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NEET CHEMISTRYEasy

The correct order of the decreasing ionic radii among the following isoelectronic species is:

A

Ca²⁺ > K⁺ > S²⁻ > Cl⁻

B

Cl⁻ > S²⁻ > Ca²⁺ > K⁺

C

S²⁻ > Cl⁻ > K⁺ > Ca²⁺

D

K⁺ > Ca²⁺ > Cl⁻ > S²⁻

Step-by-Step Solution

The species S²⁻, Cl⁻, K⁺, and Ca²⁺ are isoelectronic, meaning they all possess the same number of electrons (18 electrons).

  1. Concept: For isoelectronic ions, the ionic radius decreases as the nuclear charge (atomic number, Z) increases. A higher nuclear charge exerts a stronger electrostatic attraction on the same number of electrons, pulling the electron cloud closer to the nucleus .
  2. Analysis:
  • Sulphide ion (S²⁻): Z = 16. Lowest nuclear charge, weakest attraction → Largest Radius.
  • Chloride ion (Cl⁻): Z = 17.
  • Potassium ion (K⁺): Z = 19.
  • Calcium ion (Ca²⁺): Z = 20. Highest nuclear charge, strongest attraction → Smallest Radius.
  1. Conclusion: The order of decreasing ionic radii is S²⁻ > Cl⁻ > K⁺ > Ca²⁺.
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