The n – factor of $Mn{O}_4^{-}$ in the following change   $3Mn{O}_4^{-}\to 2M{n}^{2+}+M{n}^{+6}$

n-Factor Calculation for MnO₄^-

In the reaction:
3 MnO₄^- → 2 Mn²⁺ + Mn⁶⁺

Oxidation State Changes:

• Mn in MnO₄^-: +7
• For 2 Mn converted to Mn²⁺: 2 × (7 – 2) = 10 electrons gained
• For 1 Mn converted to Mn⁶⁺: 1 × (7 – 6) = 1 electron gained
• Total electrons transferred for 3 Mn: 10 + 1 = 11
• n-Factor per MnO₄^- ion: 11 / 3

Final answer: n-factor = 11/3