A) \[{{H}_{2}}S\]
B) \[{{H}_{2}}S{{O}_{4}}\]
C) \[S{{O}_{2}}\]
D) \[{{H}_{2}}S{{O}_{3}}\]
Correct Answer: A
Solution :
Oxidation number of sulphur in \[{{H}_{2}}S\], \[{{H}_{2}}S{{O}_{4}}\,\,{{H}_{2}}S{{O}_{3}}\] and \[S{{O}_{2}}\] are \[-2,+6,+4\] a\[+4\]. Thus, \[{{H}_{2}}S\] can act only as reducing agent.
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