The colour of the iodine solution is discharged by shaking with

Options

A. sodium sulphate
B. sodium sulphide
C. aqueous sulphur dioxide
D. sodium bromide

Explanation

To determine which reagent can discharge the color of an iodine solution, we can analyze the reactions that occur when iodine (I2) is mixed with the given options: sodium sulfate (Na2SO4), sodium sulfide (Na2S), aqueous sulfur dioxide (SO2), and sodium bromide (NaBr).

### Step-by-Step Solution:

1. **Understanding Iodine's Color**: Iodine in solution is typically brown due to the presence of I2 molecules. The discharge of this color indicates a reduction reaction where iodine is converted to a colorless species.

2. **Analyzing the Reagents**:
- **Sodium Sulfate (Na2SO4)**: This compound is a neutral salt and does not participate in redox reactions with iodine. Thus, it will not discharge the color of iodine.
- **Sodium Sulfide (Na2S)**: Sodium sulfide can react with iodine, but it primarily forms sodium iodide (NaI) and sulfur. However, the reaction is not as straightforward as with other reducing agents.
- **Aqueous Sulfur Dioxide (SO2)**: Sulfur dioxide is a well-known reducing agent. When it reacts with iodine, it reduces I2 to HI (hydroiodic acid), effectively discharging the brown color of iodine.
- **Sodium Bromide (NaBr)**: Sodium bromide contains bromide ions, which can react with iodine, but this reaction does not lead to the discharge of color. Instead, it may result in the formation of iodine monochloride (ICl) or other products without decolorization.

3. **Identifying the Correct Reagent**: Among the options, aqueous sulfur dioxide (SO2) is the most effective at discharging the color of iodine due to its ability to reduce iodine to hydroiodic acid.

4. **Conclusion**: The color of the iodine solution is discharged by shaking with **aqueous sulfur dioxide (SO2)**.

### Summary of the Reaction:
The reaction can be summarized as follows:
\[ \text{SO}_2 + \text{I}_2 + 2 \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 + 2 \text{HI} \]
In this reaction, iodine (I2) is reduced to hydroiodic acid (HI), resulting in the discharge of the brown color.