Question

The correct order of increasing ionic radii is

• A. \(Mg^{2+} < Na^+ < F^– < O^{2–} < N^{3–}\)
• B. \(N^{3–} < O^{2–} < F^{–} < Na^{+} < Mg^{2+}\)
• C. \(F^– < Na^+ < O^{2–} < Mg^{2+} < N^{3–}\)
• D. \(Na^+ < F^– < Mg^{2+ }< O^{2–} < N^{3–}\)

Answer 1

The Correct Option is A

 To determine the correct order of increasing ionic radii among the given ions, it's essential to understand how ionic size is influenced by various factors such as charge and electron configuration.

The ions provided in the question are \(Mg^{2+}\), \(Na^+\), \(F^-\), \(O^{2-}\), and \(N^{3-}\). Generally, the size of an ion is affected by the number of electrons and the effective nuclear charge.

1. Effective Nuclear Charge: Higher positive charge on the nucleus (protons) means a stronger pull on the electron cloud, reducing the ionic radius.
2. Electron-Electron Repulsion: More electrons mean increased repulsion among them, potentially increasing the size.

Let's analyze each ion:

• \(
• \(Na^+\): 11 protons, 10 electrons - Similar electron count to \(Mg^{2+}\), but one less proton provides a weaker pull, thus, it’s larger than \(Mg^{2+}\).
• \(F^-: 9\) protons, 10 electrons - More electrons for fewer protons compared to \(Na^+\), leading to greater repulsion and hence, larger size.
• \(O^{2-}\): 8 protons, 10 electrons - Larger due to more electron-electron repulsion and fewer protons to control them compared to \(F^-\).
• \(N^{3-}\): 7 protons, 10 electrons - Max electron count and lowest proton number in this group, resulting in the largest radius.

Arranging them in increasing order of ionic radii, we get:

\(Mg^{2+} < Na^+ < F^- < O^{2–} < N^{3–}\)

Therefore, the correct order is: \(Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}\).

This order accounts for the balance between proton count (and thus nuclear pull) and electron repulsions.