JEE Main · 2022 · Shift-IIeasyPERI-028

The correct order of ionic size of N3-, Na+, F-, Mg2+ and O2- is:

Classification of Elements & Periodicity · Class 11 · JEE Main Previous Year Question

Question

The correct order of ionic size of N3−^{3-}, Na+^+, F−^-, Mg2+^{2+} and O2−^{2-} is:

Options
  1. a

    \ceMg2+\ce{Mg^{2+}} < \ceNa+\ce{Na^+} < \ceF−\ce{F^-} < \ceO2−\ce{O^{2-}} < \ceN3−\ce{N^{3-}}

    ✓
  2. b

    \ceN3−\ce{N^{3-}} < \ceO2−\ce{O^{2-}} < \ceF−\ce{F^-} < \ceNa+\ce{Na^+} < \ceMg2+\ce{Mg^{2+}}

  3. c

    \ceF−\ce{F^-} < \ceNa+\ce{Na^+} < \ceO2−\ce{O^{2-}} < \ceMg2+\ce{Mg^{2+}} < \ceN3−\ce{N^{3-}}

  4. d

    \ceNa+\ce{Na^+} < \ceF−\ce{F^-} < \ceMg2+\ce{Mg^{2+}} < \ceO2−\ce{O^{2-}} < \ceN3−\ce{N^{3-}}

Correct Answera

\ceMg2+\ce{Mg^{2+}} < \ceNa+\ce{Na^+} < \ceF−\ce{F^-} < \ceO2−\ce{O^{2-}} < \ceN3−\ce{N^{3-}}

Detailed Solution

🧠 Five ions, all with 10 electrons — only Z is different \ceMg2+\ce{Mg^{2+}} (Z=12Z=12), \ceNa+\ce{Na^+} (Z=11Z=11), \ceF−\ce{F^-} (Z=9Z=9), \ceO2−\ce{O^{2-}} (Z=8Z=8), \ceN3−\ce{N^{3-}} (Z=7Z=7). Each one has 10 electrons. The size question reduces to: how strongly do the protons pull on this 10-electron cloud? More protons = tighter pull = smaller ion.

🗺️ Rank by Z (lowest Z is largest) \ceMg2+\ce{Mg^{2+}} has Z=12Z=12 pulling on 10 electrons → smallest. Drop one proton at a time: \ceNa+\ce{Na^+} (11), \ceF−\ce{F^-} (9), \ceO2−\ce{O^{2-}} (8), \ceN3−\ce{N^{3-}} (7).

Increasing size: \ceMg2+<\ceNa+<\ceF−<\ceO2−<\ceN3−\ce{Mg^{2+}} < \ce{Na^+} < \ce{F^-} < \ce{O^{2-}} < \ce{N^{3-}}.

⚠️ The trap Notice the pattern — for an isoelectronic set, the most negative ion (here \ceN3−\ce{N^{3-}}, charge −3-3) is also the largest. That is because gaining more electrons than your own proton count means a weaker overall pull. Students who try to apply the periodic trend (N is above F, so N atom is smaller) get tricked. The atom and the ion behave very differently once you start adding charges.

Answer: (a)\boxed{\text{Answer: (a)}}

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