Question

One gas bleaches the color of the flowers by reduction while the other by oxidation. The gases are:
A. ${\text{CO}}$ and ${\text{C}}{{\text{O}}_2}$
B. ${{\text{H}}_{\text{2}}}{\text{S}}$ and ${\text{B}}{{\text{r}}_{\text{2}}}$
C. ${\text{S}}{{\text{O}}_{\text{2}}}$ and ${\text{C}}{{\text{l}}_{\text{2}}}$
D. ${\text{N}}{{\text{H}}_{\text{3}}}$ and ${\text{S}}{{\text{O}}_{\text{3}}}$

Answer 1

The flowers contain water. Thus, the gases react with water causing oxidation or reduction and thus, bleaching the color of the flowers and making them colorless.

Complete step by step answer:
Step 1:
${\text{CO}}$ reacts with water as follows:
${\text{CO}} + {{\text{H}}_2}{\text{O}} \to {\text{C}}{{\text{O}}_2} + {{\text{H}}_2} \uparrow$
The ${\text{C}}{{\text{O}}_2}$ and ${{\text{H}}_2}$ gases formed in the reaction do not cause oxidation or reduction.
${\text{C}}{{\text{O}}_2}$ reacts with water as follows:
${\text{C}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}} \to {{\text{H}}_2}{\text{C}}{{\text{O}}_3} \rightleftharpoons {\text{HCO}}_3^ - + {{\text{H}}^ + }$
The carbonic acid formed in the reaction dissociates to bicarbonate and hydrogen ions. The hydrogen ion acts as a reducing agent. Thus, the hydrogen ion bleaches the colored substances making them colorless.
Thus, bleaching of color is possible only by reduction.
Step 2:
${{\text{H}}_2}{\text{S}}$ reacts with water as follows:
${{\text{H}}_2}{\text{S}} + 4{{\text{H}}_2}{\text{O}} \to {{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 4{{\text{H}}_2} \uparrow$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \updownarrow$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{HSO}}_4^ - + {{\text{H}}^ + }$
The sulfuric acid formed in the reaction dissociates to hydrogen sulfate and hydrogen ions. The hydrogen ion acts as a reducing agent. Thus, the hydrogen ion bleaches the colored substances making them colorless.
${\text{B}}{{\text{r}}_2}$ reacts with water as follows:
${\text{B}}{{\text{r}}_2} + {{\text{H}}_2}{\text{O}} \to {\text{HBr}} + {\text{HBrO}}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \updownarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\, \updownarrow$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{H}}^ + } + {\text{B}}{{\text{r}}^ - }\,\,{{\text{H}}^ + } + {\text{Br}}{{\text{O}}^ - }$
The hydrogen bromide and hypobromous acid formed in the reaction dissociate to give hydrogen ions. The hydrogen ion acts as a reducing agent. Thus, the hydrogen ion bleaches the colored substances making them colorless.
Thus, bleaching of color is possible only by reduction.
Step 3:
${\text{S}}{{\text{O}}_2}$ reacts with water as follows:
${\text{S}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}} \to {{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 2\left( {\text{H}} \right)$
The nascent hydrogen produced in the reaction acts as a reducing agent. Thus, the nascent hydrogen bleaches the colored substances making them colorless.
${\text{C}}{{\text{l}}_2}$ reacts with water as follows:
${\text{C}}{{\text{l}}_2} + {{\text{H}}_2}{\text{O}} \to 2{\text{HCl}} + \left( {\text{O}} \right)$
The nascent oxygen produced in the reaction acts as an oxidizing agent. Thus, the nascent oxygen bleaches the colored substances making them colorless.
Thus, bleaching of color is possible by reduction as well as oxidation.
Step 4:
${\text{N}}{{\text{H}}_3}$ reacts with water as follows:
${\text{N}}{{\text{H}}_3} + 2{{\text{H}}_2}{\text{O}} \to {\text{N}}{{\text{H}}_4}{\text{OH}} \rightleftharpoons {\text{NH}}_4^ + + {\text{O}}{{\text{H}}^ - }$
The ammonium hydroxide formed in the reaction dissociates to give ammonium and hydroxide ions. The ions formed do not cause oxidation or reduction.
${\text{S}}{{\text{O}}_3}$ reacts with water as follows:
${\text{S}}{{\text{O}}_3} + 2{{\text{H}}_2}{\text{O}} \to {{\text{H}}_2}{\text{S}}{{\text{O}}_4}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \updownarrow$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{HSO}}_4^ - + {{\text{H}}^ + }$
The sulfuric acid formed in the reaction dissociates to hydrogen sulfate and hydrogen ions. The hydrogen ion acts as a reducing agent. Thus, the hydrogen ion bleaches the colored substances making them colorless.
Thus, bleaching of color is possible only by reduction.
Thus, the ${\text{S}}{{\text{O}}_{\text{2}}}$ gas bleaches the color of the flowers by reduction and the ${\text{C}}{{\text{l}}_{\text{2}}}$ gas bleaches the color of the flower by oxidation.
Thus, the gases are ${\text{S}}{{\text{O}}_{\text{2}}}$ and ${\text{C}}{{\text{l}}_{\text{2}}}$.
So, the correct answer is “Option C”.

Note: The nascent hydrogen acts as a reducing agent. The nascent oxygen acts as an oxidizing agent. Thus, causing the color of flowers to bleach by oxidation as well as reduction and thus making the flowers colorless.