What is the n factor of KMnO₄ in basic medium?

In a basic (or neutral) medium, the n-factor of KMnO₄ is usually 3 because permanganate ion (MnO₄⁻) is reduced from Mn(+7) to MnO₂ with Mn(+4), involving a 3-electron change per mole. Potassium permanganate (KMnO₄) is a strong oxidizing agent. In redox reactions, the n-factor tells you the number of electrons gained or lost per mole of the substance. 

For KMnO₄, this depends on the medium: in acidic medium it gets reduced to Mn²⁺, while in basic or neutral medium it generally gets reduced to manganese dioxide (MnO₂) as a brown solid. The change in oxidation number of manganese decides the n-factor.

Oxidation number changes

• In KMnO₄, manganese is at +7.
• In MnO₂, manganese is at +4.
• Change in oxidation number = 7 − 4 = 3 per Mn atom → n-factor = 3 in basic/neutral medium.

Common n-factors of KMnO₄ by medium

Why exams usually expect “3” for basic medium:
In school/college titrations done in neutral or lightly basic conditions, the main product is MnO₂. That pathway corresponds to a 3-electron change, so the accepted n-factor is 3. If the question specifically says “strongly alkaline” and mentions formation of manganate (green solution), then the n-factor would be 1. If the question says “acidic medium,” it is 5.

Quick problem tip

• If you are balancing a redox equation in basic medium with KMnO₄, write MnO₂ on the product side and use OH⁻/H₂O to balance O and H.
• For equivalents in titration: Equivalents of KMnO₄ = moles × n-factor. In basic medium, multiply moles by 3.

Bottom line: Unless the problem clearly says “strongly alkaline giving manganate,” use n-factor = 3 for KMnO₄ in basic/neutral medium.