Concept:For isoelectronic species (ions with the exact same number of electrons), the radius decreases as the number of protons (nuclear charge) increases.
Formula:$$ \text{Ionic Radius} \propto \frac{1}{\text{Nuclear Charge (Z)}} $$
Solution:- \( \text{Na}^+ \) (11 protons), \( \text{Mg}^{2+} \) (12 protons), and \( \text{Al}^{3+} \) (13 protons) all have exactly 10 electrons (isoelectronic with Neon).
- Because Aluminum has the highest number of positive protons (13), its nucleus pulls the 10 surrounding electrons inward with the greatest force.
- This massive electrostatic attraction shrinks the electron cloud, giving \( \text{Al}^{3+} \) the smallest ionic radius.
Why other options are incorrect:- Option A & Option B: Have fewer protons, meaning weaker nuclear pull and larger radii.
- Option D: \( \text{Mg}^+ \) has 11 electrons, so it isn't fully isoelectronic and is larger than \( \text{Mg}^{2+} \).
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