Question

The correct order of ionic radii for the ions, P\(^{3-}\), S\(^{2-}\), Ca\(^{2+}\), K\(^{+}\), Cl\(^{-}\) is:

• A. K\(^{+}\) \(>\) Ca\(^{2+}\) \(>\) P\(^{3-}\) \(>\) S\(^{2-}\) \(>\) Cl\(^{-}\)
• B. P\(^{3-}\) \(>\) S\(^{2-}\) \(>\) Cl\(^{-}\) \(>\) K\(^{+}\) \(>\) Ca\(^{2+}\)
• C. P\(^{3-}\) \(>\) S\(^{2-}\) \(>\) Cl\(^{-}\) \(>\) Ca\(^{2+}\) \(>\) K\(^{+}\)
• D. Cl\(^{-}\) \(>\) S\(^{2-}\) \(>\) P\(^{3-}\) \(>\) Ca\(^{2+}\) \(>\) K\(^{+}\)

Hint:

For isoelectronic species, remember this simple rule: More protons, smaller size. Anions will always be larger than cations in an isoelectronic series because they have fewer protons pulling the same number of electrons.

Answer 1

The Correct Option is B

To determine the correct order of ionic radii for the given ions, we need to understand how ionic radii change as electrons are added or removed from an atom.

1. The ionic radius is typically larger when an atom gains electrons to form an anion and smaller when it loses electrons to form a cation. This is primarily because the addition of electrons increases electron-electron repulsion and shielding, leading to a larger radius. Conversely, losing electrons reduces electron-electron repulsion and decreases the radius.
2. We have the following ions to consider: \( \text{P}^{3-} \), \( \text{S}^{2-} \), \( \text{Cl}^{-} \), \( \text{K}^{+} \), \( \text{Ca}^{2+} \).
3. Let's analyze each ion:
   • P^{3-} : Phosphorus gains three electrons, resulting in a large ionic radius.
   • S^{2-} : Sulfur gains two electrons, which is still larger than atoms of Cl but smaller than P.
   • Cl^{-} : Chlorine gains one electron, which makes the ionic radius smaller than S and P anions.
   • K^{+} : Potassium loses one electron, leading to a smaller ionic radius.
   • Ca^{2+} : Calcium loses two electrons, giving it the smallest ionic radius among these ions.
4. Based on the above analysis, the correct order of ionic radii from largest to smallest is:
   • P^{3-} \gt S^{2-} \gt Cl^{-} \gt K^{+} \gt Ca^{2+} .

Thus, the correct answer is: P^{3-} \gt S^{2-} \gt Cl^{-} \gt K^{+} \gt Ca^{2+} .