In acidic solution perxenate is reduced to XeO3, while Mn^2+ is oxidised to MnO4-. The amount of Mn^2+ oxidised by 0.25 mol of XeO6^4- is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph148

In acidic solution perxenate is reduced to $XeO X_{3}$ , while $Mn X^{2 +}$ is oxidised to $MnO X_{4} X^{-}$ . The amount of $Mn X^{2 +}$ oxidised by $0.25 mol$ of $XeO X_{6} X^{4 -}$ is

Xe changes from +8 to +6, so each

XeO X_{6} X^{4 -}

gains 2 electrons; each

Mn X^{2 +}

(+2 to +7) loses 5 electrons. Electrons gained =

0.25 \times 2 = 0.50 mol

, so

Mn X^{2 +}

oxidised =

0.50/5 = 0.10 mol

(equation:

5 XeO X_{6} X^{4 -} + 2 Mn X^{2 +} 5 XeO X_{3} + 2 MnO X_{4} X^{-}

in acid).