One gas bleaches the color of the flowers by reduction while the other by oxidation. The gases are

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CO and CO₂

H₂S and Br₂

SO₂ and Cl₂

NH₃ and SO₃

answer is C.

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Detailed Solution

Sulphur dioxide acts as a strong reducing nature and fades the color of flower through reduction process. In the presence of moisture, it forms H₂SO₄ and nascent H.

${SO}_{2} + 2 H_{2} O \to H_{2} {SO}_{4} + 2 H$

Nascent hydrogen is highly reactive which changed the colored substance in flower into colorless substance.

$Colored substances + 2 H \to Colorless substances + H_{2}$

Chlorine gas is a strong oxidizing agent and fades the color of the flower through oxidation process. In the presence of moisture, it forms HCl and nascent oxygen.

${Cl}_{2} + H_{2} O \to 2 HCl + [O]$

Nascent oxygen is highly reactive and changed the colored substance in flower into colorless substance.

$Colored substance + [O] \to Colorless substance$

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