In acidic medium, sodium bismuthate converts manganese(II) into permanganate, itself being reduced to Bi^3+. How many moles of NaBiO3 are needed per mole of Mn^2+ oxidised?

JEE Advanced levelSingle correct optionDifficulty 3 of 3pn128

In acidic medium, sodium bismuthate converts manganese(II) into permanganate, itself being reduced to $Bi X^{3 +}$ . How many moles of $NaBiO X_{3}$ are needed per mole of $Mn X^{2 +}$ oxidised?

Mn(II) to Mn(VII) loses 5 e- per Mn; Bi(V) to Bi(III) gains 2 e- per Bi. Ratio NaBiO3 : Mn2+ = 5 : 2 = 2.5. The balanced equation is

2 Mn X^{2 +} + 5 NaBiO X_{3} + 14 H X^{+} 2 MnO X_{4} X^{-} + 5 Bi X^{3 +} + 5 Na X^{+} + 7 H X_{2} O