Which of the following metal ion can give red precipitate with `K_(2) CrO_(4)`

Options

A. `Hg_(2)^(+2)`
B. `Ag^(+)`
C. `Pb^(+2)`
D. `Hg^(2+)` and `Ag^(+)` boht

Answer

To determine which metal ion can give a red precipitate with \( K_2CrO_4 \), we will analyze the reactions of the given metal ions with potassium chromate.

### Step-by-Step Solution:

1. **Identify the Metal Ions**: The metal ions provided in the options are:
- \( Hg_2^{2+} \) (Mercury(I))
- \( Hg^{2+} \) (Mercury(II))
- \( Pb^{2+} \) (Lead(II))

2. **Reaction of \( Hg_2^{2+} \) with \( K_2CrO_4 \)**:
- When \( Hg_2^{2+} \) reacts with \( K_2CrO_4 \), the reaction can be written as:
\[ Hg_2^{2+} + K_2CrO_4 \rightarrow HgCrO_4 + 2K^{+} \]
- The product \( HgCrO_4 \) is known to form a red precipitate.

3. **Reaction of \( Hg^{2+} \) with \( K_2CrO_4 \)**:
- When \( Hg^{2+} \) reacts with \( K_2CrO_4 \), the reaction can be represented as:
\[ Hg^{2+} + K_2CrO_4 \rightarrow Hg_2CrO_4 + 2K^{+} \]
- The product \( Hg_2CrO_4 \) also forms a red precipitate.

4. **Reaction of \( Pb^{2+} \) with \( K_2CrO_4 \)**:
- When \( Pb^{2+} \) reacts with \( K_2CrO_4 \), the reaction is:
\[ Pb^{2+} + K_2CrO_4 \rightarrow PbCrO_4 + 2K^{+} \]
- The product \( PbCrO_4 \) forms a yellow precipitate, which does not meet the criteria of a red precipitate.

5. **Conclusion**:
- The metal ions \( Hg_2^{2+} \) and \( Hg^{2+} \) both produce red precipitates when reacted with \( K_2CrO_4 \).
- Therefore, the answer to the question is that both \( Hg_2^{2+} \) and \( Hg^{2+} \) can give a red precipitate with \( K_2CrO_4 \).

### Final Answer:
The metal ions that can give a red precipitate with \( K_2CrO_4 \) are \( Hg_2^{2+} \) and \( Hg^{2+} \).

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