A) \[C{{l}_{2}}\]
B) \[{{[Fe{{(CN)}_{6}}]}^{4-}}\]
C) \[N{{H}_{2}}OH\]
D) \[SO_{3}^{2-}\]
Correct Answer: A
Solution :
Hydrogen peroxide \[({{H}_{2}}{{O}_{2}})\]reduces \[C{{l}_{2}}\]into\[C{{l}^{-}}.\] \[{{H}_{2}}{{O}_{2}}+C{{l}_{2}}\xrightarrow{{}}2HCl+{{O}_{2}}\] While \[{{H}_{2}}{{O}_{2}}\]oxidises \[{{[Fe{{(CN)}_{6}}]}^{4-}},N{{H}_{2}}OH\] and \[SO_{3}^{2-}.\]
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