A) \[CO\] and \[C{{l}_{2}}\]
B) \[S{{O}_{2}}\] and \[C{{l}_{2}}\]
C) \[{{H}_{2}}\] and \[B{{r}_{2}}\]
D) \[N{{H}_{3}}\] and \[S{{O}_{2}}\]
Correct Answer: B
Solution :
\[S{{O}_{2}}\] bleaches flower by reduction \[2{{H}_{2}}O+S{{O}_{2}}\,\to \,{{H}_{2}}S{{O}_{4}}+2[H]\] \[2[H]+\underset{\text{flower}}{\mathop{\text{Coloured}}}\,\text{ }\xrightarrow{\text{Reduction}}\,\underset{\text{reduced flower}}{\mathop{\text{Colourless}}}\,\] This bleaching is temporary because reduced flower again oxidised by air to form coloured flower \[C{{l}_{2}}+{{H}_{2}}O\,\to \,2HCl+[O]\] \[[O]+\underset{\text{flower}}{\mathop{\text{Coloured}}}\,\xrightarrow{\text{Oxidation}}\,\underset{\text{Oxidised flower}}{\mathop{\text{Colourless}}}\,\text{ }\] This bleaching is permanent because oxidised flower remains colourless.
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