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 (\(Na_2S_2O_3\)) with iodine, we need to follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are sodium thiosulfate (\(Na_2S_2O_3\)) and iodine (\(I_2\)).

### Step 2: Write the Reaction
When sodium thiosulfate reacts with iodine, the products formed are sodium tetrathionate (\(Na_2S_4O_6\)) and sodium iodide (\(NaI\)). The balanced chemical equation for the reaction is:

\[ Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2NaI \]

### Step 3: Balance the Equation
To ensure that the equation is balanced, we check the number of atoms of each element on both sides of the equation. The balanced equation is:

\[ 2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2NaI \]

### Step 4: Identify the Products
From the balanced equation, we can see that the products of the reaction are sodium tetrathionate (\(Na_2S_4O_6\)) and sodium iodide (\(NaI\)).

### Conclusion
Thus, the reaction of \(Na_2S_2O_3\) with iodine gives sodium tetrathionate (\(Na_2S_4O_6\)) and sodium iodide (\(NaI\)).

### Final Answer
The correct answer is sodium tetrathionate.

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