In acidic solution, bismuthate oxidises manganese(II) according to 2Mn^2+ + 5BiO3- + 14H+ -> 2MnO4- + 5Bi^3+ + 7H2O. The number of moles of NaBiO3 required to oxidise 0.40 mol of Mn^2+ is

JEE Advanced levelSingle correct optionDifficulty 3 of 3pt203

In acidic solution, bismuthate oxidises manganese(II) according to $2 Mn X^{2 +} + 5 BiO X_{3} X^{-} + 14 H X^{+} 2 MnO X_{4} X^{-} + 5 Bi X^{3 +} + 7 H X_{2} O$ . The number of moles of $NaBiO X_{3}$ required to oxidise 0.40 mol of $Mn X^{2 +}$ is

Each

Mn X^{2 +}

gives up 5 electrons (to reach

+ 7

) and each bismuthate accepts 2 (Bi from

+ 5

+ 3

). The balanced ratio is

Mn X^{2 +} : BiO X_{3} X^{-} = 2 : 5

, so the moles needed are

0.40 \times 5/2 = 1.00

mol.