Which of the following isoelectronic series is correctly arranged in order of increasing ionic radius?

• A. F^{-} < O^{2-} < Na^{+} < Mg^{2+} < Al^{3+}
• B. Na^{+} < Mg^{2+} < Al^{3+} < F^{-} < O^{2-}
• C. O^{2-} < F^{-} < Na^{+} < Mg^{2+} < Al^{3+}
• D. Al^{3+} < Mg^{2+} < Na^{+} < F^{-} < O^{2-}

Answer: (D)

Verified step by step guidance

1. Step 1: Understand that isoelectronic species have the same number of electrons but different nuclear charges (number of protons). This means their size differences arise from the effective nuclear charge experienced by the electrons.
2. Step 2: Recognize that for isoelectronic ions, the ionic radius decreases as the positive charge on the nucleus increases because a higher nuclear charge pulls the electron cloud closer, reducing the size.
3. Step 3: Identify the ions given: \(\mathrm{Al^{3+}}\), \(\mathrm{Mg^{2+}}\), \(\mathrm{Na^{+}}\), \(\mathrm{F^{-}}\), and \(\mathrm{O^{2-}}\). All have the same number of electrons (10 electrons), making them isoelectronic.
4. Step 4: Arrange the ions in order of increasing nuclear charge: \(\mathrm{Al^{3+}}\) (13 protons), \(\mathrm{Mg^{2+}}\) (12 protons), \(\mathrm{Na^{+}}\) (11 protons), \(\mathrm{F^{-}}\) (9 protons), \(\mathrm{O^{2-}}\) (8 protons). The ionic radius decreases with increasing nuclear charge.
5. Step 5: Therefore, the correct order of increasing ionic radius is from the ion with the highest nuclear charge to the lowest: \(\mathrm{Al^{3+}} < \mathrm{Mg^{2+}} < \mathrm{Na^{+}} < \mathrm{F^{-}} < \mathrm{O^{2-}}\).