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Chemical Bonding and Molecular Structure question

2022 · 26 Jul · Shift 2 · Q2
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Chemical Bonding and Molecular Structure question

2022 · 26 Jul · Shift 2 · Q2

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Arrange the following in increasing order of their covalent character. A. CaF2\mathrm{CaF}_{2}CaF2​ B. CaCl2\mathrm{CaCl}_{2}CaCl2​ C. CaBr2\mathrm{CaBr}_{2}CaBr2​ D. CaI2\mathrm{CaI}_{2}CaI2​ Choose the correct answer from the options given below.
  1. A
    B < A < C < D
  2. B
    A < B < C < D
  3. C
    A < B < D < C
  4. D
    A < C < B < D
View written solutionFree

Correct answer: B

  1. Use Fajans' rule

    Covalent character in an ionic compound increases when:

    • the cation has high polarising power,
    • the anion has high polarisability.

    Here, the cation is the same in all compounds: Ca2+\mathrm{Ca^{2+}}Ca2+. So, the trend depends only on the halide ions: F−, Cl−, Br−, I−\mathrm{F^-},\ \mathrm{Cl^-},\ \mathrm{Br^-},\ \mathrm{I^-}F−, Cl−, Br−, I−

  2. Compare polarisability of anions

    Down the group, size of halide ions increases, so polarisability increases: F−<Cl−<Br−<I−\mathrm{F^- < Cl^- < Br^- < I^-}F−<Cl−<Br−<I−

    Greater polarisability means greater distortion by Ca2+\mathrm{Ca^{2+}}Ca2+, hence greater covalent character.

  3. Apply to the given compounds

    Therefore, covalent character increases as: CaF2<CaCl2<CaBr2<CaI2\mathrm{CaF_2 < CaCl_2 < CaBr_2 < CaI_2}CaF2​<CaCl2​<CaBr2​<CaI2​

    That is: A<B<C<DA < B < C < DA<B<C<D

  4. Check options

    • Option A: B<A<C<DB < A < C < DB<A<C<D ❌
    • Option B: A<B<C<DA < B < C < DA<B<C<D ✅
    • Option C: A<B<D<CA < B < D < CA<B<D<C ❌
    • Option D: A<C<B<DA < C < B < DA<C<B<D ❌

So, the correct answer is Option B.

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