On treating a compound with warm dil. \(\mathrm{H_2SO_4}\), gas X is evolved which turns \(\mathrm{K_2Cr_2O_7}\) paper acidified with dil. \(\mathrm{H_2SO_4}\) to a green compound Y. X and Y respectively are :
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- \(X = \mathrm{SO_2},\ Y = \mathrm{Cr_2O_3}\)
- \(X = \mathrm{SO_3},\ Y = \mathrm{Cr_2(SO_4)_3}\)
- \(X = \mathrm{SO_3},\ Y = \mathrm{Cr_2O_3}\)
- \(X = \mathrm{SO_2},\ Y = \mathrm{Cr_2(SO_4)_3}\)
The Correct Option is D
Solution and Explanation
\(\mathrm{SO_2}\) is a reducing agent. With acidified \(\mathrm{K_2Cr_2O_7}\) paper: \(\mathrm{K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \to K_2SO_4 + Cr_2(SO_4)_3 + H_2O}\).
The orange dichromate is reduced to green chromium(III) sulphate, so Y = \(\mathrm{Cr_2(SO_4)_3}\) (in acidic medium with \(\mathrm{H_2SO_4}\), the sulphate is formed, not the oxide).
Answer: (4) \(X = \mathrm{SO_2},\ Y = \mathrm{Cr_2(SO_4)_3}\)
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