Statement-1. Beilstein test can be used to detect fluorine in the organic compound.
Statement-2. CuF is not volatile and hence does not impart any colour to the flame.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is False

Explanation

To analyze the statements provided in the question, we will evaluate each statement individually and determine their correctness.

### Step-by-Step Solution:

1. **Evaluate Statement 1**:
- The Beilstein test is a qualitative test used to detect halogens (specifically chlorine, bromine, and iodine) in organic compounds.
- However, it is important to note that the Beilstein test is not effective for detecting fluorine. This is because fluorine forms stable compounds that do not release halogen atoms in a form that can be detected by this test.
- Therefore, **Statement 1 is incorrect**.

2. **Evaluate Statement 2**:
- The second statement mentions that CuF (copper(I) fluoride) is not volatile.
- In the context of the Beilstein test, when halides are heated, they can produce volatile compounds that impart color to the flame. Since CuF is not volatile, it does not vaporize to give a color to the flame.
- Thus, **Statement 2 is correct**.

### Conclusion:
- Statement 1 is incorrect because the Beilstein test cannot detect fluorine.
- Statement 2 is correct because CuF is not volatile and does not impart color to the flame.

### Final Answer:
- Statement 1: Incorrect
- Statement 2: Correct