JEE Main 2021 · 27 August · evening · Single correct · +4 / −1
The correct order of ionic radii for the ions, P³⁻, S²⁻, Ca²⁺, K⁺, Cl⁻ is :
- (a) P³⁻ > S²⁻ > Cl⁻ > K⁺ > Ca²⁺
- (b) Cl⁻ > S²⁻ > P³⁻ > Ca²⁺ > K⁺
- (c) P³⁻ > S²⁻ > Cl⁻ > Ca²⁺ > K⁺
- (d) K⁺ > Ca²⁺ > P³⁻ > S²⁻ > Cl⁻
Answer and solution
Answer (a) P³⁻ > S²⁻ > Cl⁻ > K⁺ > Ca²⁺
Solution
Chemistry ma’am · 3 steps
All five ions have 18 electrons, so the one with the fewest protons is the biggest.
- Same electron count. P, S, Cl, K and Ca have atomic numbers 15 to 20. After gaining or losing electrons, every one of these ions has 18 electrons, the argon configuration. They are isoelectronic.
- Protons decide size. With the same 18 electrons, more protons pull the electron cloud in harder. So radius falls as nuclear charge Z rises.
Ion Protons (Z) Electrons Size P³⁻ 15 18 largest S²⁻ 16 18 second Cl⁻ 17 18 third K⁺ 19 18 fourth Ca²⁺ 20 18 smallest - So. The order is P³⁻ > S²⁻ > Cl⁻ > K⁺ > Ca²⁺, which is option (a).
The idea to keep. For an isoelectronic series, line the ions up by protons. More protons for the same electrons means a smaller ion, so the most negative ion is biggest and the most positive is smallest.
Answer from the NTA final answer key. Solution by PrepWiser.