Which of the following paires can be septated by `H_(2)S` in dil `HCI`?

Options

A. `Cu^(2+)` and `Cd^(2+)`
B. `Cu^(2+)` and `Ni^(2+)`
C. `Cu^(2+)` and `Zn^(2+)`
D. `Hg^(2+)` and `Al^(2+)`

Explanation

To solve the question of which pairs can be separated by H₂S in dilute HCl, we need to understand the behavior of certain cations in the presence of H₂S and dilute HCl. Here is a step-by-step solution:

### Step 1: Identify the Cations
The question involves pairs of cations. We need to identify the cations in each option provided. Common cations that can be precipitated by H₂S in dilute HCl include:
- Lead (Pb²⁺)
- Copper (Cu²⁺)
- Bismuth (Bi³⁺)
- Mercuric (Hg²⁺)
- Cadmium (Cd²⁺)

### Step 2: Understand the Group Reagent
H₂S acts as a group reagent for Group 2 cations in qualitative analysis. When H₂S is added to a solution containing these cations in dilute HCl, it precipitates the sulfides of these cations. The sulfides that precipitate include:
- PbS (Lead sulfide)
- CuS (Copper sulfide)
- Bi₂S₃ (Bismuth sulfide)
- HgS (Mercuric sulfide)
- CdS (Cadmium sulfide)

### Step 3: Analyze Each Option
Now, we will analyze each option to see if the cations can be separated based on their behavior with H₂S:

- **Option A:** Cu²⁺ and Cd²⁺
- Both Cu²⁺ and Cd²⁺ will precipitate as CuS and CdS, respectively. Therefore, they cannot be separated.

- **Option B:** Cu²⁺ and Ni²⁺
- Cu²⁺ will precipitate as CuS, while Ni²⁺ will remain in solution. Therefore, they can be separated.

- **Option C:** Cu²⁺ and Zn²⁺
- Cu²⁺ will precipitate as CuS, while Zn²⁺ will remain in solution. Therefore, they can be separated.

- **Option D:** Hg²⁺ and Al³⁺
- Hg²⁺ will precipitate as HgS, while Al³⁺ will remain in solution. Therefore, they can be separated.

### Step 4: Conclusion
Based on the analysis, the pairs that can be separated by H₂S in dilute HCl are:
- **Option B:** Cu²⁺ and Ni²⁺
- **Option C:** Cu²⁺ and Zn²⁺
- **Option D:** Hg²⁺ and Al³⁺

Thus, the answer is options B, C, and D.

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