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

Ionic size is dictated by electron count and effective nuclear charge. Below is an analysis of specific ions:

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

Neutral fluorine (F) possesses 9 protons and 9 electrons. Upon acquiring an electron to form the anion \( \text{F}^{-} \), it gains a 10th electron, maintaining 9 protons. This augmentation in electron-electron repulsion expands the electron cloud, leading to a larger ionic radius than neutral fluorine.

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

Sodium (Na), magnesium (Mg), and aluminum (Al) are metals that form cations by losing electrons. This electron loss elevates the effective nuclear charge, as fewer electrons shield the nucleus's positive charge. The intensified nuclear attraction to the remaining electrons shrinks the ionic radius. Detailed examination follows:

• \( \text{Na}^{+} \): Neutral sodium has 11 protons and 11 electrons. Forming \( \text{Na}^{+} \) involves losing one electron, resulting in 10 electrons, isoelectronic with neon. The nucleus still has 11 protons, and its increased pull on the 10 electrons reduces the ionic radius.
• \( \text{Mg}^{2+} \): Neutral magnesium has 12 protons and 12 electrons. Losing two electrons to form \( \text{Mg}^{2+} \) leaves 10 electrons, also isoelectronic with neon. With 12 protons, the heightened nuclear charge draws the electrons closer, further diminishing the ionic radius compared to \( \text{Na}^{+} \).
• \( \text{Al}^{3+} \): Neutral aluminum has 13 protons and 13 electrons. Losing three electrons to form \( \text{Al}^{3+} \) results in 10 electrons, isoelectronic with neon. The nucleus possesses 13 protons, exerting the strongest attraction on the electrons, thus yielding the smallest ionic radius among these cations.

Conclusion

Ion size is governed by charge and electron number. Anions (e.g., \( \text{F}^{-} \)) generally exhibit larger radii due to increased electron-electron repulsion from added electrons. Cations (e.g., \( \text{Na}^{+} \), \( \text{Mg}^{2+} \), \( \text{Al}^{3+} \)) display smaller radii due to increased effective nuclear charge from electron loss, drawing remaining electrons nearer the nucleus.

Consequently, \( \text{F}^{-} \) is the largest ion, followed by \( \text{Na}^{+} \), \( \text{Mg}^{2+} \), and \( \text{Al}^{3+} \) in descending order of size.