Which is the correct order of ionic sizes ?
(Atomic no . : Ce = 58 , Sn = 50 , Yb = 70 and Lu)

Options

A. `Ce^(3+)gt Sn^(2+)gt Yb^(3+)gt Lu^(3+)`
B. `Sn^(2+)gt Yb^(3+) gt Ce^(3+) gt Lu ^(3+)`
C. `Sn^(2+)gt Ce^(3+) gt Yb^(3+) gt Lu ^(3+)`
D. `Lu^(2+)gt Yb^(3+) gt Sn^(3+) gt Ce ^(3+)`

Answer

To determine the correct order of ionic sizes for the given elements (Ce, Sn, Yb, and Lu), we will analyze their ionic states and the trends in ionic sizes based on their positions in the periodic table.

### Step-by-Step Solution:

1. **Identify the Elements and Their Atomic Numbers:**
- Cerium (Ce) - Atomic number 58
- Tin (Sn) - Atomic number 50
- Ytterbium (Yb) - Atomic number 70
- Lutetium (Lu) - Atomic number 71

2. **Determine the Oxidation States:**
- For this question, we consider the common oxidation states:
- Ce typically exhibits a +3 oxidation state.
- Sn typically exhibits a +2 oxidation state.
- Yb typically exhibits a +3 oxidation state.
- Lu typically exhibits a +3 oxidation state.

3. **Understand the Ionic Size Trends:**
- Ionic size generally decreases across a period due to increasing nuclear charge, which pulls electrons closer to the nucleus.
- In the case of lanthanides (Ce, Yb, Lu), there is a phenomenon known as lanthanide contraction, where the ionic sizes decrease as we move from Ce to Lu because of the increasing positive charge from the nucleus.

4. **Compare the Ionic Sizes:**
- **Ce³⁺**: Largest among the lanthanides due to being the first in the series.
- **Yb³⁺**: Smaller than Ce³⁺ due to higher atomic number and increased nuclear charge.
- **Lu³⁺**: Smallest among the lanthanides due to the highest nuclear charge.
- **Sn²⁺**: As a p-block element, Sn²⁺ is larger than Ce³⁺ because it has a lower positive charge compared to the +3 oxidation states of the lanthanides.

5. **Establish the Correct Order:**
- From the analysis, we can conclude the order of ionic sizes:
- Sn²⁺ > Ce³⁺ > Yb³⁺ > Lu³⁺

6. **Final Answer:**
- The correct order of ionic sizes is: **Sn²⁺ > Ce³⁺ > Yb³⁺ > Lu³⁺**