On treating a compound with warm dil. $\mathrm{H}_{2} \mathrm{SO}_{4}$, gas X is evolved which turns $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ paper acidified with dil. $\mathrm{H}_{2} \mathrm{SO}_{4}$ to a green compound Y. X and Y respectively are:

(A) $\mathrm{X}=\mathrm{SO}_{2}, \mathrm{Y}=\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$

(B) $\mathrm{X}=\mathrm{SO}_{2}, \mathrm{Y}=\mathrm{Cr}_{2} \mathrm{O}_{3}$

(C) $\mathrm{X}=\mathrm{SO}_{3}, \mathrm{Y}=\mathrm{Cr}_{2} \mathrm{O}_{3}$

(D) $\mathrm{X}=\mathrm{SO}_{3}, \mathrm{Y}=\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · p-Block Elements

On treating a compound with warm dil.

H_{2} SO_{4}

, gas X is evolved which turns

K_{2} Cr_{2} O_{7}

paper acidified with dil.

H_{2} SO_{4}

to a green compound Y. X and Y respectively are:

Answer: (A)

Warm dilute

H_{2} S O_{4}

decomposes a sulphite and the gas released is sulphur dioxide:

S O_{3}^{2 -} + 2 H^{+} \to S O_{2} + H_{2} O

Sulphur dioxide is a reducing agent. On paper moistened with dichromate acidified by dilute

H_{2} S O_{4}

it brings chromium down from the

+ 6

state of the orange

C r_{2} O_{7}^{2 -}

ion to the

+ 3

state, and in a sulphate medium the chromium(III) species formed is green chromium(III) sulphate:

K_{2} C r_{2} O_{7} + H_{2} S O_{4} + 3 S O_{2} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O

So X is

S O_{2}

and Y is

C r_{2} (S O_{4})_{3}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option