Thiosulphate reacts differently with iodine and bromine in the reactions given below :
`S_(2)O_(3)^(2-)+I_(2) rarr S_(4)O_(6)^(2-) + 2I^(-)`
`S_(2)O_(3)^(2-)+2Br_(2)+5H_(2)O rarr 2SO_(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 oxidation by bromine and reduction by iodine in these reactions.
D. Bromine undergoes oxidation and iodine undergoes reduction in these reactions.

Explanation

To solve the problem regarding the dual behavior of thiosulphate when reacting with iodine and bromine, we will analyze the oxidation states of sulfur in both reactions.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The first reaction is:
\[ S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^{-} \]
- The second reaction is:
\[ S_2O_3^{2-} + 2Br_2 + 5H_2O \rightarrow 2SO_4^{2-} + 2Br^{-} + 10H^{+} \]

2. **Determine the Oxidation State of Sulfur in Thiosulphate**:
- For thiosulphate \( S_2O_3^{2-} \):
- Let the oxidation state of sulfur be \( x \).
- The equation for the oxidation state is:
\[ 2x + 3(-2) = -2 \implies 2x - 6 = -2 \implies 2x = 4 \implies x = +2 \]
- Thus, the oxidation state of sulfur in \( S_2O_3^{2-} \) is +2.

3. **Calculate the Oxidation State of Sulfur in the Products**:
- In the product \( S_4O_6^{2-} \):
- Let the oxidation state of sulfur be \( y \).
- The equation is:
\[ 4y + 6(-2) = -2 \implies 4y - 12 = -2 \implies 4y = 10 \implies y = +2.5 \]
- Thus, in the reaction with iodine, sulfur goes from +2 to +2.5.

4. **Analyze the Reaction with Bromine**:
- In the reaction with bromine, \( S_2O_3^{2-} \) reacts to form \( SO_4^{2-} \):
- Let the oxidation state of sulfur in \( SO_4^{2-} \) be \( z \).
- The equation is:
\[ 2x + 4(-2) = -2 \implies 2z - 8 = -2 \implies 2z = 6 \implies z = +6 \]
- Thus, in the reaction with bromine, sulfur goes from +2 to +6.

5. **Compare the Changes in Oxidation States**:
- For iodine: Sulfur changes from +2 to +2.5 (a small increase).
- For bromine: Sulfur changes from +2 to +6 (a significant increase).

6. **Conclusion**:
- The dual behavior of thiosulphate can be attributed to the different strengths of iodine and bromine as oxidizing agents. Bromine is a stronger oxidizing agent than iodine, which explains why thiosulphate undergoes a larger change in oxidation state when reacting with bromine compared to iodine.

### Final Answer:
The statement that justifies the dual behavior of thiosulphate is that bromine (Br₂) is a stronger oxidizing agent than iodine (I₂), leading to a greater increase in the oxidation state of sulfur in the reaction with bromine.