Arrange `(I)Ce^(3+),(II)La^(3+),(III)Pm^(3+)` and `(IV)Yb^(3+)` in increasing order of their ionic radii.

Options

A. `IVltIIIltIltII`
B. `IltIVltIIIltII`
C. `IVltIIIltIIltI`
D. `IIIltIIltIltIV`

Explanation

To arrange the ions \( \text{Ce}^{3+}, \text{La}^{3+}, \text{Pm}^{3+}, \text{Yb}^{3+} \) in increasing order of their ionic radii, we need to consider the concept of lanthanide contraction and the general trends in ionic radii among the lanthanide series.

### Step-by-Step Solution:

1. **Understanding Lanthanide Contraction**:
- The lanthanide series consists of elements with atomic numbers from 57 (Lanthanum, La) to 71 (Lutetium, Lu). As we move from La to Lu, the effective nuclear charge increases due to the filling of the 4f orbitals. This leads to a decrease in ionic radii, known as lanthanide contraction.

2. **Identifying the Elements**:
- The ions we are considering are:
- \( \text{Ce}^{3+} \) (Cerium)
- \( \text{La}^{3+} \) (Lanthanum)
- \( \text{Pm}^{3+} \) (Promethium)
- \( \text{Yb}^{3+} \) (Ytterbium)

3. **Comparing Ionic Radii**:
- **Lanthanum (La)**: It is the first element in the series and has the largest ionic radius among the four ions.
- **Cerium (Ce)**: It is the second element and has a smaller ionic radius than La due to the increase in nuclear charge.
- **Promethium (Pm)**: It comes after Ce and has a smaller ionic radius than Ce due to the continued increase in nuclear charge.
- **Ytterbium (Yb)**: It is further down the series and has the smallest ionic radius among the four ions due to the cumulative effect of lanthanide contraction.

4. **Arranging in Increasing Order**:
- Based on the analysis, we can arrange the ions in increasing order of their ionic radii:
- \( \text{Yb}^{3+} < \text{Pm}^{3+} < \text{Ce}^{3+} < \text{La}^{3+} \)

### Final Answer:
The increasing order of ionic radii is:
\[ \text{Yb}^{3+} < \text{Pm}^{3+} < \text{Ce}^{3+} < \text{La}^{3+} \]