Consider the following sequence of tests,
`M^(n+) overset(HCl)to ` white precipitate`overset(Delta)to` water soluble.
The metal ion `(M^(n+))` would be:

Options

A. `Hg^(2+)`
B. `Ag^(+)`
C. `Pb^(2+)`
D. `Sn^(2+)`

Answer

To determine the metal ion \( M^{n+} \) based on the given sequence of tests, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Tests**: The sequence of tests indicates that when \( M^{n+} \) is treated with dilute HCl, a white precipitate is formed. This suggests that \( M^{n+} \) could be one of the cations that form insoluble chlorides.

2. **Identifying Possible Cations**: The common cations that form white precipitates with dilute HCl are:
- Silver ion \( \text{Ag}^+ \)
- Lead ion \( \text{Pb}^{2+} \)
- Mercury(I) ion \( \text{Hg}_2^{2+} \)

3. **Analyzing the Precipitate**: The white precipitate formed is likely a chloride salt. The precipitate must also be analyzed based on the next step, which indicates that upon heating, the white precipitate becomes water soluble.

4. **Testing Solubility Upon Heating**:
- **Lead(II) chloride \( \text{PbCl}_2 \)**: This salt is known to be insoluble in cold water but becomes soluble when heated.
- **Silver chloride \( \text{AgCl} \)**: This salt remains insoluble in water even upon heating.
- **Mercury(I) chloride \( \text{Hg}_2Cl_2 \)**: This salt also remains insoluble in water upon heating.

5. **Conclusion**: Since the white precipitate formed from \( M^{n+} \) becomes water soluble upon heating, the only possible cation that fits all these criteria is lead ion \( \text{Pb}^{2+} \).

Thus, the metal ion \( M^{n+} \) would be \( \text{Pb}^{2+} \).

### Final Answer:
The metal ion \( M^{n+} \) is \( \text{Pb}^{2+} \).

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