Explanation:
In \(SO _2\), sulphur has a \(+4\) oxidation state. In \(H _2 SO _4\), Sulphur exhibits its highest oxidation state of \(+6\).
On the other hand in \(H _2 S\), sulphur exists in its lowest \(- 2\) oxidation state. Thus, we can surmise that \(SO _2\) could act as both an oxidising agent as well as a reducing agent. Nitrogen exhibits its highest oxidation state of \(+5\) in \(HNO _3\)