How many of the following pairs of ions can be separated by `H_2S` in dilute HCl ?
`Bi^(3+) and Sn^(4+),Al^(3+) and Hg^(2+),Cd^(2+) and Zn^(2+),Fe^(3+) and Cu^(2+), As^(3+) and Sb^(3+)`

Explanation

To determine how many pairs of ions can be separated by `H2S` in dilute `HCl`, we need to analyze the solubility of the sulfides formed from each pair of ions. The key is to identify which ions precipitate as sulfides when `H2S` is added to a solution containing them.

### Step-by-Step Solution:

1. **Identify the pairs of ions:**
- Pair 1: `Bi^(3+)` and `Sn^(4+)`
- Pair 2: `Al^(3+)` and `Hg^(2+)`
- Pair 3: `Cd^(2+)` and `Zn^(2+)`
- Pair 4: `Fe^(3+)` and `Cu^(2+)`
- Pair 5: `As^(3+)` and `Sb^(3+)`

2. **Analyze each pair for precipitation with `H2S`:**
- **Pair 1: `Bi^(3+)` and `Sn^(4+)`**
- Both ions do not form precipitates with `H2S` in dilute `HCl`. Thus, they cannot be separated.

- **Pair 2: `Al^(3+)` and `Hg^(2+)`**
- `Hg^(2+)` forms a black precipitate of `HgS` when `H2S` is added, while `Al^(3+)` remains in solution. Thus, they can be separated.

- **Pair 3: `Cd^(2+)` and `Zn^(2+)`**
- `Cd^(2+)` forms a yellow precipitate of `CdS`, while `Zn^(2+)` remains in solution. Thus, they can be separated.

- **Pair 4: `Fe^(3+)` and `Cu^(2+)`**
- `Fe^(3+)` forms a black precipitate of `FeS`, while `Cu^(2+)` remains in solution. Thus, they can be separated.

- **Pair 5: `As^(3+)` and `Sb^(3+)`**
- Both `As^(3+)` and `Sb^(3+)` can form precipitates with `H2S`. Therefore, they cannot be separated.

3. **Count the separable pairs:**
- From the analysis, we found that pairs 2, 3, and 4 can be separated by `H2S`. Therefore, there are **3 pairs** that can be separated.

### Final Answer:
**3 pairs of ions can be separated by `H2S` in dilute `HCl`.**