Question

For Na$^+$, Mg$^{2+}$, Al$^{3+}$ and F$^-$, the CORRECT order of ionic radii is

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

Hint:

In isoelectronic ions, as nuclear charge increases, size decreases due to stronger electrostatic attraction on the same electron cloud.

Answer 1

The Correct Option is C

Step 1: Concept of isoelectronic species.
Na$^+$, Mg$^{2+}$, Al$^{3+}$, and F$^-$ are isoelectronic (each has 10 electrons). The ionic size depends mainly on the nuclear charge (Z).
Step 2: Effect of nuclear charge.
As nuclear charge increases, the same number of electrons are pulled more strongly toward the nucleus, decreasing the ionic radius.
Step 3: Order of Z (atomic number).
F (9)<Na (11)<Mg (12)<Al (13)
Hence, radius order (inverse relation): \[ \text{F}^->\text{Na}^+>\text{Mg}^{2+}>\text{Al}^{3+}. \] Step 4: Conclusion.
The correct order of ionic radii is F$^-$>Na$^+$>Mg$^{2+}$>Al$^{3+}$.