The f-block elements are those in which the differentiating electron enters the (n-2)f orbital. There are two series of f-block elements corresponding to filling of 4f and 5f-orbitals called lanthanides and actinides respectively. They show different oxidation states depending upon the stability of `f^0, f^7`and `f^14` configurations, though the principal oxidation state is +3. There is a regular decrease in size of lanthanide ions with increase in atomic number and it is known as lanthanide contraction. As a result of this, the basic character of oxides and hydroxides decreases from first element (La) to last element (Lu). All the actinides are radioactive and therefore, it is difficult to study their chemical nature.
Ce (Z = 58) and Yb (Z = 70) exhibit stable +4 and +2 oxidation states respectively. This is because

Options

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

Explanation

To answer the question regarding why Ce (Z = 58) and Yb (Z = 70) exhibit stable +4 and +2 oxidation states respectively, we can break down the explanation into a step-by-step solution.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
- For **Cerium (Ce)** with atomic number 58, the electronic configuration is:
\[ \text{Xe} \, 4f^1 \, 5d^1 \, 6s^2 \]
- For **Ytterbium (Yb)** with atomic number 70, the electronic configuration is:
\[ \text{Xe} \, 4f^{14} \, 6s^2 \]

2. **Determine the Oxidation States**:
- When Ce loses 4 electrons to form Ce\(^{4+}\), the configuration becomes:
\[ 4f^0 \, 5d^0 \, 6s^0 \]
- When Yb loses 2 electrons to form Yb\(^{2+}\), the configuration becomes:
\[ 4f^{14} \, 6s^0 \]

3. **Stability of the Configurations**:
- The configuration for Ce\(^{4+}\) (4f\(^{0}\), 5d\(^{0}\), 6s\(^{0}\)) achieves a noble gas configuration, which is highly stable.
- The configuration for Yb\(^{2+}\) (4f\(^{14}\), 6s\(^{0}\)) has a completely filled f-orbital, which is also very stable due to electron pairing and symmetry.

4. **Conclusion**:
- Ce exhibits a stable +4 oxidation state because it achieves a noble gas configuration upon losing its electrons.
- Yb exhibits a stable +2 oxidation state because it has a completely filled f-orbital, which is energetically favorable.

### Final Answer:
Ce (Z = 58) exhibits a stable +4 oxidation state because it attains a noble gas configuration (4f\(^{0}\), 5d\(^{0}\), 6s\(^{0}\)). Yb (Z = 70) exhibits a stable +2 oxidation state due to its completely filled f-orbital (4f\(^{14}\)), which is more stable than empty or partially filled orbitals.

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