Question

$ {{H}_{2}}S $ acts only as a reducing agent while $ S{{O}_{2}}, $ can act both as a reducing and oxidizing agent because

• A. S in $ {{H}_{2}}S $ has -2 oxidation state
• B. S in $ S{{O}_{2}} $ has oxidation state $ +4 $
• C. hydrogen in $ {{H}_{2}}S $ is more +ve than oxygen in $ S{{O}_{2}} $
• D. oxygen is more -ve in $ S{{O}_{2}} $

Answer 1

Answer: (C)

: $ S{{O}_{2}}+2{{H}_{2}}S\to 2{{H}_{2}}O+S $ (oxidising nature of $ S{{O}_{2}} $ ) $ S{{O}_{2}}+2{{H}_{2}}O\to {{H}_{2}}S{{O}_{4}}+2H $ (reducing nature of $ S{{O}_{2}} $ )