P-Block Elements Ques 194

Question: In which of the following pairs does the first gas bleaches flowers by reduction while the second gas does so by oxidation

[Manipal MEE 1995]

Options:

A) $ CO $ and $ Cl_2 $

B) $ SO_2 $ and $ Cl_2 $

C) $ H_2 $ and $ Br_2 $

D) $ NH_3 $ and $ SO_2 $

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Answer:

Correct Answer: B

Solution:

$ SO_2 $ bleaches flower by reduction $ 2H_2O+SO_2\to H_2SO_4+2[H] $ $ 2[H]+\underset{flower}{\mathop{Coloured}}\text{ }\xrightarrow{Reduction}\underset{\text{reduced flower}}{\mathop{Colourless}} $

This bleaching is temporary because reduced flower again oxidised by air to form coloured flower $ Cl_2+H_2O\to 2HCl+[O] $ $ [O]+\underset{flower}{\mathop{Coloured}}\xrightarrow{Oxidation}\underset{\text{Oxidised flower}}{\mathop{Colourless}} $

This bleaching is permanent because oxidised flower remains colourless.



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