A) \[I{{O}_{3}}^{-}\],\[Mn{{O}_{2}}\]respectively
B) \[{{I}_{2}}\],\[M{{n}^{2+}}\]respectively
C) \[I{{O}_{3}}^{-},M{{n}^{2+}}\]respectively
D) \[{{I}_{2}}\],\[Mn{{O}_{2}}\]respectively
Correct Answer: D
Solution :
[d] In acidic medium, \[MnO_{4}^{-}\]reduces is \[M{{n}^{2+}}\]and oxidises \[{{I}^{-}}\]to \[{{I}_{2}},\] \[\therefore KI+MnO_{4}^{-}\xrightarrow{{{H}^{+}}}{{I}_{2}}+M{{n}^{2+}}\] In basic medium, \[MnO_{4}^{-}\] reduces to \[Mn{{O}_{2}}\]and produces \[IO_{3}^{-}\]. \[\therefore KI+MnO_{4}^{-}\xrightarrow{O{{H}^{-}}}Mn{{O}_{2}}+5S+8{{H}_{2}}O\]
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