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Chemical Bonding and Molecular Structure question

2025 · 22 Jan · Shift 2 · Q9
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Chemical Bonding and Molecular Structure question

2025 · 22 Jan · Shift 2 · Q9

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
Arrange the following compounds in increasing order of their dipole moment : HBr,H2 S,NF3\mathrm{HBr}, \mathrm{H}_2 \mathrm{~S}, \mathrm{NF}_3HBr,H2​ S,NF3​ and CHCl3\mathrm{CHCl}_3CHCl3​
  1. A
    HBr<H2 S<NF3<CHCl3\mathrm{HBr}\lt \mathrm{H}_2 \mathrm{~S}\lt \mathrm{NF}_3\lt \mathrm{CHCl}_3HBr<H2​ S<NF3​<CHCl3​
  2. B
    H2 S<HBr<NF3<CHCl3\mathrm{H}_2 \mathrm{~S}\lt \mathrm{HBr}\lt \mathrm{NF}_3\lt \mathrm{CHCl}_3H2​ S<HBr<NF3​<CHCl3​
  3. C
    NF3<HBr<H2 S<CHCl3\mathrm{NF}_3\lt \mathrm{HBr}\lt \mathrm{H}_2 \mathrm{~S}\lt \mathrm{CHCl}_3NF3​<HBr<H2​ S<CHCl3​
  4. D
    CHCl3<NF3<HBr<H2S\mathrm{CHCl}_3\lt \mathrm{NF}_3\lt \mathrm{HBr}\lt \mathrm{H}_2 \mathrm{S}CHCl3​<NF3​<HBr<H2​S
View written solutionFree

Correct answer: C

  1. Recall the shapes and dipole behavior of each molecule

    • HBr\mathrm{HBr}HBr: diatomic molecule, so its dipole moment is simply due to the H−Br\mathrm{H-Br}H−Br bond.
    • H2S\mathrm{H_2S}H2​S: bent molecule, so the two S−H\mathrm{S-H}S−H bond moments do not cancel completely.
    • NF3\mathrm{NF_3}NF3​: trigonal pyramidal molecule. Here, the N−F\mathrm{N-F}N−F bond dipoles oppose the lone-pair direction, causing strong cancellation, so dipole moment is quite small.
    • CHCl3\mathrm{CHCl_3}CHCl3​: tetrahedral molecule. Three C−Cl\mathrm{C-Cl}C−Cl bonds and one C−H\mathrm{C-H}C−H bond give a substantial net dipole.

  2. Compare approximate known dipole moments

    Standard dipole moments are approximately:

    • μ(NF3)≈0.24 D\mu(\mathrm{NF_3}) \approx 0.24\,\text{D}μ(NF3​)≈0.24D
    • μ(HBr)≈0.8 D\mu(\mathrm{HBr}) \approx 0.8\,\text{D}μ(HBr)≈0.8D
    • μ(H2S)≈0.97 D\mu(\mathrm{H_2S}) \approx 0.97\,\text{D}μ(H2​S)≈0.97D
    • μ(CHCl3)≈1.0 D\mu(\mathrm{CHCl_3}) \approx 1.0\,\text{D}μ(CHCl3​)≈1.0D

  3. Arrange in increasing order

    Therefore,

    NF3<HBr<H2S<CHCl3\mathrm{NF_3} < \mathrm{HBr} < \mathrm{H_2S} < \mathrm{CHCl_3}NF3​<HBr<H2​S<CHCl3​

  4. Match with the given options

    This corresponds to Option C.

  5. Comparison with stored correct answer

    Stored correct answer: C

    Our derived answer: C

    So, they agree.

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