In the test for iodine, `I_(2)` is treated with sodium thiosulphate `(Na_(2)S_(2)O_(3))` :
`Na_(2)S_(2)O_(3)+I_(2) to NaI +..`

Options

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

Answer

To solve the question regarding the reaction between iodine (I₂) and sodium thiosulphate (Na₂S₂O₃), we will follow these steps:

### Step 1: Write the unbalanced chemical equation.
The initial reaction can be represented as:
\[ \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{NaI} + \text{?} \]

### Step 2: Identify the products of the reaction.
When sodium thiosulphate reacts with iodine, it produces sodium iodide (NaI) and sodium tetrathionate (Na₂S₄O₆). Therefore, the products 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: Balance the chemical equation.
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:
\[ 2 \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]

### Step 4: Identify the correct answer from the options.
From the balanced equation, we can see that the product formed alongside sodium iodide (NaI) is sodium tetrathionate (Na₂S₄O₆). Therefore, the correct answer is:
\[ \text{Na}_2\text{S}_4\text{O}_6 \]

### Conclusion:
The final balanced reaction is:
\[ 2 \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]
Thus, the answer is option A: Na₂S₄O₆.

---