One gas bleaches the colour of flowers by reduction and another gas by oxidation. The gases respectively are 

Options

A. `SO_2 and Cl_2`
B. `CO and Cl_2`
C. `NH_3 and SO_2`
D. `H_2S and Br_2`

Answer

To solve the question regarding the gases that bleach the color of flowers by reduction and oxidation, we can analyze the properties and reactions of the given gases.

### Step-by-Step Solution:

1. **Understanding the Question**:
- We need to identify two gases: one that bleaches color by reduction and another that does so by oxidation.

2. **Identifying Gases**:
- The options provided are:
- SO₂ and Cl₂
- CO and Cl₂
- H₂S and Cl₂

3. **Analyzing SO₂ (Sulfur Dioxide)**:
- SO₂ is known to act as a reducing agent.
- When SO₂ interacts with colored substances, it can donate electrons, leading to a reduction reaction that removes color.
- For example, SO₂ can react with colored organic compounds, reducing them and thus causing a loss of color.

4. **Analyzing Cl₂ (Chlorine)**:
- Cl₂ is known to act as an oxidizing agent.
- When Cl₂ reacts with water, it produces nascent oxygen (O), which is a highly reactive form of oxygen.
- This nascent oxygen can oxidize colored substances, leading to a bleaching effect.
- The reaction can be summarized as:
\[ Cl_2 + H_2O \rightarrow HCl + O \]
- The nascent oxygen then reacts with the colored compounds, leading to their decolorization.

5. **Conclusion**:
- The gas that bleaches color by reduction is **SO₂** (sulfur dioxide).
- The gas that bleaches color by oxidation is **Cl₂** (chlorine).
- Therefore, the gases respectively are **SO₂ and Cl₂**.

### Final Answer:
The gases respectively are **SO₂ (sulfur dioxide)** and **Cl₂ (chlorine)**.