The compound A on treatment with Na gives B, and with `PCl_(5)` gives C. B and C react together to give diethyl ether. A, B and C are in the order:

Options

A. `C_(2)H_(5)OH, C_(2)H_(6), C_(2)H_(5)Cl`
B. `C_(2)H_(5)OH, C_(2)H_(5)Cl, C_(2)H_(5)ONa`
C. `C_(2)H_(5)OH,C_(2)H_(5)ONa,C_(2)H_(5)Cl`
D. `C_(2)H_(5)Cl,C_(2)H_(6),C_(2)H_(5)OH`

Explanation

To solve the problem, we need to identify the compounds A, B, and C based on the reactions described. Let's break down the steps:

### Step 1: Identify Compound A
- **Observation**: Compound A reacts with sodium (Na) and PCl5.
- **Analysis**: PCl5 is known to chlorinate alcohols. Therefore, A is likely an alcohol.
- **Conclusion**: A can be ethanol (C2H5OH).

### Step 2: Identify Compound B
- **Observation**: When A (ethanol) reacts with sodium, it forms compound B.
- **Analysis**: The reaction of an alcohol with sodium produces an alkoxide and hydrogen gas. Thus, B is sodium ethoxide (C2H5O-Na+).
- **Conclusion**: B is sodium ethoxide (C2H5O-Na+).

### Step 3: Identify Compound C
- **Observation**: When A (ethanol) reacts with PCl5, it produces compound C.
- **Analysis**: The chlorination of ethanol by PCl5 results in the formation of ethyl chloride (C2H5Cl).
- **Conclusion**: C is ethyl chloride (C2H5Cl).

### Step 4: Confirm the Reaction between B and C
- **Observation**: B (sodium ethoxide) and C (ethyl chloride) react together to form diethyl ether.
- **Analysis**: The reaction between an alkoxide (B) and an alkyl halide (C) leads to ether formation. In this case, sodium ethoxide reacts with ethyl chloride to form diethyl ether (C2H5-O-C2H5).
- **Conclusion**: The reaction confirms the identities of B and C.

### Final Sequence
- A: Ethanol (C2H5OH)
- B: Sodium ethoxide (C2H5O-Na+)
- C: Ethyl chloride (C2H5Cl)

Thus, the order of A, B, and C is:
- A: Ethanol
- B: Sodium ethoxide
- C: Ethyl chloride