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MCQEasyJEE 2023Trends in Atomic Radius & Ionisation Energy

JEE Chemistry 2023 Question with Solution

The correct increasing order of the ionic radii is:

  • A

    Cl−<Ca2+<K+<S2−Cl^- < Ca^{2+} < K^+ < S^{2-}

  • B

    K+<S2−<Ca2+<Cl−K^+ < S^{2-} < Ca^{2+} < Cl^-

  • C

    S2−<Cl−<Ca2+<K+S^{2-} < Cl^- < Ca^{2+} < K^+

  • D

    Ca2+<K+<Cl−<S2−Ca^{2+} < K^+ < Cl^- < S^{2-}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The ions Ca2+Ca^{2+}, K+K^+, Cl−Cl^- and S2−S^{2-} are to be arranged in increasing order of ionic radii.

Find: The correct increasing order of ionic size.

All four ions are isoelectronic species. In isoelectronic species, ionic size decreases as nuclear charge increases. Hence, ionic size is inversely proportional to nuclear charge:

ionic radius∝1Z\text{ionic radius} \propto \frac{1}{Z}
Diagram showing the order Ca2+ less than K+ less than Cl- less than S2- in size, with nuclear charges 20, 19, 17 and 16 indicated respectively.

Among these ions, the nuclear charge is highest for Ca2+Ca^{2+} and lowest for S2−S^{2-}. Therefore, the ionic radius increases in the order:

Ca2+<K+<Cl−<S2−Ca^{2+} < K^+ < Cl^- < S^{2-}

Therefore, the correct option is D. The solution also contains a conflicting label "The Correct Option is A", but the worked order clearly matches option D.

Using Isoelectronic Trend

Given: Ca2+,K+,Cl−,Ca^{2+}, K^+, Cl^-, and S2−S^{2-} are isoelectronic.

Find: Their increasing order of ionic radii.

For isoelectronic species, the number of electrons is the same, so the deciding factor is the nuclear charge. Greater nuclear charge pulls the same electron cloud more strongly and decreases size.

Here, the relevant nuclear charges are:

  • Ca2+:Z=20Ca^{2+}: Z = 20
  • K+:Z=19K^+: Z = 19
  • Cl−:Z=17Cl^-: Z = 17
  • S2−:Z=16S^{2-}: Z = 16

So, size increases as ZZ decreases:

Ca2+<K+<Cl−<S2−Ca^{2+} < K^+ < Cl^- < S^{2-}

Hence, the correct increasing order is Ca2+<K+<Cl−<S2−Ca^{2+} < K^+ < Cl^- < S^{2-}.

Common mistakes

  • Assuming cations are always smaller and anions are always larger without checking that the ions are isoelectronic. Here, the correct comparison must be based on nuclear charge for species having the same number of electrons.

  • Using atomic number order incorrectly. For isoelectronic species, larger ZZ means stronger attraction on the same electron cloud, so the radius becomes smaller, not larger.

  • Missing the discrepancy in the displayed solution label. The text says option A, but the worked order shown is Ca2+<K+<Cl−<S2−Ca^{2+} < K^+ < Cl^- < S^{2-}, which matches option D. Always trust the actual working over a mislabeled option tag.

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