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

Answer: (D)

Cu+2 oxidises Iodide toiodine.

Answer 2

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.