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Periodic Table and Periodicity question

2020 · 4 Sep · Shift 1 · Q17
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Periodic Table and Periodicity question

2020 · 4 Sep · Shift 1 · Q17

JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1
The ionic radii of O2−O^{2-}O2−, F−F^-F−, Na+Na^+Na+ and Mg2+Mg^{2+}Mg2+ are in the order :
  1. A
    F– > O2– > Na+ > Mg2+
  2. B
    Mg2+ > Na+ > F– > O2–
  3. C
    O2– > F– > Mg2+ > Na+
  4. D
    O2– > F– > Na+ > Mg2+
View written solutionFree

Correct answer: D

  1. Identify the species

The ions are:

  • O2−O^{2-}O2−
  • F−F^-F−
  • Na+Na^+Na+
  • Mg2+Mg^{2+}Mg2+
  1. Check electron count

All these ions have 10 electrons each:

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

So, they are isoelectronic species.

  1. Rule for isoelectronic species

For isoelectronic ions, radius decreases as nuclear charge ZZZ increases.

This is because all ions have the same number of electrons, but a higher number of protons pulls the electron cloud more strongly.

Their atomic numbers are:

  • Oxygen: Z=8Z=8Z=8
  • Fluorine: Z=9Z=9Z=9
  • Sodium: Z=11Z=11Z=11
  • Magnesium: Z=12Z=12Z=12

Thus, increasing nuclear charge order is: O<F<Na<MgO < F < Na < MgO<F<Na<Mg

Hence, ionic radius order will be the reverse: O2−>F−>Na+>Mg2+O^{2-} > F^- > Na^+ > Mg^{2+}O2−>F−>Na+>Mg2+

  1. Compare with options
  • A: F−>O2−>Na+>Mg2+F^- > O^{2-} > Na^+ > Mg^{2+}F−>O2−>Na+>Mg2+ ❌
  • B: Mg2+>Na+>F−>O2−Mg^{2+} > Na^+ > F^- > O^{2-}Mg2+>Na+>F−>O2− ❌
  • C: O2−>F−>Mg2+>Na+O^{2-} > F^- > Mg^{2+} > Na^+O2−>F−>Mg2+>Na+ ❌
  • D: O2−>F−>Na+>Mg2+O^{2-} > F^- > Na^+ > Mg^{2+}O2−>F−>Na+>Mg2+ ✅

Therefore, the correct option is D.

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