Element furnishing coloured ions in the aqueous medium is

Options

A. `Zn`
B. `Hg`
C. `Cu`
D. `Al`

Answer

To determine which element furnishes colored ions in an aqueous medium, we need to analyze the electronic configurations and the nature of the ions formed by the given elements: zinc, mercury, copper, and aluminum.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
- **Zinc (Zn)**: The electronic configuration is \( \text{[Ar]} \, 3d^{10} \, 4s^2 \).
- **Mercury (Hg)**: The electronic configuration is \( \text{[Xe]} \, 4f^{14} \, 5d^{10} \, 6s^2 \).
- **Copper (Cu)**: The electronic configuration is \( \text{[Ar]} \, 3d^{10} \, 4s^1 \).
- **Aluminum (Al)**: The electronic configuration is \( \text{[Ne]} \, 3s^2 \, 3p^1 \).

2. **Determine the Ions Formed**:
- **Zinc Ion (Zn²⁺)**: Loses 2 electrons to form \( \text{Zn}^{2+} \) with configuration \( 3d^{10} \).
- **Mercury Ion (Hg²⁺)**: Loses 2 electrons to form \( \text{Hg}^{2+} \) with configuration \( 5d^{10} \).
- **Copper Ion (Cu²⁺)**: Loses 2 electrons to form \( \text{Cu}^{2+} \) with configuration \( 3d^{9} \).
- **Aluminum Ion (Al³⁺)**: Loses 3 electrons to form \( \text{Al}^{3+} \) with no d electrons.

3. **Analyze the d-Orbital Configurations**:
- **Zinc (Zn²⁺)**: \( 3d^{10} \) - completely filled d-orbitals, no color.
- **Mercury (Hg²⁺)**: \( 5d^{10} \) - completely filled d-orbitals, no color.
- **Copper (Cu²⁺)**: \( 3d^{9} \) - incompletely filled d-orbitals, can exhibit color due to d-d transitions.
- **Aluminum (Al³⁺)**: No d electrons, hence no color.

4. **Conclusion**:
- The only element that furnishes colored ions in an aqueous medium is **Copper (Cu)**, specifically its ion \( \text{Cu}^{2+} \) which has an incomplete d-orbital configuration allowing for d-d transitions that result in color.

### Final Answer:
The element furnishing colored ions in the aqueous medium is **Copper (Cu)**.