Which of the following lanthanoids ions is diamagnetic?

Options

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

Explanation

To determine which of the lanthanoid ions is diamagnetic, we need to analyze the electronic configurations of each ion and check for the presence of unpaired electrons. Diamagnetic ions have all their electrons paired, while paramagnetic ions have unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Lanthanoid Ions**: The ions we need to analyze are:
- Sm²⁺ (Samarium ion)
- Eu²⁺ (Europium ion)
- Cr²⁺ (Chromium ion)
- Yb²⁺ (Ytterbium ion)

2. **Write the Electronic Configuration of Each Lanthanoid**:
- **Samarium (Sm)**: The atomic number of Sm is 62. Its ground state electronic configuration is:
\[ \text{Xe} 4f^6 5d^0 6s^2 \]
For Sm²⁺, we remove 2 electrons (from the 6s orbital):
\[ \text{Sm}^{2+}: \text{Xe} 4f^6 \]

- **Europium (Eu)**: The atomic number of Eu is 63. Its ground state electronic configuration is:
\[ \text{Xe} 4f^7 5d^0 6s^2 \]
For Eu²⁺, we remove 2 electrons (from the 6s orbital):
\[ \text{Eu}^{2+}: \text{Xe} 4f^7 \]

- **Chromium (Cr)**: The atomic number of Cr is 24. Its ground state electronic configuration is:
\[ \text{Ar} 3d^5 4s^1 \]
For Cr²⁺, we remove 2 electrons (1 from 4s and 1 from 3d):
\[ \text{Cr}^{2+}: \text{Ar} 3d^4 \]

- **Ytterbium (Yb)**: The atomic number of Yb is 70. Its ground state electronic configuration is:
\[ \text{Xe} 4f^{14} 5d^0 6s^2 \]
For Yb²⁺, we remove 2 electrons (from the 6s orbital):
\[ \text{Yb}^{2+}: \text{Xe} 4f^{14} \]

3. **Determine the Electron Pairing**:
- **Sm²⁺ (Xe 4f⁶)**: The 4f subshell has 6 electrons, which means there are unpaired electrons. Thus, Sm²⁺ is **paramagnetic**.
- **Eu²⁺ (Xe 4f⁷)**: The 4f subshell has 7 electrons, which means there are unpaired electrons. Thus, Eu²⁺ is **paramagnetic**.
- **Cr²⁺ (Ar 3d⁴)**: The 3d subshell has 4 electrons, which means there are unpaired electrons. Thus, Cr²⁺ is **paramagnetic**.
- **Yb²⁺ (Xe 4f¹⁴)**: The 4f subshell has 14 electrons, which means all electrons are paired. Thus, Yb²⁺ is **diamagnetic**.

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

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