The paramagnetic lanthanoid ion among the following is

Options

A. `Ce^(4+)`
B. `Yb^(2+)`
C. `Lu^(3+)`
D. `Sm^(2+)`

Answer

To determine the paramagnetic lanthanoid ion among the given options, we need to analyze the electronic configurations of each lanthanoid ion and identify if they have unpaired electrons. A paramagnetic substance has unpaired electrons, which contribute to its magnetic properties.

### Step-by-Step Solution:

1. **Identify the Lanthanoid Ions**:
We need to analyze the given lanthanoid ions. Let's assume the options are Cerium (Ce), Ytterbium (Yb), Lutetium (Lu), and Samarium (Sm).

2. **Determine the Atomic Numbers and Electronic Configurations**:
- **Cerium (Ce)**: Atomic number = 58
- Electronic configuration: [Xe] 4f¹ 5d¹ 6s²
- **Ytterbium (Yb)**: Atomic number = 70
- Electronic configuration: [Xe] 4f¹⁴ 6s²
- **Lutetium (Lu)**: Atomic number = 71
- Electronic configuration: [Xe] 4f¹⁴ 5d¹ 6s²
- **Samarium (Sm)**: Atomic number = 62
- Electronic configuration: [Xe] 4f⁶ 6s²

3. **Determine the Oxidation States**:
- **Cerium in +4 Oxidation State**:
- Remove 4 electrons: [Xe] 4f⁰ (no unpaired electrons)
- **Ytterbium in +2 Oxidation State**:
- Remove 2 electrons: [Xe] 4f¹⁴ (no unpaired electrons)
- **Lutetium in +3 Oxidation State**:
- Remove 3 electrons: [Xe] 4f¹⁴ 5d⁰ (no unpaired electrons)
- **Samarium in +2 Oxidation State**:
- Remove 2 electrons: [Xe] 4f⁶ (unpaired electrons present)

4. **Identify Unpaired Electrons**:
- **Cerium**: 4f⁰ → No unpaired electrons
- **Ytterbium**: 4f¹⁴ → No unpaired electrons
- **Lutetium**: 4f¹⁴ 5d⁰ → No unpaired electrons
- **Samarium**: 4f⁶ → Contains unpaired electrons

5. **Conclusion**:
The only lanthanoid ion among the options that is paramagnetic (having unpaired electrons) is **Samarium (Sm)** in the +2 oxidation state.

### Final Answer:
The paramagnetic lanthanoid ion among the following is **Samarium (Sm)**.