1. Reaction of excess $KI$ with $CuSO_4$

Copper(II) ions oxidize iodide ions to iodine, while themselves getting reduced to copper(I), which precipitates as cuprous iodide.

The reaction is:

$2CuSO_4 + 4KI \rightarrow 2CuI\downarrow + I_2 + 2K_2SO_4$

So in this step:

• $Cu^{2+} \to Cu^+$ : reduction
• $I^- \to I_2$ : oxidation
• The precipitate formed is $CuI$ (sometimes written as $Cu_2I_2$ as dimeric form)

Thus, statement A: "$Cu_2I_2$ is reduced" is acceptable, because $Cu^{2+}$ ultimately gets reduced to cuprous iodide.

2. Addition of $Na_2S_2O_3$ to the liberated iodine

Sodium thiosulfate reacts with iodine as:

$I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$

Here:

• $I_2 \to I^-$ : reduction
• $S_2O_3^{2-} \to S_4O_6^{2-}$ : oxidation

So:

• Statement B "$Evolved\, I_2$ is reduced" is correct.
• Statement C "$Na_2S_2O_3$ is oxidized" is correct.

3. Check statement about $CuI_2$

In the reaction of $CuSO_4$ with excess $KI$, $CuI_2$ is not formed as a stable product. Instead, $Cu^{2+}$ oxidizes iodide and itself gets reduced to $CuI$.

So statement D "$CuI_2$ is formed" is incorrect.

4. Option-wise conclusion

• A: Correct
• B: Correct
• C: Correct
• D: Incorrect

Therefore, the incorrect statement is:

$\boxed{D}$