Reduction of the metal centre in aqueous permanganate ion involves
A
3 electrons in neutral medium
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B
5 electrons in neutral medium
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C
3 electrons in alkaline medium
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D
5 electrons in acidic medium
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Solution
The correct option is D 5 electrons in acidic medium In acidic medium: MnO⊝4+5e⊝→Mn2+(′n′factor=5)
In strongly alkaline medium: MnO⊝4+e⊝→MnO2−4(′n′factor=1)
In weakly basic and neutral medium: MnO⊝4+3e⊝→MnO2(′n′factor=3)