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 explain why all Cu(II) halides are known except for the iodide, we can break down the reasoning into several steps:

### Step 1: Understanding Copper Halides
Copper can exist in two oxidation states: +1 and +2. The halides of copper can be represented as CuX2, where X is a halogen (like Cl, Br, or I).

### Step 2: Formation of CuI2
For CuI2 to exist, copper would need to be in the +2 oxidation state, and iodine would be in the -1 oxidation state. The formula would imply that two iodide ions (I-) combine with one copper ion (Cu^2+).

### Step 3: Stability of CuI2
If CuI2 were to exist, it would be expected to be stable. However, it is not stable and tends to decompose. The decomposition reaction can be represented as:
\[ \text{CuI}_2 \rightarrow \text{CuI} + \text{I}_2 \]
In this reaction, iodine gas (I2) is liberated, and copper iodide (CuI) is formed.

### Step 4: Oxidation Process
In the decomposition of CuI2, the Cu^2+ ion oxidizes the iodide ions (I-) to form elemental iodine (I2). This is an important point because it indicates that the presence of Cu^2+ leads to the oxidation of I- to I2, which is not a characteristic of stable halides.

### Step 5: Conclusion
The instability of CuI2 is primarily due to the oxidation of iodide ions by Cu^2+, leading to the formation of CuI and the liberation of iodine gas. Therefore, CuI2 does not exist as a stable compound.

### Final Answer
All Cu(II) halides are known except the iodide because CuI2 is unstable and decomposes to form CuI and I2, with Cu^2+ oxidizing the iodide ions.

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