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Periodic Table and Periodicity question

2010 · Shift 0 · Q24
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Periodic Table and Periodicity question

2010 · Shift 0 · Q24

JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1
The correct sequence which shows decreasing order of the ionic radii of the elements is
  1. A
    Al3+Al^{3+}Al3+ > Mg2+Mg^{2+}Mg2+ > Na+Na^+Na+ > F−F^-F− > O2−O^{2-}O2−
  2. B
    Na+ > Mg2+ > Al3+ > O2– > F–
  3. C
    Na+ > F– > Mg2+ > O2– > Al3+
  4. D
    O2– > F– > Na+ > Mg2+ > Al3+
View written solutionFree

Correct answer: D

  1. Identify the species

The ions given are:

O2−, F−, Na+, Mg2+, Al3+O^{2-},\ F^-,\ Na^+,\ Mg^{2+},\ Al^{3+}O2−, F−, Na+, Mg2+, Al3+

  1. Check whether they are isoelectronic

Number of electrons in each ion:

  • 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
  • Al3+Al^{3+}Al3+: 13−3=1013 - 3 = 1013−3=10

So, all are isoelectronic with 10 electrons.

  1. Use the rule for isoelectronic species

For isoelectronic ions, as nuclear charge increases, the electrons are pulled more strongly, so the ionic radius decreases.

Their nuclear charges are:

  • OOO: Z=8Z=8Z=8
  • FFF: Z=9Z=9Z=9
  • NaNaNa: Z=11Z=11Z=11
  • MgMgMg: Z=12Z=12Z=12
  • AlAlAl: Z=13Z=13Z=13

Thus, radius decreases in the order of increasing nuclear charge:

O2−>F−>Na+>Mg2+>Al3+O^{2-} > F^- > Na^+ > Mg^{2+} > Al^{3+}O2−>F−>Na+>Mg2+>Al3+

  1. Match with the options
  • A: Al3+>Mg2+>Na+>F−>O2−Al^{3+} > Mg^{2+} > Na^+ > F^- > O^{2-}Al3+>Mg2+>Na+>F−>O2− ❌ exactly reverse
  • B: Na+>Mg2+>Al3+>O2−>F−Na^+ > Mg^{2+} > Al^{3+} > O^{2-} > F^-Na+>Mg2+>Al3+>O2−>F− ❌ incorrect
  • C: Na+>F−>Mg2+>O2−>Al3+Na^+ > F^- > Mg^{2+} > O^{2-} > Al^{3+}Na+>F−>Mg2+>O2−>Al3+ ❌ incorrect
  • D: O2−>F−>Na+>Mg2+>Al3+O^{2-} > F^- > Na^+ > Mg^{2+} > Al^{3+}O2−>F−>Na+>Mg2+>Al3+ ✅ correct

Therefore, the correct decreasing order of ionic radii is:

O2−>F−>Na+>Mg2+>Al3+\boxed{O^{2-} > F^- > Na^+ > Mg^{2+} > Al^{3+}}O2−>F−>Na+>Mg2+>Al3+​

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