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

Options

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

Answer

To determine the type of reaction represented by the equation \(2 \text{CuI} \rightarrow \text{Cu} + \text{CuI}_2\), we can analyze the changes in oxidation states of the elements involved.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactant is \(2 \text{CuI}\) (Copper(I) iodide).
- The products are \(\text{Cu}\) (Copper) and \(\text{CuI}_2\) (Copper(II) iodide).

2. **Assign Oxidation States**:
- In \( \text{CuI} \), the oxidation state of Cu is +1 (since Iodine is -1).
- In the product \(\text{Cu}\), the oxidation state of Cu is 0.
- In \(\text{CuI}_2\), the oxidation state of Cu is +2 (since there are two Iodine atoms, each with a -1 charge).

3. **Determine Changes in Oxidation States**:
- The Cu in \( \text{CuI} \) changes from +1 to 0 (reduction).
- The Cu in the product \(\text{CuI}_2\) changes from +1 to +2 (oxidation).

4. **Identify the Type of Reaction**:
- Since the same element (Copper) is undergoing both oxidation (increase in oxidation state) and reduction (decrease in oxidation state), this reaction is classified as a **disproportionation reaction**.

5. **Conclusion**:
- Therefore, the given reaction \(2 \text{CuI} \rightarrow \text{Cu} + \text{CuI}_2\) is an example of a **disproportionation reaction**.

### Final Answer:
The given reaction is an example of **disproportionation**.