Question

Arrange the isoelectronic species in increasing order of ionic radius:
\(\text{Al}^{3+}\), \(\text{Na}^{+}\), \(\text{F}^{-}\), \(\text{O}^{2-}\)

• A. \(\text{Al}^{3+}<\text{Na}^{+}<\text{F}^{-}<\text{O}^{2-}\)
• B. \(\text{O}^{2-}<\text{F}^{-}<\text{Na}^{+}<\text{Al}^{3+}\)
• C. \(\text{Na}^{+}<\text{F}^{-}<\text{Al}^{3+}<\text{O}^{2-}\)
• D. \(\text{F}^{-}<\text{Na}^{+}<\text{O}^{2-}<\text{Al}^{3+}\)

Hint:

For isoelectronic ions, simply arrange them in order of decreasing atomic number to get the increasing order of ionic radius.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Isoelectronic species are those that possess the exact same number of electrons. The size of an ion in an isoelectronic series is inversely proportional to its effective nuclear charge (which corresponds to the number of protons or the atomic number, \(Z\)).
Step 2: Key Formula or Approach:
For a given isoelectronic series, the theoretical approach is based on the relation:
\[ \text{Ionic Radius} \propto \frac{1}{\text{Nuclear Charge (Z)}} \]
Step 3: Detailed Explanation:
First, let's verify that all species are isoelectronic by calculating their total number of electrons:
- \(\text{Al}^{3+}\): Atomic number (\(Z\)) = 13. Electrons = \(13 - 3 = 10\)
- \(\text{Na}^{+}\): Atomic number (\(Z\)) = 11. Electrons = \(11 - 1 = 10\)
- \(\text{F}^{-}\): Atomic number (\(Z\)) = 9. Electrons = \(9 + 1 = 10\)
- \(\text{O}^{2-}\): Atomic number (\(Z\)) = 8. Electrons = \(8 + 2 = 10\)
All species have 10 electrons.
Now, compare their nuclear charge (\(Z\)):
\(Z_{\text{Al}} (13)>Z_{\text{Na}} (11)>Z_{\text{F}} (9)>Z_{\text{O}} (8)\)
A higher nuclear charge exerts a greater pull on the same number of electrons, shrinking the electron cloud. Therefore, the ion with the highest positive charge will be the smallest, and the ion with the highest negative charge will be the largest.
Thus, the increasing order of ionic radius is:
\(\text{Al}^{3+}<\text{Na}^{+}<\text{F}^{-}<\text{O}^{2-}\)
Step 4: Final Answer:
The correct option is i) \(\text{Al}^{3+}<\text{Na}^{+}<\text{F}^{-}<\text{O}^{2-}\).