The correct order of ionic radii of the following species is

Options

A. `Se^(2-) gt I^(-) gt Br^(-) gt O^(2-) gt F^(-)`
B. `I^(-) gt Se^(2-) gt O^(2-) gt F^(-)`
C. Se^(2-) gt I^(-) gt Br^(-) gt F^(-) gt O^(2-)
D. `I^(-) gt Se^(2-) gt Br^(-) gt O^(2-) gt F^(-)`

Answer

To determine the correct order of ionic radii for the species selenium (Se²⁻), iodine (I⁻), bromine (Br⁻), oxygen (O²⁻), and fluorine (F⁻), we will analyze their positions in the periodic table and the factors affecting ionic size.

### Step-by-Step Solution:

1. **Identify the Groups and Periods**:
- Selenium (Se) is in Group 16.
- Iodine (I) is in Group 17.
- Bromine (Br) is in Group 17.
- Oxygen (O) is in Group 16.
- Fluorine (F) is in Group 17.

2. **Consider the Ionic Charges**:
- Se²⁻: Selenium gains two electrons.
- I⁻: Iodine gains one electron.
- Br⁻: Bromine gains one electron.
- O²⁻: Oxygen gains two electrons.
- F⁻: Fluorine gains one electron.

3. **Analyze the Trend in Ionic Size**:
- As you move down a group in the periodic table, ionic size increases due to the addition of electron shells.
- For ions with the same charge, the ionic radius increases with increasing atomic number (more protons and electrons).

4. **Compare the Ions**:
- **For Se²⁻ and Br⁻**: Both are in the same period (4th period for Se and 4th period for Br). Se²⁻ has a greater negative charge (2-) compared to Br⁻ (1-), making Se²⁻ larger than Br⁻.
- **For O²⁻ and F⁻**: Both are in the same period (2nd period for O and F). O²⁻ has a greater negative charge (2-) compared to F⁻ (1-), making O²⁻ larger than F⁻.
- **For I⁻**: Iodine is larger than both Br⁻ and Se²⁻ due to being in a lower period (5th period).

5. **Final Order of Ionic Radii**:
- From largest to smallest, the order is:
- I⁻ > Se²⁻ > Br⁻ > O²⁻ > F⁻

### Conclusion:
The correct order of ionic radii is:
**I⁻ > Se²⁻ > Br⁻ > O²⁻ > F⁻**