The lanthanide ion that would show colour is :

Options

A. `Gd^(3+)`
B. `Sm^(3+)`
C. `La^(3+)`
D. `Lu^(3+)`

Answer

To determine which lanthanide ion shows color, we need to analyze the electronic configurations of the given ions and identify the presence of unpaired electrons. The presence of unpaired electrons is crucial for color because they allow for electronic transitions that can absorb visible light.

### Step-by-Step Solution:

1. **Identify the Neutral Atoms and Their Configurations**:
- Gadolinium (Gd): [Xe] 4f^7 5d^1 6s^2
- Samarium (Sm): [Xe] 4f^6 6s^2
- Lanthanum (La): [Xe] 5d^1 6s^2 (4f^0)
- Lutetium (Lu): [Xe] 4f^14 5d^1 6s^2

2. **Determine the Configurations for the +3 Oxidation State**:
- Gd^3+: [Xe] 4f^7 (loses 5d and 6s electrons)
- Sm^3+: [Xe] 4f^5 (loses 2 electrons from 6s and 1 from 4f)
- La^3+: [Xe] (loses all 5d and 6s electrons, 4f^0)
- Lu^3+: [Xe] 4f^14 (loses 5d and 6s electrons)

3. **Count the Unpaired Electrons**:
- Gd^3+: 7 unpaired electrons (4f^7)
- Sm^3+: 5 unpaired electrons (4f^5)
- La^3+: 0 unpaired electrons (4f^0)
- Lu^3+: 0 unpaired electrons (4f^14)

4. **Determine Which Ions Show Color**:
- **Gd^3+**: Although it has unpaired electrons, the half-filled stability of 4f^7 means it does not participate in electronic transitions that absorb visible light, making it colorless.
- **Sm^3+**: With 5 unpaired electrons, it can undergo electronic transitions, thus it will show color.
- **La^3+**: No unpaired electrons, hence colorless.
- **Lu^3+**: No unpaired electrons, hence colorless.

5. **Conclusion**:
- The only lanthanide ion that shows color in its +3 state is **Samarium (Sm^3+)**.

### Final Answer:
The lanthanide ion that would show color is **Samarium (Sm^3+)**.