JEE Main · 2021 · Shift-ImediumPB12-088

On treating a compound with warm dil. H2SO4, gas X is evolved which turns K2Cr2O7 paper acidified with dil. H2SO4 to a…

p-Block Elements (Class 12) · Class 12 · JEE Main Previous Year Question

Question

On treating a compound with warm dil. H2SO4\mathrm{H_2SO_4}, gas X is evolved which turns K2Cr2O7\mathrm{K_2Cr_2O_7} paper acidified with dil. H2SO4\mathrm{H_2SO_4} to a green compound Y. X and Y respectively are:

Options
  1. a

    X=SO2, Y=Cr2O3\mathrm{X = SO_2,\ Y = Cr_2O_3}

  2. b

    X=SO2, Y=Cr2(SO4)3\mathrm{X = SO_2,\ Y = Cr_2(SO_4)_3}

  3. c

    X=SO3, Y=Cr2(SO4)3\mathrm{X = SO_3,\ Y = Cr_2(SO_4)_3}

    ✓
  4. d

    X=SO3, Y=Cr2O3\mathrm{X = SO_3,\ Y = Cr_2O_3}

Correct Answerc

X=SO3, Y=Cr2(SO4)3\mathrm{X = SO_3,\ Y = Cr_2(SO_4)_3}

Detailed Solution

Strategy: Identify the gas evolved from sulphite reduction and its reaction with dichromate.

Step 1: Evolution of X Treating a sulphite or similar compound with dilute acid yields Sulphur dioxide gas (\ceSO2)(\ce{SO2}). \ceSO32−+2H+→SO2(g)+H2O\ce{SO3^{2-} + 2H^+ \to SO_2(g) + H2O} Thus, X is \ceSO2\ce{SO2}.

Step 2: Reaction with Dichromate \ceSO2\ce{SO2} reduces acidified Potassium dichromate (\ceK2Cr2O7(\ce{K2Cr2O7}, orange)) to Chromium(III) sulphate (\ceCr2(SO4)3(\ce{Cr2(SO4)_3}, green)). \ceK2Cr2O7+3SO2+H2SO4→K2SO4+Cr2(SO4)3+H2O\ce{K2Cr2O7 + 3SO2 + H2SO4 \to K2SO4 + Cr2(SO4)_3 + H2O} The green compound Y is \ceCr2(SO4)3\ce{Cr2(SO4)3}.

Answer: (Option) [b] \ceX=SO2,Y=Cr2(SO4)3\boxed{\text{Answer: (Option) [b] $\ce{X = SO2, Y = Cr2(SO4)_3}$}}

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