Question

The ions $O ^{2-}, F ^{-}, Na ^{+}, Mg ^{2+}$ and $Al ^{3+}$ are isoelecronic. Their ionic radii show :

• A. An increase from $O ^{2-}$ to $F ^{-}$ and then decrease from $Na ^{+}$ to $Al ^{3+}$
• B. An decrease from $O ^{2-}$ to $F ^{-}$ and then increase from $N a ^{+}$ to $Al ^{3+}$
• C. A significant increase from $O ^{2-}$ to $Al ^{3+}$
• D. A significant decrease from $O ^{2-}$ to $Al ^{3+}$

Answer 1

The Correct Option is D

The ions \(O ^{2-}\), \(F ^{-}\), \(Na ^{+}\), \(Mg ^{2+}\), and \(Al ^{3+}\) are isoelectronic, meaning they all have the same number of electrons. However, their ionic radii differ due to the differing nuclear charges (i.e., the number of protons) in their respective nuclei.

1. Understanding Nuclear Charge: As the positive charge of the nucleus (i.e., atomic number) increases, the electrons are pulled closer to the nucleus. This results in a decrease in the ionic radius.
   • \(O ^{2-}\) has 8 protons.
   • \(F ^{-}\) has 9 protons.
   • \(Na ^{+}\) has 11 protons.
   • \(Mg ^{2+}\) has 12 protons.
   • \(Al ^{3+}\) has 13 protons.
2. Trend in Ionic Radii: Since all these ions are isoelectronic (same number of electrons), the size of the ion largely depends on the nuclear charge pulling on the same electron cloud:
   • \(O ^{2-}\) has the fewest protons, so it will have the largest radius.
   • As we move from \(O ^{2-}\) to \(Al ^{3+}\), the nuclear charge increases, pulling the electrons closer, thereby decreasing the radius.
3. Conclusion: Thus, the ionic radii show a significant decrease as we move from \(O ^{2-}\) to \(Al ^{3+}\).

Therefore, the correct answer is: A significant decrease from \(O ^{2-}\) to \(Al ^{3+}\).