To determine which of the ions \( Na^+ \), \( Ne \), and \( F^- \) is the largest in size, we need to analyze their electronic configurations and the concept of isoelectronic species.
### Step-by-Step Solution:
1. **Identify the electronic configurations:**
- Sodium (Na) has an atomic number of 11, so its electronic configuration is \( 1s^2 2s^2 2p^6 3s^1 \).
- When sodium loses one electron to form \( Na^+ \), its electronic configuration becomes \( 1s^2 2s^2 2p^6 \) (10 electrons).
- Neon (Ne) has an atomic number of 10, so its electronic configuration is \( 1s^2 2s^2 2p^6 \) (10 electrons).
- Fluorine (F) has an atomic number of 9, and when it gains one electron to form \( F^- \), its electronic configuration also becomes \( 1s^2 2s^2 2p^6 \) (10 electrons).
2. **Determine that all species are isoelectronic:**
- Since \( Na^+ \), \( Ne \), and \( F^- \) all have the same number of electrons (10), they are isoelectronic.
3. **Understand the trend in ionic size:**
- In isoelectronic species, the size of the ions decreases with increasing nuclear charge. This is because a higher nuclear charge pulls the electrons closer to the nucleus.
- The nuclear charges are as follows:
- \( Na^+ \): +11 (11 protons)
- \( Ne \): +10 (10 protons)
- \( F^- \): +9 (9 protons)
4. **Compare the sizes:**
- Since \( F^- \) has the least nuclear charge, it will have the largest size due to less effective nuclear pull on the electrons.
- \( Ne \) has a higher nuclear charge than \( F^- \) but lower than \( Na^+ \), so it will be smaller than \( F^- \) but larger than \( Na^+ \).
- \( Na^+ \) has the highest nuclear charge and will be the smallest among the three.
5. **Conclusion:**
- Therefore, the order of size from largest to smallest is: \( F^- > Ne > Na^+ \).
- The largest ion among \( Na^+ \), \( Ne \), and \( F^- \) is \( F^- \).
### Final Answer:
The largest in size out of \( Na^+ \), \( Ne \), and \( F^- \) is \( F^- \).
