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Periodic Table and Periodicity question

2018 · 15 Apr · Shift 1 · Q17
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Periodic Table and Periodicity question

2018 · 15 Apr · Shift 1 · Q17

JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1
For Na+Na^+Na+, Mg2+Mg^{2+}Mg2+, F−F^-F− and O2−O^{2-}O2−; the correct order of increasing ionic radii is :
  1. A
    O2−O^{2-}O2− < F−F^-F− < Na+Na^+Na+ < Mg2+Mg^{2+}Mg2+
  2. B
    Na+Na^+Na+ < Mg2+Mg^{2+}Mg2+ < F−F^-F− < O2−O^{2-}O2−
  3. C
    Mg2+Mg^{2+}Mg2+ < Na+Na^+Na+ < F−F^-F− < O2−O^{2-}O2−
  4. D
    Mg2+Mg^{2+}Mg2+ < O2−O^{2-}O2− < Na+Na^+Na+ < F−F^-F−
View written solutionFree

Correct answer: C

  1. Identify the species and electrons

All four ions are isoelectronic:

  • O2−O^{2-}O2− : 8+2=108 + 2 = 108+2=10 electrons
  • F−F^-F− : 9+1=109 + 1 = 109+1=10 electrons
  • Na+Na^+Na+ : 11−1=1011 - 1 = 1011−1=10 electrons
  • Mg2+Mg^{2+}Mg2+ : 12−2=1012 - 2 = 1012−2=10 electrons

So each ion has the electronic configuration of Ne, i.e. 10 electrons.

  1. Rule for isoelectronic species

For an isoelectronic series, ionic radius decreases as nuclear charge ZZZ increases.

This is because the same number of electrons are attracted more strongly when the number of protons is larger.

Here the nuclear charges are:

  • O2−O^{2-}O2− : Z=8Z = 8Z=8
  • F−F^-F− : Z=9Z = 9Z=9
  • Na+Na^+Na+ : Z=11Z = 11Z=11
  • Mg2+Mg^{2+}Mg2+ : Z=12Z = 12Z=12

Thus, radius order from largest to smallest is:

O2−>F−>Na+>Mg2+O^{2-} > F^- > Na^+ > Mg^{2+}O2−>F−>Na+>Mg2+

Therefore, the increasing order of ionic radii is:

Mg2+<Na+<F−<O2−Mg^{2+} < Na^+ < F^- < O^{2-}Mg2+<Na+<F−<O2−

  1. Check options
  • A: O2−<F−<Na+<Mg2+O^{2-} < F^- < Na^+ < Mg^{2+}O2−<F−<Na+<Mg2+ ❌ reversed
  • B: Na+<Mg2+<F−<O2−Na^+ < Mg^{2+} < F^- < O^{2-}Na+<Mg2+<F−<O2− ❌ Mg2+Mg^{2+}Mg2+ should be smaller than Na+Na^+Na+
  • C: Mg2+<Na+<F−<O2−Mg^{2+} < Na^+ < F^- < O^{2-}Mg2+<Na+<F−<O2− ✅ correct
  • D: Mg2+<O2−<Na+<F−Mg^{2+} < O^{2-} < Na^+ < F^-Mg2+<O2−<Na+<F− ❌ wrong middle order
  1. Final answer

The correct order of increasing ionic radii is:

Mg2+<Na+<F−<O2−Mg^{2+} < Na^+ < F^- < O^{2-}Mg2+<Na+<F−<O2−

So, Option C is correct.

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