Oxidation of thiosulphate by iodine gives

Options

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

Explanation

To solve the question regarding the oxidation of thiosulphate by iodine, we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are thiosulphate and iodine. Thiosulphate is represented as \( \text{S}_2\text{O}_3^{2-} \) and iodine as \( \text{I}_2 \).

### Step 2: Write the Oxidation Reaction
In this reaction, thiosulphate (\( \text{S}_2\text{O}_3^{2-} \)) undergoes oxidation. The balanced reaction can be written as:
\[ 2 \text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2 \text{I}^- \]
Here, \( \text{S}_4\text{O}_6^{2-} \) is formed, which is known as the tetrathionate ion.

### Step 3: Understand the Oxidation Process
During the reaction, each thiosulphate ion loses electrons, which is characteristic of oxidation. Specifically, two electrons are lost in the formation of the tetrathionate ion.

### Step 4: Write the Reduction Reaction
Iodine (\( \text{I}_2 \)) undergoes reduction in this reaction. The balanced reduction half-reaction is:
\[ \text{I}_2 + 2 \text{e}^- \rightarrow 2 \text{I}^- \]
This shows that iodine gains electrons to form iodide ions.

### Step 5: Combine the Half-Reactions
Combining the oxidation and reduction half-reactions gives us the overall balanced equation:
\[ 2 \text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2 \text{I}^- \]

### Step 6: Identify the Product
The product of the oxidation of thiosulphate by iodine is the tetrathionate ion, \( \text{S}_4\text{O}_6^{2-} \).

### Final Answer
The oxidation of thiosulphate by iodine gives the tetrathionate ion, \( \text{S}_4\text{O}_6^{2-} \).

---