An aqueous solution contains `Hg^(2+),Hg_(2)^(2+),Pb^(2+) and Cd^(2+)` Out of these how many ions will produce white precipitate with dilute HCl ?

Explanation

To determine how many ions will produce a white precipitate with dilute HCl from the given ions \(Hg^{2+}\), \(Hg_2^{2+}\), \(Pb^{2+}\), and \(Cd^{2+}\), we can analyze each ion's reaction with dilute HCl.

### Step-by-Step Solution:

1. **Identify the ions present**: The ions in the solution are \(Hg^{2+}\), \(Hg_2^{2+}\), \(Pb^{2+}\), and \(Cd^{2+}\).

2. **Determine the reaction of each ion with dilute HCl**:
- **For \(Pb^{2+}\)**:
- The reaction is:
\[ Pb^{2+} + 2HCl \rightarrow PbCl_2 \downarrow + 2H^+ \]
- Lead(II) chloride (\(PbCl_2\)) is a white precipitate.

- **For \(Hg^{2+}\)**:
- The reaction is:
\[ Hg^{2+} + 2HCl \rightarrow HgCl_2 \downarrow + 2H^+ \]
- Mercury(II) chloride (\(HgCl_2\)) is also a white precipitate.

- **For \(Hg_2^{2+}\)**:
- The reaction is:
\[ Hg_2^{2+} + 2HCl \rightarrow Hg_2Cl_2 \downarrow + 2H^+ \]
- Mercury(I) chloride (\(Hg_2Cl_2\)) is a white precipitate.

- **For \(Cd^{2+}\)**:
- The reaction is:
\[ Cd^{2+} + 2HCl \rightarrow CdCl_2 + 2H^+ \]
- Cadmium chloride (\(CdCl_2\)) is soluble in water and does not produce a precipitate.

3. **Count the ions producing white precipitate**:
- From the above reactions, \(Pb^{2+}\), \(Hg^{2+}\), and \(Hg_2^{2+}\) produce white precipitates.
- \(Cd^{2+}\) does not produce a precipitate.

4. **Conclusion**:
- The ions that produce a white precipitate with dilute HCl are \(Pb^{2+}\), \(Hg^{2+}\), and \(Hg_2^{2+}\).
- Therefore, the total number of ions that produce a white precipitate is **3**.

### Final Answer:
**3 ions will produce a white precipitate with dilute HCl.**