Question

Arrange the following ions in the increasing order of their radii:
\[ O^{2-}, \; N^{3-}, \; F^{-}, \; Mg^{2+} \]

• A. IV < II < I < III
• B. III < IV < II < I
• C. IV < III < I < II
• D. IV < II < I < IV

Hint:

For isoelectronic species, ionic radius decreases with increasing atomic number (Z).

Answer 1

The Correct Option is C

Step 1: Understand the concept of ionic radius trend.
Ionic radius depends on effective nuclear charge and number of electrons. For isoelectronic species, the ion with higher nuclear charge has smaller radius because electrons are pulled more strongly toward the nucleus. Here, all ions are isoelectronic (10 electrons each), so comparison is purely based on nuclear charge.

Step 2: Determine electron configuration similarity.
Each ion has 10 electrons: \[ N^{3-}, O^{2-}, F^{-}, Mg^{2+} \rightarrow 10 \; \text{electrons (Ne-like configuration)} \] So all are isoelectronic with Neon. Therefore, radius depends only on number of protons (atomic number).

Step 3: Arrange based on nuclear charge (Z).
Atomic numbers: \[ N=7,\; O=8,\; F=9,\; Mg=12 \] Higher nuclear charge pulls electrons more strongly, reducing radius. So: \[ Mg^{2+} \text{ is smallest (Z=12)} \] followed by: \[ F^{-} (Z=9), \quad O^{2-} (Z=8), \quad N^{3-} (Z=7) \]

Step 4: Determine increasing order of radius.
Since radius increases as nuclear charge decreases: \[ Mg^{2+} \lt F^{-} \lt O^{2-} \lt N^{3-} \] So in terms of given labels: \[ IV \lt III \lt I \lt II \]

Step 5: Physical reasoning behind trend.
Cations are always smaller than anions due to loss of electrons and increased effective nuclear charge per electron. Among anions, more negative charge increases electron-electron repulsion, increasing size. Hence \(N^{3-}\) is largest and \(Mg^{2+}\) is smallest.

Step 6: Final conclusion.
Thus, the correct increasing order of ionic radii is: \[ \boxed{IV \lt III \lt I \lt II} \]