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Periodic Table and Periodicity question

2020 · 5 Sep · Shift 2 · Q7
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  5. /2020 · 5 Sep · Shift 2 · Q7

Periodic Table and Periodicity question

2020 · 5 Sep · Shift 2 · Q7

JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1
The correct order of the ionic radii of O2−O^{2-}O2−, N3−N^{3-}N3−, F−F^-F− , Mg2+Mg^{2+}Mg2+, Na+Na^+Na+ and Al3+Al^{3+}Al3+ is :
  1. A
    N3– < O2– < F– < Na+ < Mg2+ < Al3+
  2. B
    N3– < F– < O2– < Mg2+ < Na+ < Al3+
  3. C
    Al3+ < Na+ < Mg2+ < O2– < F– < N3–
  4. D
    Al3+ < Mg2+ < Na+ < F– < O2– < N3–
View written solutionFree

Correct answer: D

  1. Identify the species

The ions are: O2−,  N3−,  F−,  Mg2+,  Na+,  Al3+O^{2-},\; N^{3-},\; F^-,\; Mg^{2+},\; Na^+,\; Al^{3+}O2−,N3−,F−,Mg2+,Na+,Al3+

  1. Check electron count

Each of these ions has 10 electrons:

  • N3−N^{3-}N3−: 7+3=107+3=107+3=10
  • 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 species.

  1. Rule for isoelectronic species

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

Reason: all ions have the same number of electrons, but a higher number of protons pulls the electron cloud more strongly.

Their atomic numbers are:

  • N=7N = 7N=7
  • O=8O = 8O=8
  • F=9F = 9F=9
  • Na=11Na = 11Na=11
  • Mg=12Mg = 12Mg=12
  • Al=13Al = 13Al=13

Thus, among these isoelectronic ions:

  • smallest radius →\to→ highest ZZZ →Al3+\to Al^{3+}→Al3+
  • largest radius →\to→ lowest ZZZ →N3−\to N^{3-}→N3−

So increasing order of ionic radii is: Al3+<Mg2+<Na+<F−<O2−<N3−Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}Al3+<Mg2+<Na+<F−<O2−<N3−

  1. Match with options

Option D says: Al3+<Mg2+<Na+<F−<O2−<N3−Al^{3+} < Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}Al3+<Mg2+<Na+<F−<O2−<N3−

This matches exactly.

Therefore, the correct option is D.

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