JEE Main 2023 · 31 January · morning · Single correct · +4 / −1
The correct increasing order of the ionic radii is
- (a) Cl⁻ < Ca²⁺ < K⁺ < S²⁻
- (b) K⁺ < S²⁻ < Ca²⁺ < Cl⁻
- (c) Ca²⁺ < K⁺ < Cl⁻ < S²⁻
- (d) S²⁻ < Cl⁻ < Ca²⁺ < K⁺
Answer and solution
Answer (c) Ca²⁺ < K⁺ < Cl⁻ < S²⁻
Solution
Chemistry ma’am · 3 steps
All four ions have 18 electrons, so just compare how many protons are pulling on them.
- Same electron count. S²⁻ (16 + 2), Cl⁻ (17 + 1), K⁺ (19 − 1) and Ca²⁺ (20 − 2) each have 18 electrons, the argon configuration. Ions like this are called isoelectronic.
- Protons decide size. With the same 18 electrons, more protons in the nucleus pull the electron cloud in harder. So the larger the nuclear charge, the smaller the ion.
Ion Protons Electrons Pull per electron Ca²⁺ 20 18 strongest, smallest K⁺ 19 18 strong Cl⁻ 17 18 weak S²⁻ 16 18 weakest, largest - So. Increasing radius follows decreasing proton number: Ca²⁺ < K⁺ < Cl⁻ < S²⁻, which is option (c).
The idea to keep. For isoelectronic ions, size falls as nuclear charge rises. Cations with high positive charge are smallest, and anions with high negative charge are largest.
Answer from the NTA final answer key. Solution by PrepWiser.