JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
The ionic radii of O 2– , F – , Na⁺ and Mg²⁺ are in the order :
- (a) F – > O 2– > Na⁺ > Mg²⁺
- (b) Mg²⁺ > Na⁺ > F – > O 2–
- (c) O 2– > F – > Mg²⁺ > Na⁺
- (d) O 2– > F – > Na⁺ > Mg²⁺
Answer and solution
Answer (d) O 2– > F – > Na⁺ > Mg²⁺
Solution
Chemistry ma’am · 3 steps
All four ions have 10 electrons, so the one with fewer protons holding them is the bigger ion.
- Same electron count. O²⁻, F⁻, Na⁺ and Mg²⁺ each have 10 electrons, the neon configuration. They form an isoelectronic series.
- Compare nuclear charge. With the electron cloud fixed, only the number of protons changes. More protons pull the same 10 electrons in more tightly, so the ion gets smaller as Z rises.
Ion Protons (Z) Electrons Ionic radius O²⁻ 8 10 140 pm, largest F⁻ 9 10 133 pm Na⁺ 11 10 102 pm Mg²⁺ 12 10 72 pm, smallest - So. The radius falls as Z goes up: O²⁻ > F⁻ > Na⁺ > Mg²⁺, which is option (d).
The idea to keep. In an isoelectronic series, the electron count is the same, so radius depends only on the nuclear charge: the higher Z, the smaller the ion.
Answer from the NTA final answer key. Solution by PrepWiser.