`Na_(2)S_(2)O_(3)` is oxidised by `I_(2)` to

Options

A. `Na_(2)S`
B. `Na_(2)SO_(4)`
C. `NaHSO_(3)`
D. `Na_(2)S_(4)O_(6)`

Explanation

To solve the question regarding the oxidation of \( \text{Na}_2\text{S}_2\text{O}_3 \) (sodium thiosulfate) by \( \text{I}_2 \) (iodine), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are sodium thiosulfate \( \text{Na}_2\text{S}_2\text{O}_3 \) and iodine \( \text{I}_2 \).

### Step 2: Write the Reaction
When sodium thiosulfate is oxidized by iodine, it produces sodium iodide and sodium tetrathionate. The balanced chemical equation for this reaction can be written as:
\[ \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]

### Step 3: Identify the Products
From the balanced equation, the products of the reaction are:
- Sodium iodide \( \text{NaI} \)
- Sodium tetrathionate \( \text{Na}_2\text{S}_4\text{O}_6 \)

### Step 4: Conclusion
Thus, when \( \text{Na}_2\text{S}_2\text{O}_3 \) is oxidized by \( \text{I}_2 \), the products formed are \( \text{NaI} \) and \( \text{Na}_2\text{S}_4\text{O}_6 \).

### Final Answer
The final products of the oxidation of \( \text{Na}_2\text{S}_2\text{O}_3 \) by \( \text{I}_2 \) are \( \text{NaI} \) and \( \text{Na}_2\text{S}_4\text{O}_6 \).

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