When `SO_2` is passed through an aqueous solution of `I_2` , it becomes colourless . This is due to

Options

A. bleaching reaction of `SO_2`
B. formation of `HIO_2`
C. combination of `SO_2 and I_2`
D. formation of `NHO_3`

Answer

To solve the question of why the aqueous solution of iodine (I₂) becomes colorless when sulfur dioxide (SO₂) is passed through it, we can break down the process into the following steps:

### Step-by-Step Solution:

1. **Understanding the Reactants**:
- We have sulfur dioxide (SO₂), which is a gas, and iodine (I₂), which is a diatomic molecule and typically appears as a brownish color in solution.

2. **Reaction with Water**:
- When SO₂ is passed through an aqueous solution of I₂, it interacts with water (H₂O) present in the solution. The presence of water is crucial as it facilitates the reaction.

3. **Formation of Acids**:
- The SO₂ can react with water to form sulfurous acid (H₂SO₃). However, in the presence of iodine, it can also lead to the formation of sulfuric acid (H₂SO₄) and hydroiodic acid (HI):
\[ SO_2 + I_2 + H_2O \rightarrow H_2SO_4 + 2 HI \]

4. **Color Change**:
- The iodine (I₂) in the solution is responsible for the brown color. When it reacts with SO₂, it forms hydroiodic acid (HI), which is colorless. The sulfuric acid (H₂SO₄) formed is also colorless. Therefore, the overall solution becomes colorless due to the consumption of the brown iodine and the formation of colorless products.

5. **Conclusion**:
- The solution becomes colorless as a result of the combination of SO₂ with I₂, leading to the formation of colorless compounds (H₂SO₄ and HI).

### Final Answer:
The solution becomes colorless due to the combination of SO₂ and I₂, resulting in the formation of colorless sulfuric acid (H₂SO₄) and hydroiodic acid (HI).

---