`Ce(Z = 58)` and `Yb(Z = 70)` exhibits stable +4 and 2 oxidation states respectively. This is because

Options

A. `Ce^(4+)` and `Yb^(2+)`acquire `f^(7)` configuration
B. `Ce^(4+)` and `Yb^(2+)` acquire `f^(0)` configuration
C. `Ce^(4+)` and `Yb^(2+)` acquire `f_(o)` and `f^(14)` configuration
D. `Ce^(4+)` and `Yb^(2+)` acquire `f^(7)` and `f^(14)` configuration

Answer

To understand why cerium (Ce) exhibits a stable +4 oxidation state and ytterbium (Yb) exhibits a stable +2 oxidation state, we need to analyze their electronic configurations and the stability associated with certain electron arrangements.

### Step 1: Determine the electronic configurations of Ce and Yb
- **Cerium (Ce, Z = 58)**: The electronic configuration is:
\[ [Xe] \, 4f^1 \, 5d^1 \, 6s^2 \]
- **Ytterbium (Yb, Z = 70)**: The electronic configuration is:
\[ [Xe] \, 4f^{14} \, 6s^2 \]

### Step 2: Analyze the oxidation states
- **For Cerium (Ce)**:
- In the +4 oxidation state, cerium loses 4 electrons. This involves losing:
- 2 electrons from the 6s subshell
- 1 electron from the 5d subshell
- 1 electron from the 4f subshell
- After losing these electrons, the configuration becomes:
\[ [Xe] \, 4f^0 \, 5d^0 \, 6s^0 \]
- The resulting configuration (4f^0) is stable because there are no electrons in the f-orbital, leading to no electron-electron repulsion.

- **For Ytterbium (Yb)**:
- In the +2 oxidation state, ytterbium loses 2 electrons from the 6s subshell:
\[ [Xe] \, 4f^{14} \, 6s^0 \]
- The resulting configuration (4f^{14}) is stable because the f-orbital is fully filled, which is a stable arrangement.

### Step 3: Conclusion
- Cerium exhibits a stable +4 oxidation state due to the 4f^0 configuration (no electrons in the f-orbital).
- Ytterbium exhibits a stable +2 oxidation state due to the fully filled 4f^{14} configuration.

### Final Answer:
Cerium (Ce) exhibits a stable +4 oxidation state because it achieves a stable 4f^0 configuration after losing 4 electrons, while Ytterbium (Yb) exhibits a stable +2 oxidation state because it achieves a fully filled 4f^{14} configuration after losing 2 electrons.

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