When a copper(II) salt (containing Cu²⁺) is treated with potassium iodide (KI), the cupric ions (Cu²⁺) are reduced to copper(I) (Cu⁺), and iodine (I₂) is released. The solid product formed in the reaction is copper(I) iodide, CuI.

The simplified reaction is:

$ \text{2 Cu}^{2+} + 4 \text{I}^{-} \;\longrightarrow\; 2 \,\text{CuI} \!\downarrow + \text{I}_{2} $

So, the correct answer is:

Option C: CuI.