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Periodic Table and Periodicity question

2021 · 27 Aug · Shift 2 · Q5
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  5. /2021 · 27 Aug · Shift 2 · Q5

Periodic Table and Periodicity question

2021 · 27 Aug · Shift 2 · Q5

JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1
The correct order of ionic radii for the ions, P3−P^{3-}P3−, S2−S^{2-}S2−, Ca2+Ca^{2+}Ca2+, K+K^+K+, Cl −-− is :
  1. A
    P3−P^{3-}P3− > S2−S^{2-}S2− > Cl −-− > K+K^+K+ > Ca2+Ca^{2+}Ca2+
  2. B
    Cl −-− > S2−S^{2-}S2− > P3−P^{3-}P3− > Ca2+Ca^{2+}Ca2+ > K+K^+K+
  3. C
    P3−P^{3-}P3− > S2−S^{2-}S2− > Cl −-− > Ca2+Ca^{2+}Ca2+ > K+K^+K+
  4. D
    K+K^+K+ > Ca2+Ca^{2+}Ca2+ > P3−P^{3-}P3− > S2−S^{2-}S2− > Cl −-−
View written solutionFree

Correct answer: A

  1. Identify the nature of the ions

The ions are: P3−, S2−, Cl−, K+, Ca2+P^{3-},\ S^{2-},\ Cl^-,\ K^+,\ Ca^{2+}P3−, S2−, Cl−, K+, Ca2+

All of these ions are isoelectronic because each has 18 electrons:

  • P3−P^{3-}P3−: 15+3=1815 + 3 = 1815+3=18
  • S2−S^{2-}S2−: 16+2=1816 + 2 = 1816+2=18
  • Cl−Cl^-Cl−: 17+1=1817 + 1 = 1817+1=18
  • K+K^+K+: 19−1=1819 - 1 = 1819−1=18
  • Ca2+Ca^{2+}Ca2+: 20−2=1820 - 2 = 1820−2=18

So all have the electron configuration of Ar.

  1. Rule for isoelectronic species

For isoelectronic ions, ionic radius depends on nuclear charge:

  • Greater nuclear charge ⇒\Rightarrow⇒ stronger attraction on the same number of electrons
  • Hence, radius decreases as atomic number increases

Atomic numbers are:

  • P=15P = 15P=15
  • S=16S = 16S=16
  • Cl=17Cl = 17Cl=17
  • K=19K = 19K=19
  • Ca=20Ca = 20Ca=20

Thus, increasing nuclear charge gives decreasing ionic size:

P3−>S2−>Cl−>K+>Ca2+P^{3-} > S^{2-} > Cl^- > K^+ > Ca^{2+}P3−>S2−>Cl−>K+>Ca2+

  1. Compare with the options
  • A: P3−>S2−>Cl−>K+>Ca2+P^{3-} > S^{2-} > Cl^- > K^+ > Ca^{2+}P3−>S2−>Cl−>K+>Ca2+ ✅
  • B: Incorrect
  • C: Incorrect, because K+K^+K+ should be larger than Ca2+Ca^{2+}Ca2+
  • D: Incorrect
  1. Final Answer

The correct order is: P3−>S2−>Cl−>K+>Ca2+P^{3-} > S^{2-} > Cl^- > K^+ > Ca^{2+}P3−>S2−>Cl−>K+>Ca2+

So the correct option is A.

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