JEE Main 2021 — Salt Analysis Question with Solution

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:

${\text{2 Cu}}^{2+}+4{\text{I}}^{-}⟶2\text{CuI}\downarrow +{\text{I}}_2$

So, the correct answer is:

Option C: CuI.