Question

The ionic radii of $K^+$, $Na^+$, $Al^{3+}$ and $Mg^{2+}$ are in the order :

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

Hint:

For isoelectronic species, the greater the positive charge, the smaller the ionic radius because the nucleus pulls the same number of electrons more strongly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Ionic radius depends on the number of shells and the effective nuclear charge.
For isoelectronic species (ions with the same number of electrons), the radius decreases as the atomic number increases.
Step 2: Key Formula or Approach:
1. Identify isoelectronic ions.
2. Apply the rule: Radius \(\propto \frac{1}{\text{Atomic Number (Z)}}\).
3. Compare ions from different periods based on shell count.
Step 3: Detailed Explanation:
1. \(Na^+\), \(Mg^{2+}\), and \(Al^{3+}\) all have 10 electrons (isoelectronic).
- Atomic numbers: \(Na(11), Mg(12), Al(13)\).
- Order: \(Al^{3+}<Mg^{2+}<Na^+\).
2. \(K^+\) has 18 electrons and belongs to a higher period (3rd shell for the ion, vs 2nd shell for the others).
- Ions with more shells are significantly larger.
- Thus, \(K^+\) is the largest among the four.
Combining these, we get: \(Al^{3+}<Mg^{2+}<Na^+<K^+\).
Step 4: Final Answer:
The correct order is \(Al^{3+}<Mg^{2+}<Na^+<K^+\).