A) \[C{{l}_{2}}\]
B) \[KCl\]
C) \[HCl\]
D) \[MnC{{l}_{2}}\]
Correct Answer: D
Solution :
\[\overset{+7}{\mathop{KMn{{O}_{4}}}}\,+\overset{-1}{\mathop{HCl}}\,\xrightarrow{{}}\overset{0}{\mathop{C{{l}_{2}}}}\,+\overset{+2}{\mathop{MnC{{l}_{2}}}}\,+KCl+{{H}_{2}}O\] Oxidation number of \[Mn\] decreases, hence \[KMn{{O}_{4}}\]has been reduced to \[MnC{{l}_{2}}.\] Thus the reduced product is \[MnC{{l}_{2}}.\]You need to login to perform this action.
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