Answer:
In \[S{{O}_{2}}\] die oxidation state of
sulphur is - 2. It can only increase its oxidation state but cannot decrease it
Therefore it can act only as a reducing agent. In \[S{{O}_{2}}\]. The oxidation
state of sulphur is + 4. It is in a position to undergo a crease as well as an
increase in the oxidation state. Thus, \[S{{O}_{2}}\] can act both as an
oxidising agent and a reducing agent.
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