Question

For $Na^{+}, Mg^{2+}, F^{-}$ and $O^{2-}$; the correct order of increasing ionic radii is :

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

Answer 1

The Correct Option is C

To determine the correct order of increasing ionic radii among Na^+, Mg^{2+}, F^-, \text{and } O^{2-}, we need to consider their electron configurations and effective nuclear charge.

Ionic radii depend on the number of electrons and the nuclear charge of an ion. Among isoelectronic species (species with the same number of electrons), those with a higher positive charge have a smaller ionic radius. Conversely, those with a higher negative charge have a larger ionic radius due to increased electron-electron repulsion.

Let's analyze these ions, which all have the electron configuration of Neon (1s² 2s² 2p⁶, 10 electrons):

1. Mg^{2+}: This ion has lost two electrons compared to its neutral magnesium atom, leaving it with a higher positive nuclear charge relative to the number of electrons, resulting in the smallest radius among the group.
2. Na^+: This ion has lost one electron compared to its neutral sodium atom. It is larger than Mg^{2+} but smaller than the anions because it has a lesser positive charge.
3. F^−: This ion has gained an electron compared to its neutral fluorine atom, increasing the electron-electron repulsion slightly. It is larger than both of the cations.
4. O^{2-}: This ion has gained two electrons compared to its neutral oxygen atom, leading to the greatest increase in electron-electron repulsion, making it the largest among these ions.

Based on this analysis, the correct order of increasing ionic radii is:

Mg^{2+} < Na^+ < F^- < O^{2-}

This conclusion is consistent with the given correct answer.