`2CuIrarr Cu+CuI_2` , the given reaction is an example of

Options

A. Disproportion
B. Neutralisation
C. Oxidation
D. Reduction

Answer

To solve the question regarding the reaction \(2CuI \rightarrow Cu + CuI_2\) and identify the type of reaction it represents, we can follow these steps:

### Step 1: Analyze the Reaction
The reaction involves the decomposition of copper(I) iodide (CuI) into elemental copper (Cu) and copper(II) iodide (CuI2).

### Step 2: Determine the Oxidation States
- In \(CuI\), copper (Cu) has an oxidation state of +1 and iodine (I) has an oxidation state of -1.
- In the products:
- The elemental copper (Cu) has an oxidation state of 0.
- In \(CuI_2\), copper (Cu) has an oxidation state of +2 (since each iodine is -1 and there are two iodines).

### Step 3: Identify Changes in Oxidation States
- The oxidation state of copper changes from +1 in \(CuI\) to 0 in elemental copper (reduction).
- The oxidation state of copper changes from +1 in \(CuI\) to +2 in \(CuI_2\) (oxidation).

### Step 4: Conclusion on the Type of Reaction
Since one species (Cu) is both oxidized (from +1 to +2) and reduced (from +1 to 0) in the same reaction, this is classified as a **disproportionation reaction**.

### Final Answer
The given reaction \(2CuI \rightarrow Cu + CuI_2\) is an example of a **disproportionation reaction**.

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