The size of isoelectronic species `F^(ɵ)`, `Ne`, and `Na^(o+)` is affected by

Options

A. Nuclear charge `(Z)`
B. Valence principal quantum number `(n)`
C. Electron-electron interaction in the outer orbitals
D. None of the factors because their size the same

Answer

To determine the factors affecting the size of isoelectronic species \( F^{-} \), \( Ne \), and \( Na^{+} \), we need to analyze their electronic configurations and the nuclear charge associated with each species.

### Step-by-Step Solution:

1. **Identify the Isoelectronic Species**:
- Isoelectronic species are atoms or ions that have the same number of electrons.
- In this case, \( F^{-} \), \( Ne \), and \( Na^{+} \) all have 10 electrons.

2. **Determine the Number of Protons**:
- **Fluorine (\( F \))**: Atomic number = 9, thus \( F^{-} \) has 9 protons.
- **Neon (\( Ne \))**: Atomic number = 10, thus \( Ne \) has 10 protons.
- **Sodium (\( Na \))**: Atomic number = 11, thus \( Na^{+} \) has 11 protons (after losing one electron).

3. **Analyze Nuclear Charge**:
- The nuclear charge is the total charge of the nucleus due to protons.
- The more protons present in the nucleus, the greater the nuclear charge, which affects the attraction between the nucleus and the electrons.

4. **Compare the Nuclear Charges**:
- For \( F^{-} \): 9 protons, 10 electrons → Nuclear charge = +9
- For \( Ne \): 10 protons, 10 electrons → Nuclear charge = +10
- For \( Na^{+} \): 11 protons, 10 electrons → Nuclear charge = +11

5. **Determine the Size Based on Nuclear Charge**:
- As the nuclear charge increases, the size of the species decreases due to a stronger attraction between the nucleus and the electrons.
- Therefore, the order of size will be:
- \( F^{-} \) (largest, due to the lowest nuclear charge)
- \( Ne \) (intermediate size)
- \( Na^{+} \) (smallest, due to the highest nuclear charge)

6. **Conclusion**:
- The size of the isoelectronic species is primarily affected by the **nuclear charge**. The greater the nuclear charge, the smaller the size of the ion or atom.

### Final Order of Sizes:
- \( F^{-} > Ne > Na^{+} \)