Which of the following lanthanoid ions is diamagnetic ?
(At nos . `Ce = 58 , Sm = 62, Eu = 63 , Yb =70)

Options

A. `Sm^(2+)`
B. `Eu^(2+)`
C. `Yb^(2+)`
D. `Ce^(2+)`

Answer

To determine which of the given lanthanoid ions is diamagnetic, we need to analyze the electronic configurations of the lanthanoids provided: Ce (Cerium), Sm (Samarium), Eu (Europium), and Yb (Ytterbium). We will look at their electronic configurations and the number of unpaired electrons in their respective ions.

### Step 1: Determine the electronic configurations of the lanthanoids.

1. **Cerium (Ce, atomic number 58)**:
- Electronic configuration: \( \text{Xe} \, 4f^1 \, 5d^1 \, 6s^2 \)
- For \( \text{Ce}^{2+} \): Remove 2 electrons (from 6s and 5d):
- \( \text{Ce}^{2+} \): \( \text{Xe} \, 4f^1 \)

2. **Samarium (Sm, atomic number 62)**:
- Electronic configuration: \( \text{Xe} \, 4f^6 \, 6s^2 \)
- For \( \text{Sm}^{2+} \): Remove 2 electrons (from 6s):
- \( \text{Sm}^{2+} \): \( \text{Xe} \, 4f^6 \)

3. **Europium (Eu, atomic number 63)**:
- Electronic configuration: \( \text{Xe} \, 4f^7 \, 6s^2 \)
- For \( \text{Eu}^{2+} \): Remove 2 electrons (from 6s):
- \( \text{Eu}^{2+} \): \( \text{Xe} \, 4f^7 \)

4. **Ytterbium (Yb, atomic number 70)**:
- Electronic configuration: \( \text{Xe} \, 4f^{14} \, 6s^2 \)
- For \( \text{Yb}^{2+} \): Remove 2 electrons (from 6s):
- \( \text{Yb}^{2+} \): \( \text{Xe} \, 4f^{14} \)

### Step 2: Analyze the number of unpaired electrons in each ion.

1. **Ce²⁺**:
- Configuration: \( 4f^1 \) (1 unpaired electron) → Paramagnetic

2. **Sm²⁺**:
- Configuration: \( 4f^6 \) (6 unpaired electrons) → Paramagnetic

3. **Eu²⁺**:
- Configuration: \( 4f^7 \) (7 unpaired electrons) → Paramagnetic

4. **Yb²⁺**:
- Configuration: \( 4f^{14} \) (14 paired electrons) → Diamagnetic

### Step 3: Conclusion

The only lanthanoid ion that is diamagnetic is **Ytterbium (Yb²⁺)**.

### Final Answer:
**Yb²⁺** is the diamagnetic ion among the given options.

---