The lanthanide ion that would show colour is :

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the Lanthanide Ions**:
We need to analyze the following lanthanide ions: Lutetium (Lu³⁺), Lanthanum (La³⁺), Gadolinium (Gd³⁺), and Samarium (Sm³⁺).

2. **Determine the Electronic Configurations**:
- **Lutetium (Lu)**: Atomic number 71
- Electronic configuration: [Xe] 4f¹⁴ 5d¹ 6s²
- For Lu³⁺: Remove 3 electrons (from 5d and 6s)
- Configuration: [Xe] 4f¹⁴ (fully filled, no unpaired electrons)

- **Lanthanum (La)**: Atomic number 57
- Electronic configuration: [Xe] 5d¹ 6s²
- For La³⁺: Remove 3 electrons (from 5d and 6s)
- Configuration: [Xe] 5d⁰ (fully filled, no unpaired electrons)

- **Gadolinium (Gd)**: Atomic number 64
- Electronic configuration: [Xe] 4f⁷ 5d¹ 6s²
- For Gd³⁺: Remove 3 electrons (from 5d and 6s)
- Configuration: [Xe] 4f⁷ (contains unpaired electrons, but stable half-filled configuration)

- **Samarium (Sm)**: Atomic number 62
- Electronic configuration: [Xe] 4f⁶ 5d⁰ 6s²
- For Sm³⁺: Remove 3 electrons (from 5d and 6s)
- Configuration: [Xe] 4f⁵ (contains unpaired electrons)

3. **Analyze the Presence of Unpaired Electrons**:
- **Lu³⁺**: No unpaired electrons (colorless)
- **La³⁺**: No unpaired electrons (colorless)
- **Gd³⁺**: Unpaired electrons present, but stable half-filled configuration (may not show color)
- **Sm³⁺**: Unpaired electrons present (shows color)

4. **Conclusion**:
The lanthanide ion that shows color is **Samarium (Sm³⁺)** due to the presence of unpaired electrons in its 4f orbital.

### Final Answer:
The lanthanide ion that would show color is **Samarium (Sm³⁺)**.