Oxidation of thiosulphate with iodine gives 

Options

A. Sulphate ion
B. Sulphite ion
C. Tetrathionate
D. Sulphide ion.

Answer

To solve the question regarding the oxidation of thiosulfate with iodine, we can follow these steps:

### Step 1: Identify the Reactants
The reactant in this reaction is thiosulfate, which has the chemical formula \( \text{S}_2\text{O}_3^{2-} \).

### Step 2: Identify the Oxidizing Agent
Iodine (\( \text{I}_2 \)) is the oxidizing agent in this reaction. It will accept electrons and get reduced.

### Step 3: Write the Reaction
When thiosulfate reacts with iodine, the products formed are tetrathionate (\( \text{S}_4\text{O}_6^{2-} \)) and iodide ions (\( \text{I}^- \)). The balanced chemical equation for this reaction is:

\[ 2 \text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2 \text{I}^- \]

### Step 4: Identify the Products
From the balanced equation, we can see that the products of the reaction are:
- Tetrathionate ion (\( \text{S}_4\text{O}_6^{2-} \))
- Iodide ions (\( \text{I}^- \))

### Step 5: Conclusion
Thus, the oxidation of thiosulfate with iodine gives tetrathionate and iodide ions. Therefore, the correct answer to the question is tetrathionate.

### Final Answer
The oxidation of thiosulfate with iodine gives tetrathionate (\( \text{S}_4\text{O}_6^{2-} \)).

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