Which of the following are colourless?

Options

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

Explanation

To determine which of the given metal ions are colorless, we need to analyze the electronic configurations of each ion and check for the presence of unpaired electrons in the d orbitals. A metal ion is typically colorless if it has no unpaired electrons in its d orbitals, as color arises from d to d electronic transitions involving unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Metal Ions**: The ions given are:
- Cerium (Ce³⁺)
- Lanthanum (La³⁺)
- Lutetium (Lu³⁺)
- Gadolinium (Gd³⁺)

2. **Determine the Electronic Configuration of Each Ion**:
- **Cerium (Ce)**: The atomic number is 58.
- Ground state configuration: [Xe] 4f¹ 5d¹ 6s²
- Ce³⁺ loses three electrons (two from 6s and one from 5d):
- Ce³⁺ configuration: [Xe] 4f¹
- **Unpaired Electrons**: 1 in 4f (not in d), hence **colorless**.

- **Lanthanum (La)**: The atomic number is 57.
- Ground state configuration: [Xe] 5d¹ 6s²
- La³⁺ loses three electrons (two from 6s and one from 5d):
- La³⁺ configuration: [Xe]
- **Unpaired Electrons**: None, hence **colorless**.

- **Lutetium (Lu)**: The atomic number is 71.
- Ground state configuration: [Xe] 4f¹⁴ 5d¹ 6s²
- Lu³⁺ loses three electrons (two from 6s and one from 5d):
- Lu³⁺ configuration: [Xe] 4f¹⁴
- **Unpaired Electrons**: None, hence **colorless**.

- **Gadolinium (Gd)**: The atomic number is 64.
- Ground state configuration: [Xe] 4f⁷ 5d¹ 6s²
- Gd³⁺ loses three electrons (two from 6s and one from 5d):
- Gd³⁺ configuration: [Xe] 4f⁷
- **Unpaired Electrons**: 7 in 4f (not in d), hence **colorless**.

3. **Conclusion**: All the given metal ions (Ce³⁺, La³⁺, Lu³⁺, Gd³⁺) are colorless because they do not have unpaired electrons in their d orbitals.

### Final Answer:
All of the given metal ions (Ce³⁺, La³⁺, Lu³⁺, Gd³⁺) are colorless.