`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 solve the question regarding why cerium (Ce) exhibits a stable +4 oxidation state and yttrium (Yb) exhibits a stable +2 oxidation state, we can follow these steps:

### Step 1: Determine the Electronic Configuration
- **Cerium (Ce, Z = 58)** has the electronic configuration:
\[ \text{Xe} \, 4f^1 \, 5d^1 \, 6s^2 \]

- **Ytterbium (Yb, Z = 70)** has the electronic configuration:
\[ \text{Xe} \, 4f^{14} \, 6s^2 \]

### Step 2: Analyze the Oxidation States
- For **Cerium (Ce)** to achieve a +4 oxidation state, it loses 4 electrons:
- The electrons lost come from the 6s and 5d orbitals:
\[ \text{Ce}^{4+} \rightarrow \text{Xe} \, 4f^0 \, 5d^0 \, 6s^0 \]
- This results in a configuration of \( 4f^0 \), which is stable because it has no electrons in the f-orbital.

- For **Ytterbium (Yb)** to achieve a +2 oxidation state, it loses 2 electrons:
- The electrons lost come from the 6s orbital:
\[ \text{Yb}^{2+} \rightarrow \text{Xe} \, 4f^{14} \, 6s^0 \]
- This results in a configuration of \( 4f^{14} \), which is stable because it is fully filled.

### Step 3: Stability of the Configurations
- The stability of oxidation states is often associated with achieving noble gas configurations or stable electron configurations (like half-filled or fully filled subshells).
- **Cerium (+4)** achieves a stable configuration of \( 4f^0 \) (empty f-orbital).
- **Ytterbium (+2)** achieves a stable configuration of \( 4f^{14} \) (fully filled f-orbital).

### Conclusion
- Thus, the reason why cerium exhibits a stable +4 oxidation state and yttrium exhibits a stable +2 oxidation state is that:
- **Cerium (Ce) achieves a stable \( 4f^0 \) configuration.**
- **Ytterbium (Yb) achieves a stable \( 4f^{14} \) configuration.**

### Final Answer
The correct option is: **Cerium \( 4^+ \) acquires \( F^0 \) configuration and Ytterbium \( 2^+ \) acquires \( F^{14} \) configuration.**

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