`p-Cl-C_(6)H_(4)NH_(2)" and "PhNH_(3)""^(+)Cl^(-)` can be distinguished by

Options

A. `NaOH`
B. `AgNO_(3)`
C. `LiAlH_(4)`
D. `Zn`

Answer

To distinguish between para-chloroaniline (p-Cl-C6H4NH2) and anilinium chloride (PhNH3+Cl-), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compounds**:
- Para-chloroaniline (p-Cl-C6H4NH2) is an amine with a chlorine substituent at the para position.
- Anilinium chloride (PhNH3+Cl-) is the protonated form of aniline, which exists as a cation with a chloride ion.

2. **Understand the Reactivity**:
- Para-chloroaniline does not release chloride ions in solution because it is a neutral amine.
- Anilinium chloride, being a salt, can dissociate in solution to release chloride ions.

3. **Choose a Suitable Reagent**:
- Silver nitrate (AgNO3) is a reagent that can react with chloride ions to form a precipitate of silver chloride (AgCl).

4. **Perform the Test**:
- When you add AgNO3 to a solution containing anilinium chloride, the chloride ions will react with AgNO3 to form a white precipitate of AgCl:
\[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl (s)} \]
- In contrast, when you add AgNO3 to a solution of para-chloroaniline, no reaction occurs, and no precipitate is formed because para-chloroaniline does not release chloride ions.

5. **Observe the Results**:
- If a white precipitate forms, it indicates the presence of anilinium chloride.
- If no precipitate forms, it indicates the presence of para-chloroaniline.

### Conclusion:
Thus, the two compounds can be distinguished by adding AgNO3 to the solutions. The formation of a white precipitate indicates the presence of anilinium chloride, while the absence of a precipitate indicates para-chloroaniline.