Excess of `KI` reacts with `CuSO_(4)` solution and `Na_(2)SO_(3)` solution is added to it. Which of the following statements in incorrect for the reaction?

Options

A. Evolved `I_(2)` is reduced
B. `CuI_(2)` is formed.
C. `Na_(2)S_(2)O_(3)` is oxidised.
D. `Cu_(2)I_(2)` is formed.

Explanation

To solve the question regarding the reaction between excess KI and CuSO4 followed by the addition of Na2SO3, we need to analyze the reactions that take place and identify the incorrect statement.

### Step-by-Step Solution:

1. **Identify the Initial Reaction**:
The first step is to write the reaction between copper(II) sulfate (CuSO4) and potassium iodide (KI). The reaction can be represented as:
\[ \text{CuSO}_4 + 2 \text{KI} \rightarrow \text{Cu}_2\text{I}_2 + \text{I}_2 + \text{K}_2\text{SO}_4 \]
Here, copper(I) iodide (CuI) is formed along with iodine (I2) and potassium sulfate (K2SO4).

2. **Balancing the Reaction**:
The balanced equation for the reaction is:
\[ 2 \text{CuSO}_4 + 4 \text{KI} \rightarrow 2 \text{CuI} + \text{I}_2 + 2 \text{K}_2\text{SO}_4 \]
This shows that for every 2 moles of CuSO4, 4 moles of KI are required to produce 2 moles of CuI and 1 mole of I2.

3. **Addition of Sodium Thiosulfate (Na2SO3)**:
When sodium thiosulfate (Na2SO3) is added to the solution containing I2, it reacts with iodine to form sodium tetrathionate (Na2S4O6) and sodium iodide (NaI):
\[ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2 \text{NaI} \]
In this reaction, iodine (I2) is reduced to iodide (I-) and sodium thiosulfate is oxidized.

4. **Determine the Incorrect Statement**:
Now we analyze the statements provided in the question:
- **Statement A**: Sodium thiosulfate is oxidized. (This is correct)
- **Statement B**: CuI2 is formed. (This is incorrect; Cu2I2 is formed, not CuI2)
- **Statement C**: I2 is reduced. (This is correct)
- **Statement D**: Cu2I2 is formed. (This is correct)

Therefore, the incorrect statement is **Statement B**.

### Final Answer:
The incorrect statement regarding the reaction is **Option B: CuI2 is formed.**