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JEE Main levelSingle correct optionDifficulty 2 of 3ph022

$HI$ cannot be prepared in a pure form by heating $NaI$ with concentrated $H X_{2} SO X_{4}$ because

A $HI$ is a strong reducing agent and is oxidised to $I X_{2}$ by concentrated $H X_{2} SO X_{4}$
B $HI$ is so stable that it does not leave the mixture
C $NaI$ does not react with concentrated $H X_{2} SO X_{4}$
D $NaHSO X_{4}$ coats the crystals and stops the reaction

Answer: A

NaI + H X_{2} SO X_{4} NaHSO X_{4} + HI

occurs first, but

HI

then reduces the concentrated acid, giving

I X_{2}

and

SO X_{2}

(and some

H X_{2} S

). Hence non-oxidising

H X_{3} PO X_{4}

is used instead.

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