Which of the following pairs of cations cannot be separated by using dilute HCl?

Options

A. `Hg_(2)^(2+),Pb^(2+)`
B. `Hg^(2+),Ag^(+)`
C. `Ag^(+),Cu^(2+)`
D. `Hg_(2)^(2+),Bi^(3+)`

Answer

To determine which pairs of cations cannot be separated using dilute HCl, we need to analyze the behavior of each cation in the presence of dilute hydrochloric acid (HCl).

### Step-by-Step Solution:

1. **Identify the Cations**: The pairs of cations given are:
- Pair 1: Hg2²⁺ (Mercury(I)) and Pb²⁺ (Lead)
- Pair 2: Hg²⁺ (Mercury(II)) and Ag⁺ (Silver)
- Pair 3: Ag⁺ (Silver) and Cu²⁺ (Copper)
- Pair 4: Hg2²⁺ (Mercury(I)) and Bi³⁺ (Bismuth)

2. **Understand Group Reagents**: In qualitative analysis, certain cations are grouped based on their reactions with specific reagents. Dilute HCl is a group reagent for certain cations, particularly for the first group which includes Hg²⁺, Hg2²⁺, and Pb²⁺.

3. **Reactions with Dilute HCl**:
- When **dilute HCl** is added to **Hg2²⁺**, it forms **Hg2Cl2** (Mercury(I) chloride), which is a white precipitate.
- When **dilute HCl** is added to **Pb²⁺**, it forms **PbCl2** (Lead(II) chloride), which is also a white precipitate.

4. **Separation Analysis**:
- **Pair 1 (Hg2²⁺ and Pb²⁺)**: Both cations precipitate when dilute HCl is added, hence they cannot be separated.
- **Pair 2 (Hg²⁺ and Ag⁺)**: Ag⁺ precipitates as **AgCl** (Silver chloride) while Hg²⁺ does not precipitate with dilute HCl, hence they can be separated.
- **Pair 3 (Ag⁺ and Cu²⁺)**: Ag⁺ precipitates as **AgCl**, while Cu²⁺ does not precipitate with dilute HCl, hence they can be separated.
- **Pair 4 (Hg2²⁺ and Bi³⁺)**: Hg2²⁺ precipitates as **Hg2Cl2**, while Bi³⁺ does not precipitate with dilute HCl, hence they can be separated.

5. **Conclusion**: The only pair of cations that cannot be separated using dilute HCl is **Hg2²⁺ and Pb²⁺** because both precipitate out of the solution when HCl is added.

### Final Answer:
**The pair of cations that cannot be separated by using dilute HCl is Hg2²⁺ and Pb²⁺.**