JEE Main 2020 · 5 September · evening · Single correct · +4 / −1

The correct order of the ionic radii of
O 2– , N 3– , F – , Mg²⁺, Na⁺ and Al³⁺ is :

  1. (a) N 3– < O 2– < F – < Na⁺ < Mg²⁺ < Al³⁺
  2. (b) N 3– < F – < O 2– < Mg²⁺ < Na⁺ < Al³⁺
  3. (c) Al³⁺ < Na⁺ < Mg²⁺ < O 2– < F – < N 3–
  4. (d) Al³⁺ < Mg²⁺ < Na⁺ < F – < O 2– < N 3–
Answer and solution

Answer (d) Al³⁺ < Mg²⁺ < Na⁺ < F – < O 2– < N 3–

Solution

Chemistry ma’am · 3 steps

All six ions carry 10 electrons, so just count protons: more protons pull the same cloud in tighter.

  1. Same electron count. Each ion here has 10 electrons, the neon configuration 1s² 2s² 2p⁶. These ions form an isoelectronic series, so the only thing that changes from one to the next is the nuclear charge.
  2. Nuclear charge decides. More protons pull the same 10 electrons in harder, so the ion shrinks. Fewer protons hold the cloud loosely, so the ion swells.
    IonProtonsElectronsRadius
    Al³⁺1310smallest, about 54 pm
    Mg²⁺1210about 72 pm
    Na⁺1110about 102 pm
    F⁻910about 133 pm
    O²⁻810about 140 pm
    N³⁻710largest, about 171 pm
  3. So. Radius rises as the proton count falls: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻, which is option (d).

The idea to keep. In an isoelectronic series, the electron count is fixed, so size depends only on the protons. More protons mean a smaller ion. Fewer protons mean a bigger ion.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 120 times in JEE Main since 2020. This question is from 2020.