When the following five anions are arranged in order of decreasing ionic radius , the correct sequence is :

Options

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

Explanation

To arrange the given anions in order of decreasing ionic radius, we need to consider their positions in the periodic table and the general trends in ionic size.

### Step-by-Step Solution:

1. **Identify the Anions**: The anions provided are:
- O²⁻ (Oxide ion)
- F⁻ (Fluoride ion)
- Se²⁻ (Selenide ion)
- Br⁻ (Bromide ion)
- I⁻ (Iodide ion)

2. **Understand Ionic Size Trends**:
- Anions are larger than their corresponding neutral atoms because they gain electrons, which increases electron-electron repulsion.
- As you move down a group in the periodic table, ionic size increases due to the addition of electron shells.
- As you move across a period from left to right, ionic size generally decreases due to increasing effective nuclear charge.

3. **Analyze the Anions**:
- **Iodide ion (I⁻)**: Located in Group 17 (Halogens), Period 5. It is the largest because it has the most electron shells.
- **Bromide ion (Br⁻)**: Located in Group 17, Period 4. It is smaller than I⁻ but larger than the other anions listed.
- **Selenide ion (Se²⁻)**: Located in Group 16, Period 4. It is larger than Br⁻ because it is in the same period but belongs to a group with larger anions.
- **Oxide ion (O²⁻)**: Located in Group 16, Period 2. It is smaller than Se²⁻ and Br⁻ due to being in a higher period.
- **Fluoride ion (F⁻)**: Located in Group 17, Period 2. It is the smallest of all the anions listed.

4. **Order the Anions by Size**: Based on the analysis, we can arrange the anions in order of decreasing ionic radius:
- I⁻ > Br⁻ > Se²⁻ > O²⁻ > F⁻

5. **Final Sequence**: The correct sequence of the anions in order of decreasing ionic radius is:
- I⁻ > Se²⁻ > Br⁻ > O²⁻ > F⁻