Question

Which of the following isoelectronic species has the smallest radius?

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

Hint:

In isoelectronic species, the size of the species decreases as the nuclear charge increases. This is because a higher nuclear charge results in a stronger attraction between the nucleus and the electrons, pulling the electrons closer to the nucleus.

Answer 1

Answer: (E)

To determine which isoelectronic species has the smallest radius, we need to understand that isoelectronic species are atoms or ions that have the same number of electrons but different numbers of protons. The key factor that determines the size of these species is the nuclear charge, which is the number of protons in the nucleus.
When comparing isoelectronic species:
- The species with the highest nuclear charge (Z) will have the smallest radius because the greater number of protons pulls the electrons closer to the nucleus.
Let's examine the species:
1. \( {Mg}^{2+} \) has 10 electrons and 12 protons.
2. \( {F}^- \) has 10 electrons and 9 protons.
3. \( {Na}^+ \) has 10 electrons and 11 protons.
4. \( {O}^{2-} \) has 10 electrons and 8 protons.
5. \( {Al}^{3+} \) has 10 electrons and 13 protons.
Since \( {Al}^{3+} \) has the highest number of protons (13), it will have the smallest radius among these species.
Thus, the correct answer is: \[ \boxed{{E) } {Al}^{3+}} \]