Question

Which of the following has the maximum size?

• A. \( \text{Al}^{3+} \)
• B. \( \text{Mg}^{2+} \)
• C. \( \text{F}^{-} \)
• D. \( \text{Na}^{+} \)

Hint:

In general, cations are smaller than their neutral atoms, and anions are larger due to the extra electron-electron repulsion.

Answer 1

The Correct Option is C

The ionic size of an ion is determined by a combination of the number of electrons and the effective nuclear charge (the charge felt by the electrons from the nucleus). Here's a more detailed explanation of the given ions:

Step 1: Ionic Size of \( \text{F}^{-} \)

Fluorine (F) in its neutral form has 9 protons and 9 electrons. When it gains one electron to form the anion \( \text{F}^{-} \), the number of electrons increases to 10, while the number of protons remains the same. This results in an increased electron-electron repulsion, which causes the electron cloud to expand. As a result, \( \text{F}^{-} \) has a larger ionic radius compared to its neutral form.

Step 2: Ionic Size of Cations \( \text{Na}^{+} \), \( \text{Mg}^{2+} \), and \( \text{Al}^{3+} \)

On the other hand, sodium (\( \text{Na} \)), magnesium (\( \text{Mg} \)), and aluminum (\( \text{Al} \)) are all metals that lose electrons to form cations. When an atom loses electrons, it results in a higher effective nuclear charge because there are fewer electrons to shield the positive charge of the nucleus. This increased attraction between the nucleus and the remaining electrons causes the ionic radius to decrease. Here's a closer look at each cation:

• \( \text{Na}^{+} \): Sodium has 11 protons and 11 electrons in its neutral state. When it loses one electron to form \( \text{Na}^{+} \), it becomes isoelectronic with neon (with 10 electrons), but the nucleus still has 11 protons. The greater nuclear charge pulls the electrons closer, resulting in a smaller ionic radius.
• \( \text{Mg}^{2+} \): Magnesium has 12 protons and 12 electrons in its neutral state. When it loses two electrons to form \( \text{Mg}^{2+} \), it becomes isoelectronic with neon, but with 12 protons. The increased positive charge pulls the electrons even closer, further reducing the ionic radius compared to \( \text{Na}^{+} \).
• \( \text{Al}^{3+} \): Aluminum has 13 protons and 13 electrons in its neutral state. When it loses three electrons to form \( \text{Al}^{3+} \), it becomes isoelectronic with neon, but with 13 protons. The higher nuclear charge results in the greatest attraction to the electrons, leading to the smallest ionic radius among the three cations.

Conclusion

The size of an ion is determined by its charge and the number of electrons. Anions (like \( \text{F}^{-} \)) tend to have larger ionic radii than cations because the addition of electrons increases electron-electron repulsion, expanding the electron cloud. In contrast, cations (such as \( \text{Na}^{+} \), \( \text{Mg}^{2+} \), and \( \text{Al}^{3+} \)) have smaller ionic radii because the removal of electrons increases the effective nuclear charge, pulling the electrons closer to the nucleus.

Therefore, \( \text{F}^{-} \) has the largest size among the given ions, followed by \( \text{Na}^{+} \), \( \text{Mg}^{2+} \), and \( \text{Al}^{3+} \) in decreasing order of size.