Consider the following transformation :
`2CuX_(2) overset("Room temperature")(to) 2CuX+X_(2) (uarr)`
Then `X^(-)` can be:

Options

A. `F^(-), Br^(-)`
B. `Cl^(-), Br^(-)`
C. `CN^(-), I^(-)`
D. `Cl^(-), F^(-)`

Explanation

To solve the question, we need to analyze the given transformation and determine what the anion \(X^-\) can be in the reaction:

\[ 2CuX_2 \overset{\text{Room temperature}}{\rightarrow} 2CuX + X_2 \]

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The reaction involves the decomposition of \(CuX_2\) into \(CuX\) and \(X_2\). This indicates that \(X\) must be a halogen or a similar anion that can form a stable compound with copper.

2. **Identifying Possible Anions**:
- The possible anions given are: \(F^-\), \(Br^-\), \(Cl^-\), \(CN^-\), and \(I^-\).
- We need to determine which of these anions can form a stable \(CuX\) compound and decompose to release \(X_2\) at room temperature.

3. **Analyzing Each Anion**:
- **For \(F^-\)**: \(CuF_2\) is stable, but \(F_2\) is a very reactive molecule and is unlikely to form under these conditions.
- **For \(Br^-\)**: \(CuBr_2\) is stable and can decompose to \(CuBr\) and \(Br_2\) at room temperature.
- **For \(Cl^-\)**: \(CuCl_2\) is stable and can decompose to \(CuCl\) and \(Cl_2\) at room temperature.
- **For \(CN^-\)**: \(Cu(CN)_2\) is known to be unstable and can decompose to \(CuCN\) and \(CN^-\) at room temperature.
- **For \(I^-\)**: \(CuI_2\) is also unstable and can decompose to \(CuI\) and \(I_2\) at room temperature.

4. **Conclusion**:
- The anions \(CN^-\) and \(I^-\) are unstable and can decompose to produce \(CuCN\) and \(CuI\) respectively, which aligns with the reaction given. Therefore, the correct answer is:
\[ X^- = CN^- \text{ and } I^- \]

### Final Answer:
The possible values for \(X^-\) are \(CN^-\) and \(I^-\).

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