Question

All Cu(II) halides are known, except ithe iodide, the reason for it is that

• A. Cu⁺² has much more negative hydration enthalpy.
• B. Iodide is bulky ion.
• C. Cu⁺² ion has smaller size.
• D. Cu⁺² oxidises Iodide toiodine.

Answer 1

The Correct Option is D

Copper (II) ion \(\text{(Cu}^{2+}\text{)}\) has a high oxidizing ability. In the presence of iodide ion (I^-), \(\text{(Cu}^{2+}\text{)}\) readily oxidizes iodide to form iodine \((\text{I}_2)\). The reaction can be represented as follows:
\(2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{Cu}^+ + \text{I}_2\)
As a result of this oxidation reaction, \(\text{(Cu}^{2+}\text{)}\) is reduced to \(\text{(Cu}^{1+}\text{)}\), and iodide is oxidized to iodine.
This chemical behavior prevents the stable formation of \(\text{Cu(II)}\) iodide \(\text{(CuI}_2\text{)}\) because the \(\text{(Cu}^{2+}\text{)}\)ion oxidizes the iodide ions to form iodine. Instead, copper (I) iodide\( (CuI)\) is formed, where the \(\text{(Cu}^{1+}\text{)}\) ion is stabilized by the iodide ions.
Therefore, the correct reason why \(\text{Cu(II)}\) iodide is not known is option (D) \(\text{(Cu}^{2+}\text{)}\) oxidizes iodide to iodine.

Answer 2

Correct answer: Cu²⁺ oxidises iodide to iodine. 

All Cu(II) halides are known except CuI₂ (copper(II) iodide). The reason is:

\[ 2Cu^{2+} + 4I^- \rightarrow 2CuI \downarrow + I_2 \]

In this reaction, Cu²⁺ is reduced to Cu⁺, and iodide (I⁻) is oxidized to iodine (I₂). Hence, instead of forming CuI₂, the reaction leads to formation of:

• CuI (copper(I) iodide) — a white precipitate
• I₂ (iodine) — liberated as a brown vapor or solution

This redox behavior prevents the existence of stable Cu(II) iodide.

Answer 3

All Cu(II) halides are known, except the iodide. The reason for this is that Cu⁺² oxidizes iodide to iodine.

Explanation:

• Cu²⁺ is a good oxidizing agent. Iodide (I^-) is a relatively good reducing agent.
• The Cu²⁺ ion can oxidize I^- to iodine (I₂), and in the process, Cu²⁺ is reduced to Cu⁺.
• The reaction is: \(2Cu^{2+} + 4I^- \rightarrow 2CuI(s) + I_2(aq)\) Note that \(CuI\) (copper(I) iodide) is formed, not \(CuI_2\) (copper(II) iodide).
• While Cu⁺² does have a more negative hydration enthalpy than \(Cu^+\), this is not the primary reason for the instability of CuI₂.
• The size of the iodide ion is also not the primary reason.

Therefore, the correct answer is:

Cu⁺² oxidises Iodide to iodine.