Question

$Na^+, Mg^{2+}, Al^3+ \, and \, Si^{4+}$ are isoelectronic. The order of their ionic size is

• A. $Na^+ > Mg^{2+} < Al^{3+} < Si^{4+}$
• B. $Na^+ < Mg^{2+} > Al^{3+} > Si^{4+}$
• C. $Na^+ > Mg^{2+} > Al^{3+} > Si^{4+}$
• D. $Na^+ < Mg^{2+} > Al^{3+} < Si^{4+}$

Answer 1

The Correct Option is C

 To determine the order of ionic size for the isoelectronic ions \( \text{Na}^+, \text{Mg}^{2+}, \text{Al}^{3+}, \text{and Si}^{4+} \), we need to consider the nuclear charge effect on these ions. Here's a step-by-step breakdown of the reasoning:

1. Definition of Isoelectronic Ions: Isoelectronic species have the same number of electrons. In this case, all these ions have 10 electrons (same as the neon atom).
2. Understanding Nuclear Charge: Though these ions have the same number of electrons, they differ in the number of protons in their nuclei. The increasing nuclear charge (from \( \text{Na}^+ \) to \( \text{Si}^{4+} \)) attracts the shared electron cloud more strongly, reducing the ion's radius.
3. Identify the Proton Numbers:
   • \( \text{Na}^+ \) has 11 protons.
   • \( \text{Mg}^{2+} \) has 12 protons.
   • \( \text{Al}^{3+} \) has 13 protons.
   • \( \text{Si}^{4+} \) has 14 protons.
4. Analysis of Ionic Sizes: As the number of protons increases, the effective nuclear charge also increases, pulling the electron cloud closer to the nucleus. Therefore, as the number of protons increases from \( \text{Na}^+ \) to \( \text{Si}^{4+} \), the ionic radii decrease.
   • \( \text{Na}^+ \) has the largest ionic radius because of the lowest nuclear charge among the given ions.
   • \( \text{Si}^{4+} \) has the smallest ionic radius due to the highest nuclear charge.
5. Conclusion: The order of their ionic sizes is \( \text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+} > \text{Si}^{4+} \).

The correct answer is therefore: \(\text{Na}^+ > \text{Mg}^{2+} > \text{Al}^{3+} > \text{Si}^{4+}\).