Thiosulphate reacts differently with iodine and bromine in the reaction given below
`2S_(2)O_(3)^(2-) to S_(4)O_(6)^(2-)+2I^(-)`
`S_(2)O_(3)^(2-)+2Br_(2)+5H_(2)Oto2SO_(4)^(2-)+2Br^(-)+10H^(+)`
Which of the following statements justifies the above dual behaviour of thiosulphate?

Options

A. Bromine is a stronger oxidant than iodine
B. Bromine is a weaker oxidant than iodine
C. Thiosulphate undergoes by bromine and reduction by iodine in these reactions
D. Bromine undergoes oxidation and iodine undergoes reduction in these reactions

Explanation

To understand the dual behavior of thiosulphate (S₂O₃²⁻) when reacting with iodine (I₂) and bromine (Br₂), we need to analyze the reactions and the changes in oxidation states of sulfur and the halogens involved.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The first reaction is:
\[ 2S_{2}O_{3}^{2-} + I_{2} \rightarrow S_{4}O_{6}^{2-} + 2I^{-} \]
- The second reaction is:
\[ S_{2}O_{3}^{2-} + 2Br_{2} + 5H_{2}O \rightarrow 2SO_{4}^{2-} + 2Br^{-} + 10H^{+} \]

2. **Determine Oxidation States**:
- For the first reaction:
- In \( S_{2}O_{3}^{2-} \), let the oxidation state of sulfur be \( x \):
\[ 2x + 3(-2) = -2 \implies 2x - 6 = -2 \implies 2x = 4 \implies x = +2 \]
- In \( S_{4}O_{6}^{2-} \):
\[ 4x + 6(-2) = -2 \implies 4x - 12 = -2 \implies 4x = 10 \implies x = +2.5 \]
- Therefore, sulfur goes from +2 to +2.5 (oxidation).

- For iodine, it is in elemental form (I₂), so its oxidation state is 0.

- For the second reaction:
- In \( S_{2}O_{3}^{2-} \), the oxidation state of sulfur is still +2.
- In \( SO_{4}^{2-} \):
\[ x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6 \]
- Therefore, sulfur goes from +2 to +6 (oxidation).

- For bromine (Br₂), it is also in elemental form, so its oxidation state is 0.

3. **Analyze the Oxidation Process**:
- In the first reaction, thiosulphate is oxidized to tetrathionate (S₄O₆²⁻) by iodine, which means iodine acts as an oxidizing agent.
- In the second reaction, thiosulphate is oxidized to sulfate (SO₄²⁻) by bromine, which also acts as an oxidizing agent.

4. **Compare the Strength of Oxidizing Agents**:
- The change in oxidation state for sulfur in the first reaction is from +2 to +2.5 (a change of 0.5).
- The change in oxidation state for sulfur in the second reaction is from +2 to +6 (a change of 4).
- This indicates that bromine is a stronger oxidizing agent than iodine because it causes a greater change in the oxidation state of sulfur.

5. **Conclusion**:
- The dual behavior of thiosulphate can be justified by the fact that bromine is a stronger oxidizing agent than iodine, leading to different products in the reactions.

### Final Answer:
The correct statement that justifies the dual behavior of thiosulphate is: **Bromine is a stronger oxidant than iodine.**