• # question_answer 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] 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 $S{{O}_{2}}$

$S{{O}_{2}}$ 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.