question_answer

\[{{H}_{2}}S,\]acts only as reducing agent while \[S{{O}_{2}}\] can act both as a reducing agent and as an oxidizing agent. How will you account for it?

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.