Incorrect Statement: b) CuI2 is formed

Explanation:
- The given reaction involves the reaction between excess KI (potassium iodide) and CuSO4 (copper sulfate) solution followed by the addition of Na2S2O3 (sodium thiosulfate) solution.
- When excess KI is added to CuSO4 solution, a redox reaction takes place. Cu2+ ions in CuSO4 are reduced to Cu+ ions, while I- ions in KI are oxidized to I2 (elemental iodine).
- The overall reaction can be represented as follows:
CuSO4 + 2KI -> CuI2 + K2SO4

Analysis of the given options:
a) Na2S2O3 is oxidized:
- Na2S2O3 is a reducing agent and can undergo oxidation. However, in this reaction, Na2S2O3 is not oxidized. It acts as a reducing agent to reduce the CuI2 formed in the previous step.
- Na2S2O3 reduces CuI2 to form Cu2+ ions and I- ions.
- The reaction can be represented as follows:
CuI2 + 2Na2S2O3 -> 2Cu2+ + S4O6^2- + 2NaI

b) CuI2 is formed:
- This statement is incorrect. CuI2 is not formed in this reaction. Instead, CuI (copper(I) iodide) is formed.
- CuI is a yellow precipitate that forms when the Cu+ ions from CuSO4 react with the I- ions from excess KI.
- The reaction can be represented as follows:
2Cu+ + 4I- -> 2CuI

c) Cu2I2 is formed:
- This statement is correct. Cu2I2 (copper(II) iodide) is formed as a result of the reaction between CuI and I2.
- CuI reacts with I2 to form Cu2I2, which is a white precipitate.
- The reaction can be represented as follows:
2CuI + I2 -> Cu2I2

d) Evolved I2 is reduced:
- This statement is correct. The I2 that is evolved during the reaction is reduced by Na2S2O3.
- Na2S2O3 acts as a reducing agent and reduces I2 to I- ions.
- The reaction can be represented as follows:
I2 + 2Na2S2O3 -> 2NaI + Na2S4O6

In conclusion, the incorrect statement in this reaction is b) CuI2 is formed. The correct product formed is CuI (copper(I) iodide).