Ionic Radius Comparison: Na⁺, Mg⁺², Al⁺³, Si⁺⁴

The question asks to identify the ion with the smallest ionic radius among the given options: Mg⁺², Na⁺, Si⁺⁴, and Al⁺³.

Isoelectronic Ions Analysis

All the given ions are isoelectronic, meaning they have the same number of electrons. Let's determine the number of electrons and protons (atomic number, Z) for each:

• Na⁺: Atomic number (Z) = 11, Electrons = 10 (11 - 1). Configuration: $1s^2 2s^2 2p^6$.
• Mg⁺²: Atomic number (Z) = 12, Electrons = 10 (12 - 2). Configuration: $1s^2 2s^2 2p^6$.
• Al⁺³: Atomic number (Z) = 13, Electrons = 10 (13 - 3). Configuration: $1s^2 2s^2 2p^6$.
• Si⁺⁴: Atomic number (Z) = 14, Electrons = 10 (14 - 4). Configuration: $1s^2 2s^2 2p^6$.

Reasoning for Ionic Radius

For isoelectronic species, the ionic radius decreases as the nuclear charge (number of protons) increases. A higher nuclear charge pulls the electron shells closer to the nucleus more effectively.

• Na⁺ has Z=11.
• Mg⁺² has Z=12.
• Al⁺³ has Z=13.
• Si⁺⁴ has Z=14.

Comparing the nuclear charges, Si⁺⁴ has the highest nuclear charge (Z=14). Therefore, it attracts its 10 electrons most strongly, resulting in the smallest ionic radius.

Conclusion

The ion with the least ionic radius is Si⁺⁴ due to its highest nuclear charge among the isoelectronic species.