The set representing the correct order of ionic radius is

Options

A. `Li^(+) gt Be^(2+) gt Na^(+) gt Mg^(2+)`
B. `Na^(+) gt Li^(+) Mg^(2+) gt Be^(2+)`
C. `Li^(+) gt Na^(+) gt Mg^(2+) gt Be^(2+)`
D. `Mg^(2+) gt Be^(2+) gt Li^(+) gt Na^(+)`

Answer

To determine the correct order of ionic radii for the given ions, we will analyze the trends in ionic radii based on their positions in the periodic table.

### Step-by-Step Solution:

1. **Identify the Ions**: The ions we are comparing are Li\(^+\), Na\(^+\), Mg\(^{2+}\), and Be\(^{2+}\).

2. **Understand the Periodic Trends**:
- **Down a Group**: Ionic radius increases as we move down a group in the periodic table. This is due to the addition of electron shells.
- **Across a Period**: Ionic radius generally decreases as we move from left to right across a period. This is due to the increasing nuclear charge, which pulls the electrons closer to the nucleus.

3. **Analyze the Ions**:
- **Li\(^+\)**: Lithium is in the second period. As a cation, it has lost one electron.
- **Na\(^+\)**: Sodium is in the third period and has lost one electron.
- **Mg\(^{2+}\)**: Magnesium is also in the third period but has lost two electrons.
- **Be\(^{2+}\)**: Beryllium is in the second period and has lost two electrons.

4. **Comparing Ionic Radii**:
- **Na\(^+\) vs. Li\(^+\)**: Na\(^+\) (3rd period) will have a larger radius than Li\(^+\) (2nd period) because Na\(^+\) is further down the group.
- **Mg\(^{2+}\) vs. Na\(^+\)**: Mg\(^{2+}\) has a higher positive charge (+2) than Na\(^+\) (+1), which means it pulls its electrons closer, resulting in a smaller radius. Thus, Na\(^+\) > Mg\(^{2+}\).
- **Be\(^{2+}\) vs. Mg\(^{2+}\)**: Be\(^{2+}\) is in the second period and has a higher positive charge than Mg\(^{2+}\), making it smaller than Mg\(^{2+}\). Thus, Mg\(^{2+}\) > Be\(^{2+}\).

5. **Final Order**:
- From the comparisons, we can establish the following order:
- Na\(^+\) > Li\(^+\) > Mg\(^{2+}\) > Be\(^{2+}\)

### Conclusion:
The correct order of ionic radii is:
**Na\(^+\) > Li\(^+\) > Mg\(^{2+}\) > Be\(^{2+}\)**

### Answer:
The final answer for this question is option B.