Which of the following statements are correct?
A. \(CuI_2\) is not known
B. \(Cu_2I_2\) is known
C. \(Cu^{2+}\) ion oxidizes \(I^-\) to \(I_2\)
D. \(Cu^{2+}\) ion does not oxidize \(I^-\) to \(I_2\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A, B and C only
\( \ce{Cu^2+} \) is a strong enough oxidising agent to oxidise iodide ions to iodine, and is itself reduced to copper(I). So copper(II) iodide cannot exist; copper(I) iodide forms instead.
\( \ce{2Cu^2+ + 4I- -> Cu2I2(s) + I2} \)
(A) CuI2 is not known — True. All copper(II) halides are known except the iodide, because \( \ce{Cu^2+} \) oxidises \( \ce{I-} \) as soon as they meet.
(B) Cu2I2 is known — True. Copper(I) iodide is the solid product of the reaction above.
(C) Cu2+ ion oxidizes I− to I2 — True. Cu goes from +2 to +1 (gains one electron) and iodine from −1 to 0 (loses one electron), as in the equation above.
(D) Cu2+ ion does not oxidize I− to I2 — False. It is the opposite of (C); the reaction above does take place.
NCERT (2026-27 edition):
Trends in Stability of Higher Oxidation States – “… Another feature of fluorides is their instability in the low oxidation states e.g., VX2 (X = Cl, Br or I) and the same applies to CuX. On the other hand, all Cu2+ halides are known except the iodide. In this case, Cu2+ oxidises I– to I2:”
\( \ce{2Cu^2+ + 4I- -> Cu2I2(s) + I2} \)