JEE Main 26 Feb 2021 Shift 1 Solved Paper

Gas, X tums acidified dichromate (K2Cr2O7) green that means it is a reducing agent.
Cr (VI) from compound K2‌2Cr2O7 on reduction changes its colour from orange to green which is the colour of Cr (III) compound Y.
SO2 can show both reducing and oxidising properties whereas SO3 cannot show reducing property because of highest group number 16, oxidation state (+6) of sulphur in SO3.
So, X is SO2.
As, X reacts with K2Cr2O7 in dil. H2SO4 medium, the green coloured Cr (III) compound, Y must be Cr (III) sulphate or Cr2(SO4)3.
So, Y is Cr2(SO4)3.
The overall reactions can be shown as :

(Compound)

Water‌soluble‌SO3−‌salt

‌ Dil. ‌

───────▶

H2SO4

SO2

(X)

(g)

K2Cr2O7

───────▶

dil.H2SO4

Cr2

(SO4)3 (Y)

Green

(i) SO32−+H2SO4⟶‌

SO2

(X)

↑+H2O+SO42−
(ii)

K2 +6 Cr2 O7

(Orange)

+

+4

SO2

+H2SO4⟶

+3

Cr2

(SO4)3 (Y)

(Green)

+K2

+6

SO4

+H2O