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

Options

A. iodide is a bulky ion
B. `Cu^(2+)` oxidizes iodide to iodine
C. `Cu_((aq))^(2+)` has much negative hydration enthalpy
D. `Cu^(2+)` ion has smaller size.

Explanation

To solve the question regarding why all Cu(II) halides are known except the iodide, we can break it down into a step-by-step explanation.

### Step 1: Understanding Copper Halides
Copper can exist in two oxidation states: Cu(I) and Cu(II). The halides of copper are formed when copper reacts with halogens (F, Cl, Br, I).

### Step 2: Reaction of Cu(II) with Iodide
When Cu(II) ions (Cu²⁺) react with iodide ions (I⁻), a potential reaction can occur:
\[ \text{Cu}^{2+} + 4 \text{I}^- \rightarrow 2 \text{CuI} + \text{I}_2 \]

### Step 3: Analyzing the Reaction
In this reaction, the iodide ions (I⁻) are oxidized to iodine (I₂), while the Cu²⁺ ions are reduced to Cu(I) ions (Cu⁺). This indicates that Cu²⁺ is capable of oxidizing iodide ions.

### Step 4: Oxidation and Reduction
- **Oxidation**: Iodide (I⁻) is oxidized to iodine (I₂) as it loses electrons.
- **Reduction**: Copper (Cu²⁺) is reduced to Cu(I) as it gains electrons.

### Step 5: Conclusion
Due to the oxidation of iodide to iodine, Cu(II) cannot effectively form a stable iodide compound. This is why all Cu(II) halides are known except for the iodide.

### Final Answer
The reason all Cu(II) halides are known except the iodide is that Cu²⁺ oxidizes iodide ions (I⁻) to iodine (I₂), making it difficult to form Cu(II) iodide.

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