The reaction of `Na_2 S_2 O_3` with iodine gives

Options

A. Sodium sulphide
B. Sodium sulphite
C. Sodium sulphate
D. Sodium tetrathionate

Explanation

To solve the question regarding the reaction of sodium thiosulfate (Na₂S₂O₃) with iodine (I₂), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are sodium thiosulfate (Na₂S₂O₃) and iodine (I₂).

2. **Write the Reaction**:
When sodium thiosulfate reacts with iodine, the reaction can be represented as:
\[ \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Products} \]

3. **Determine the Products**:
The reaction of sodium thiosulfate with iodine produces sodium tetrathionate (Na₂S₄O₆) and sodium iodide (NaI). The balanced chemical equation for the reaction is:
\[ 2 \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2 \text{NaI} \]

4. **Identify the Main Product**:
The main product of this reaction is sodium tetrathionate (Na₂S₄O₆), which is a key compound formed during the reaction.

5. **Final Answer**:
Therefore, the reaction of Na₂S₂O₃ with iodine gives sodium tetrathionate (Na₂S₄O₆) and sodium iodide (NaI).

### Summary of the Reaction:
- **Reactants**: Sodium thiosulfate (Na₂S₂O₃) and iodine (I₂)
- **Products**: Sodium tetrathionate (Na₂S₄O₆) and sodium iodide (NaI)