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Periodic Table and Periodicity question

2023 · 31 Jan · Shift 1 · Q6
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Periodic Table and Periodicity question

2023 · 31 Jan · Shift 1 · Q6

JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1
The correct increasing order of the ionic radii is
  1. A
    Cl−<Ca2+<K+<S2−\mathrm{Cl^- \lt Ca^{2+} \lt K^+ \lt S^{2-}}Cl−<Ca2+<K+<S2−
  2. B
    K+<S2−<Ca2+<Cl−\mathrm{K^+ \lt S^{2-} \lt Ca^{2+} \lt Cl^-}K+<S2−<Ca2+<Cl−
  3. C
    Ca2+<K+<Cl−<S2−\mathrm{Ca^{2+} \lt K^+ \lt Cl^- \lt S^{2-}}Ca2+<K+<Cl−<S2−
  4. D
    S2−<Cl−<Ca2+<K+\mathrm{S^{2-} \lt Cl^- \lt Ca^{2+} \lt K^+}S2−<Cl−<Ca2+<K+
View written solutionFree

Correct answer: C

  1. Identify the ions and their electron counts

All four ions are isoelectronic with argon:

S2−:16+2=18 electrons\mathrm{S^{2-}}: 16 + 2 = 18 \text{ electrons}S2−:16+2=18 electrons Cl−:17+1=18 electrons\mathrm{Cl^-}: 17 + 1 = 18 \text{ electrons}Cl−:17+1=18 electrons K+:19−1=18 electrons\mathrm{K^+}: 19 - 1 = 18 \text{ electrons}K+:19−1=18 electrons Ca2+:20−2=18 electrons\mathrm{Ca^{2+}}: 20 - 2 = 18 \text{ electrons}Ca2+:20−2=18 electrons

So, all are isoelectronic species.

  1. Rule for isoelectronic species

For isoelectronic ions, radius decreases as nuclear charge increases.

That is, among species with the same number of electrons:

  • higher atomic number ZZZ ⇒\Rightarrow⇒ stronger attraction on electrons
  • hence smaller ionic radius
  1. Arrange by atomic number

The atomic numbers are:

S=16,Cl=17,K=19,Ca=20\mathrm{S} = 16, \quad \mathrm{Cl} = 17, \quad \mathrm{K} = 19, \quad \mathrm{Ca} = 20S=16,Cl=17,K=19,Ca=20

Since radius is inversely related to nuclear charge for isoelectronic species:

Ca2+<K+<Cl−<S2−\mathrm{Ca^{2+} < K^+ < Cl^- < S^{2-}}Ca2+<K+<Cl−<S2−
  1. Match with the options

This corresponds to Option C.

  1. Final Answer
Ca2+<K+<Cl−<S2−\boxed{\mathrm{Ca^{2+} < K^+ < Cl^- < S^{2-}}}Ca2+<K+<Cl−<S2−​
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