H_{2}O$
In $H_{2}S$, sulphur has an oxidation number of-2. In $S{O}_2$ Sulphur has an oxidation number of +4.
On the right-hand side, we only have elemental sulphur whoseoxidation number is 0. The sulphur in $S{O}_2$ changes its oxidation number from +4 to 0.
This means that the sulphur is gaining electrons. We know that when a reactantgains electrons,it gets reduced. Hence, $S{O}_2$ is getting reduced.
Also, the reactant that gets reduced is the oxidising agent. Hence, here, $S{O}_2$ acts as an oxidising agent.