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 how the size of isoelectronic species \( F^{-} \), \( Ne \), and \( Na^{+} \) is affected, we need to analyze the factors influencing atomic size and the nature of these 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 Nuclear Charge**:
- The nuclear charge refers to the total charge of the nucleus due to protons.
- The number of protons in each species is:
- \( F^{-} \): 9 protons (atomic number 9)
- \( Ne \): 10 protons (atomic number 10)
- \( Na^{+} \): 11 protons (atomic number 11)

3. **Effect of Nuclear Charge on Size**:
- As the number of protons (nuclear charge) increases, the attraction between the nucleus and the electrons also increases.
- This increased attraction pulls the electrons closer to the nucleus, resulting in a smaller atomic size.

4. **Compare the Sizes**:
- Since \( F^{-} \) has the least nuclear charge (9 protons), it will be the largest.
- \( Ne \) has a higher nuclear charge (10 protons), making it smaller than \( F^{-} \).
- \( Na^{+} \), with the highest nuclear charge (11 protons), will be the smallest of the three.

5. **Conclusion**:
- The size of the isoelectronic species \( F^{-} \), \( Ne \), and \( Na^{+} \) is affected by the nuclear charge. The correct order of size from largest to smallest is:
\[ F^{-} > Ne > Na^{+} \]

### Final Answer:
The size of isoelectronic species \( F^{-} \), \( Ne \), and \( Na^{+} \) is affected by **nuclear charge**.

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