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\[S{{O}_{2}}\]
Correct Answer: C
Solution :
: \[S{{O}_{2}}+2{{H}_{2}}S\to 2{{H}_{2}}O+S\] (oxidising nature of\[S{{O}_{2}}\]) \[S{{O}_{2}}+2{{H}_{2}}O\to {{H}_{2}}S{{O}_{4}}+2H\] (reducing nature of\[S{{O}_{2}}\])
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