Which of the following lanthanoid ions is diamagnetic? (Atomic number of `Ce=58,Sm=62,Eu=63,Yb=70`]

Options

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

Explanation

To determine which of the given lanthanoid ions is diamagnetic, we need to analyze the electronic configurations of each ion and check for unpaired electrons. Diamagnetic substances have all their electrons paired, resulting in no net magnetic moment.

### Step-by-Step Solution:

1. **Identify the Ions and Their Atomic Numbers:**
- Ce²⁺ (Cerium) - Atomic number 58
- Sm²⁺ (Samarium) - Atomic number 62
- Eu²⁺ (Europium) - Atomic number 63
- Yb²⁺ (Ytterbium) - Atomic number 70

2. **Determine the Electronic Configuration of Each Ion:**
- For **Ce²⁺**:
- Atomic number of Ce = 58
- Neutral cerium (Ce) has the configuration: [Xe] 6s² 4f¹
- For Ce²⁺, we remove two electrons (from 6s²):
- Configuration: [Xe] 4f²
- For **Sm²⁺**:
- Atomic number of Sm = 62
- Neutral samarium (Sm) has the configuration: [Xe] 6s² 4f⁶
- For Sm²⁺, we remove two electrons (from 6s²):
- Configuration: [Xe] 4f⁶
- For **Eu²⁺**:
- Atomic number of Eu = 63
- Neutral europium (Eu) has the configuration: [Xe] 6s² 4f⁷
- For Eu²⁺, we remove two electrons (from 6s²):
- Configuration: [Xe] 4f⁷
- For **Yb²⁺**:
- Atomic number of Yb = 70
- Neutral ytterbium (Yb) has the configuration: [Xe] 6s² 4f¹⁴
- For Yb²⁺, we remove two electrons (from 6s²):
- Configuration: [Xe] 4f¹⁴

3. **Count the Number of Unpaired Electrons:**
- **Ce²⁺**: 4f² has 2 unpaired electrons (not diamagnetic).
- **Sm²⁺**: 4f⁶ has 6 unpaired electrons (not diamagnetic).
- **Eu²⁺**: 4f⁷ has 7 unpaired electrons (not diamagnetic).
- **Yb²⁺**: 4f¹⁴ has 0 unpaired electrons (diamagnetic).

4. **Conclusion:**
- The only ion that is diamagnetic is **Yb²⁺**.

### Final Answer:
The lanthanoid ion that is diamagnetic is **Yb²⁺**.