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 $
Show Answer
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.
