Reduction of the metal centre in aqueous permanganate ion involves

3 electrons in neutral medium

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5 electrons in neutral medium

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3 electrons in alkaline medium

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5 electrons in acidic medium

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Solution

The correct options are
A 3 electrons in neutral medium
D 5 electrons in acidic medium
In acidic medium:
$M n O_{4}^{⊝} + 5 e^{⊝} \to M n^{2 +} (^{'} n^{'} f a c t o r = 5)$
In strongly alkaline medium:
$M n O_{4}^{⊝} + e^{⊝} \to M n O_{4}^{2 -} (^{'} n^{'} f a c t o r = 1)$
In weakly basic and neutral medium:
$M n O_{4}^{⊝} + 3 e^{⊝} \to M n O_{2} (^{'} n^{'} f a c t o r = 3)$

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