JEE Main 2022 · 27 June · evening · Single correct · +4 / −1

The correct order of increasing ionic radii is

  1. (a) Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻
  2. (b) N³⁻ < O²⁻ < F⁻ < Na⁺ < Mg²⁺
  3. (c) F⁻ < Na⁺ < O²⁻ < Mg²⁺ < N³⁻
  4. (d) Na⁺ < F⁻ < Mg²⁺ < O²⁻ < N³⁻
Answer and solution

Answer (a) Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻

Solution

Chemistry ma’am · 3 steps

All five ions have 10 electrons, so just compare protons: more protons pull the cloud in tighter.

  1. Spot the series. Each ion has exactly 10 electrons, the neon configuration 1s² 2s² 2p⁶. Ions with the same number of electrons form an isoelectronic series.
  2. What decides size. With the electron count fixed, only the nuclear charge changes. More protons pull the same 10 electrons closer, so the ion gets smaller.
    IonProtonsElectronsSize
    Mg²⁺1210smallest
    Na⁺1110small
    F⁻910medium
    O²⁻810large
    N³⁻710largest
  3. So. Going from 12 protons down to 7 protons, the radius increases: Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻. That is option (a).

The idea to keep. In an isoelectronic series, the ion with the most protons is the smallest. Count protons, not just the charge sign.

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 2022.