Arrange the given species `F, F^(-), O and O^(-2)` in the order of their radii

Options

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

Answer

To arrange the species \( F, F^-, O, \) and \( O^{2-} \) in order of their radii, we need to consider the number of protons and electrons in each species, as well as the effective nuclear charge.

### Step-by-Step Solution:

1. **Identify the Species and Their Electron Configurations:**
- **Fluorine (F)**: Atomic number = 9, Electron configuration = \( 1s^2 2s^2 2p^5 \) (Total electrons = 9)
- **Fluoride Ion (F^-)**: Gains one electron, Total electrons = 10
- **Oxygen (O)**: Atomic number = 8, Electron configuration = \( 1s^2 2s^2 2p^4 \) (Total electrons = 8)
- **Oxide Ion (O^{2-})**: Gains two electrons, Total electrons = 10

2. **Determine the Number of Protons and Electrons:**
- For \( F \): 9 protons, 9 electrons
- For \( F^- \): 9 protons, 10 electrons
- For \( O \): 8 protons, 8 electrons
- For \( O^{2-} \): 8 protons, 10 electrons

3. **Analyze the Effective Nuclear Charge:**
- The effective nuclear charge (Z_eff) is the net positive charge experienced by electrons in an atom. As we move from left to right in the periodic table, the Z_eff increases due to the increase in the number of protons.
- \( F \) has a higher Z_eff than \( O \) because it has more protons (9 vs 8) while having the same number of electron shells.

4. **Compare the Radii:**
- **Neutral Atoms vs Anions**: Anions are larger than their parent atoms because the addition of electrons increases electron-electron repulsion, which expands the electron cloud.
- **Comparison of \( F \) and \( O \)**: Since \( F \) has a higher Z_eff than \( O \), \( F \) is smaller than \( O \).
- **Comparison of Anions**:
- \( F^- \) has 10 electrons and 9 protons, while \( O^{2-} \) has 10 electrons and 8 protons. The greater number of protons in \( F^- \) pulls the electrons closer, making \( F^- \) smaller than \( O^{2-} \).
- Therefore, \( O^{2-} \) is larger than both \( F^- \) and \( O \).

5. **Final Order of Radii**:
- From the analysis, we can conclude the order of increasing radius is:
\[ F < O < F^- < O^{2-} \]

### Final Answer:
The order of the species by increasing radius is:
\[ F < O < F^- < O^{2-} \]