question_answer

In the reaction, \[2KMn{{O}_{4}}+16HCl\to 5C{{l}_{2}}+2MnC{{l}_{2}}+2KCl+8{{H}_{2}}O\] the reduction product is [Kerala (Med.) 2003]

A) \[C{{l}_{2}}\]

B) \[MnC{{l}_{2}}\]

C) \[{{H}_{2}}O\]

D) \[KCl\]