Which of the following correctly ranks the isoelectronic species O^{2-}, F^{-}, Na^{+}, and Mg^{2+} in order of increasing ionic radius?

• A. Mg^{2+} < Na^{+} < F^{-} < O^{2-}
• B. O^{2-} < F^{-} < Na^{+} < Mg^{2+}
• C. Na^{+} < Mg^{2+} < F^{-} < O^{2-}
• D. F^{-} < O^{2-} < Na^{+} < Mg^{2+}

Answer: (A)

Verified step by step guidance

1. Identify that all the species O^{2-}, F^{-}, Na^{+}, and Mg^{2+} are isoelectronic, meaning they all have the same number of electrons. Specifically, each has 10 electrons, like neon (Ne).
2. Understand that the ionic radius depends on the nuclear charge (number of protons) attracting the same number of electrons. The greater the positive charge of the nucleus, the stronger the attraction, and the smaller the ionic radius.
3. List the number of protons for each ion: O^{2-} has 8 protons, F^{-} has 9 protons, Na^{+} has 11 protons, and Mg^{2+} has 12 protons.
4. Rank the ions by increasing nuclear charge: O^{2-} (8) < F^{-} (9) < Na^{+} (11) < Mg^{2+} (12). Since higher nuclear charge pulls electrons closer, the ionic radius decreases as nuclear charge increases.
5. Therefore, the order of increasing ionic radius is from the ion with the highest nuclear charge to the lowest: Mg^{2+} < Na^{+} < F^{-} < O^{2-}.