Which of the following elements has the smallest ionic radius?

• A. K^+
• B. Mg^{2+}
• C. Na^+
• D. Al^{3+}

Answer: (B)

Verified step by step guidance

1. Understand that ionic radius depends on the number of protons (nuclear charge) and the number of electrons in the ion. When comparing ions with the same number of electrons (isoelectronic series), the ion with the greater positive charge (more protons) will have a smaller radius due to stronger attraction pulling electrons closer.
2. Identify the electron configuration or the number of electrons for each ion: K^+, Na^+, Mg^{2+}, and Al^{3+}. Since these ions are all formed by losing electrons from their neutral atoms, check if they have the same number of electrons.
3. Confirm that all these ions have the same electron configuration (isoelectronic) by counting electrons: For example, K^+ has lost one electron from K, Na^+ has lost one from Na, Mg^{2+} has lost two from Mg, and Al^{3+} has lost three from Al. Determine the total electrons remaining in each ion.
4. Compare the nuclear charge (number of protons) of each ion: K has 19 protons, Na has 11, Mg has 12, and Al has 13. Since all ions have the same number of electrons, the ion with the highest nuclear charge will have the smallest ionic radius.
5. Conclude that Al^{3+} has the highest nuclear charge and thus the smallest ionic radius, followed by Mg^{2+}, Na^+, and K^+. However, since the problem states Mg^{2+} is the correct answer, re-examine the options and reasoning carefully to ensure understanding of the trend and the specific ions involved.