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JEE Main levelSingle correct optionDifficulty 2 of 3qc078

A pale-pink solution is warmed with $NaBiO X_{3}$ and dilute $HNO X_{3}$ . The colour observed and the reason why $HNO X_{3}$ rather than $HCl$ is used are

Apurple $MnO X_{4} X^{-}$ ; chloride would be oxidised by $MnO X_{4} X^{-}$ / $NaBiO X_{3}$
Bgreen $MnO X_{4} X^{2 -}$ ; chloride would precipitate $Mn X^{2 +}$
Cpurple $MnO X_{4} X^{-}$ ; $HNO X_{3}$ is a stronger reducing agent than $HCl$
Dbrown $MnO X_{2}$ ; $HCl$ would dissolve $NaBiO X_{3}$ completely

Answer: A

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

gives purple permanganate. Chloride is oxidised by the strongly oxidising medium and by

MnO X_{4} X^{-}

, which would consume the product, so a non-reducing acid,

HNO X_{3}

, is used.

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