A metal chloride original solution `(i.e. O.S)` on mixing with
`K_(2)CrO_(4)` solution gives a yellow precipitate soluble in aqueous
sodium hydroxide. The metal may be `:`

Options

A. mercury
B. iron
C. silver
D. lead

Explanation

To solve the problem step by step, we will analyze the information provided and apply our knowledge of qualitative analysis.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a metal chloride solution (original solution) that reacts with potassium chromate (K₂CrO₄) to form a yellow precipitate. This precipitate is soluble in aqueous sodium hydroxide (NaOH).

2. **Identifying the Yellow Precipitate**:
The yellow precipitate formed when a metal chloride reacts with K₂CrO₄ is typically lead chromate (PbCrO₄). This is a well-known reaction in qualitative analysis.

3. **Chemical Reaction**:
The reaction can be represented as:
\[ \text{PbCl}_2 (aq) + \text{K}_2\text{CrO}_4 (aq) \rightarrow \text{PbCrO}_4 (s) + 2\text{KCl} (aq) \]
Here, lead(II) chloride reacts with potassium chromate to form lead chromate, which is a yellow precipitate.

4. **Solubility in Sodium Hydroxide**:
The problem states that the yellow precipitate is soluble in aqueous sodium hydroxide. Lead chromate (PbCrO₄) can dissolve in NaOH to form a soluble complex:
\[ \text{PbCrO}_4 (s) + 2\text{NaOH} (aq) \rightarrow \text{Na}_2\text{Pb(OH)}_4 (aq) + \text{CrO}_4^{2-} (aq) \]
This confirms that the metal in question is lead (Pb).

5. **Conclusion**:
Based on the above reactions and the properties of the precipitate, we conclude that the metal in the original solution is lead (Pb).

### Final Answer:
The metal may be **Lead (Pb)**.

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