Arrange `Ce^(3+),La^(3+), Pm^(3)` and `Yb^(3+)` in increasing order of their size -

Options

A. `Ce^(+3)ltYb^(+3)ltPm^(+3)ltLa^(+3)`
B. `Yb^(+3)ltPm^(+3)ltCe^(+3)ltLa^(+3)`
C. `Yb^(+3)ltPm^(+3)ltLa^(+3)ltCl^(+3)`
D. `Pm^(+3)ltLa^(3+)ltCe^(+3)ltYb^(+3)`

Answer

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

### Step-by-Step Solution:

1. **Understand the Elements**:
- We have four ions: Cerium (\( \text{Ce}^{3+} \)), Lanthanum (\( \text{La}^{3+} \)), Promethium (\( \text{Pm}^{3+} \)), and Ytterbium (\( \text{Yb}^{3+} \)).
- Their atomic numbers are:
- \( \text{Ce} = 58 \)
- \( \text{La} = 57 \)
- \( \text{Pm} = 61 \)
- \( \text{Yb} = 70 \)

2. **Consider Ionic Charge**:
- All ions have a +3 charge. This means they have lost three electrons, which affects their size.

3. **Lanthanide Contraction**:
- As we move from Lanthanum to Lutetium in the lanthanide series, the size of the ions decreases due to the increasing nuclear charge without a significant increase in shielding effect. This phenomenon is known as lanthanide contraction.

4. **Order of Sizes**:
- **Lanthanum (\( \text{La}^{3+} \))**: Being the first in the series, it has the largest ionic radius.
- **Cerium (\( \text{Ce}^{3+} \))**: Comes next, slightly smaller than Lanthanum due to the lanthanide contraction.
- **Promethium (\( \text{Pm}^{3+} \))**: Is smaller than Cerium and Lanthanum due to the continued effect of lanthanide contraction.
- **Ytterbium (\( \text{Yb}^{3+} \))**: Has the smallest ionic radius among these ions due to the highest atomic number and the effects of lanthanide contraction.

5. **Final Arrangement**:
- Therefore, the increasing order of their sizes is:
\[ \text{La}^{3+} < \text{Ce}^{3+} < \text{Pm}^{3+} < \text{Yb}^{3+} \]