A solution of potassium iodide is added to an aqueous solution of a c…

The correct answer is Option (2) → \( \ce{Cu2I2} \) and \( \ce{I2} \)

The Cu²⁺ ion is an oxidising agent strong enough to oxidise iodide ions to iodine. In doing so, copper is reduced from +2 to +1 and comes out as the solid copper(I) iodide, Cu₂I₂:

    \( \displaystyle \ce{2Cu^2+ + 4I^- -> Cu2I2(s) + I2} \)

Two of the four iodide ions lose one electron each and form I₂; the two Cu²⁺ ions take these two electrons. This is why all copper(II) halides are known except the iodide.

Why the other options are incorrect

Option (1) \( \ce{CuI2} \) only — Incorrect. Copper(II) iodide is not formed: as soon as Cu²⁺ meets I⁻, it oxidises the iodide to iodine and is itself reduced to copper(I).

Option (3) \( \ce{Cu} \) and \( \ce{I2} \) — Incorrect. Copper is reduced only to the +1 state, which is held as the insoluble Cu₂I₂; no copper metal (oxidation state 0) is formed.

Option (4) \( \ce{Cu2I2} \) only — Incorrect. Copper can be reduced only if something is oxidised at the same time. Here the iodide is oxidised, so iodine must form along with Cu₂I₂.

NCERT (2026-27 edition):
4.3.7 Trends in Stability of Higher Oxidation States – “…On the other hand, all Cu^II halides are known except the iodide. In this case, Cu²⁺ oxidises I^– to I₂: 2Cu²⁺ + 4I⁻ → Cu₂I₂(s) + I₂”