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.
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