The order of size is :

Options

A. `S^(-2) lt Cl^(-) lt O^(2-) lt F^(-)`
B. `Cl^(-) gt S^(-2) gt O^(-2) gt F^(-)`
C. `S^(-2) gt O^(-2) gt Cl^(-) gt F^(-)`
D. `S^(-2) gt O^(-2) gt F^(-) gt Cl^(-)`

Answer

To determine the order of size for the given anions (S²⁻, O²⁻, Cl⁻, and F⁻), we need to consider the following factors:

### Step-by-Step Solution:

1. **Identify the Anions**: The anions we are comparing are:
- Sulfide (S²⁻)
- Oxide (O²⁻)
- Chloride (Cl⁻)
- Fluoride (F⁻)

2. **Understand the Charge Effect**: Anions are formed when atoms gain electrons. The more electrons an atom gains, the larger the anion becomes due to increased electron-electron repulsion. Thus, anions with higher negative charges will generally be larger.

3. **Consider the Periodic Trends**:
- As we move down a group in the periodic table, the size of the elements increases due to the addition of electron shells.
- As we move across a period from left to right, the size decreases due to an increase in effective nuclear charge, which pulls the electrons closer to the nucleus.

4. **Analyze Each Anion**:
- **S²⁻**: Sulfur gains 2 electrons. It has a higher negative charge than the others, leading to a larger size.
- **O²⁻**: Oxygen also gains 2 electrons but has a smaller atomic radius than sulfur, so O²⁻ is smaller than S²⁻.
- **Cl⁻**: Chlorine gains 1 electron. It is larger than F⁻ due to being lower in the group (Group 17).
- **F⁻**: Fluorine gains 1 electron and is the smallest among these anions.

5. **Order the Sizes**: Based on the above analysis, we can arrange the anions in order of size:
- **Largest**: S²⁻
- O²⁻
- Cl⁻
- **Smallest**: F⁻

Thus, the final order of size from largest to smallest is:
**S²⁻ > O²⁻ > Cl⁻ > F⁻**