Which of the following sequences is correct for decreasing order of ionic radius :

Options

A. `Se^(-2) gt I^(-) gt Br^(-) gt O^(-2) gt F^(-)`
B. `I^(-) gt Se^(2-) gt O^(-2) gt Br^(-) 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 sequence for the decreasing order of ionic radius among the given ions, we need to analyze the ionic sizes based on their electronic configurations and the trends in ionic radii.

### Step-by-Step Solution:

1. **Identify the Ions**: The ions given are:
- Iodide (I⁻)
- Bromide (Br⁻)
- Oxide (O²⁻)
- Fluoride (F⁻)
- Selenium (Se²⁻)

2. **Determine the Electronic Configurations**:
- **Iodide (I⁻)**: Atomic number = 53, gains 1 electron → 54 electrons. Configuration: [Kr] 4d¹⁰ 5s² 5p⁶ (Noble gas configuration)
- **Bromide (Br⁻)**: Atomic number = 35, gains 1 electron → 36 electrons. Configuration: [Ar] 3d¹⁰ 4s² 4p⁶ (Noble gas configuration)
- **Oxide (O²⁻)**: Atomic number = 8, gains 2 electrons → 10 electrons. Configuration: [He] 2s² 2p⁶ (Noble gas configuration)
- **Fluoride (F⁻)**: Atomic number = 9, gains 1 electron → 10 electrons. Configuration: [He] 2s² 2p⁶ (Noble gas configuration)
- **Selenide (Se²⁻)**: Atomic number = 34, gains 2 electrons → 36 electrons. Configuration: [Kr] 4d¹⁰ 5s² 5p⁶ (Noble gas configuration)

3. **Analyze Ionic Radii Trends**:
- **General Trend**: Ionic radius increases down a group and decreases across a period. For ions with the same number of electrons (isoelectronic), the ionic radius decreases with increasing nuclear charge.
- **Comparing Ions**:
- Iodide (I⁻) has the largest ionic radius due to having the most electron shells (5 shells).
- Selenide (Se²⁻) and Bromide (Br⁻) both have 4 shells, but Se²⁻ has a greater negative charge, leading to a larger radius than Br⁻.
- Oxide (O²⁻) and Fluoride (F⁻) both have 2 shells, but O²⁻ has a greater negative charge than F⁻, making it larger.

4. **Order of Ionic Radii**:
- I⁻ > Se²⁻ > Br⁻ > O²⁻ > F⁻

5. **Conclusion**: The correct sequence for decreasing order of ionic radius is:
- **I⁻ > Se²⁻ > Br⁻ > O²⁻ > F⁻**