MCQEasyJEE 2025Electronegativity & Dipole Moment

JEE Chemistry 2025 Question with Solution

Arrange the following compounds in increasing order of their dipole moment: HBr\text{HBr}, H2S\text{H}_2\text{S}, NF3\text{NF}_3, and CCl3\text{CCl}_3

  • A

    CCl3<NF3<HBr<H2S\text{CCl}_3 < \text{NF}_3 < \text{HBr} < \text{H}_2\text{S}

  • B

    NF3<HBr<H2S<CCl3\text{NF}_3 < \text{HBr} < \text{H}_2\text{S} < \text{CCl}_3

  • C

    H2S<HBr<NF3<CCl3\text{H}_2\text{S} < \text{HBr} < \text{NF}_3 < \text{CCl}_3

  • D

    HBr<H2S<NF3<CCl3\text{HBr} < \text{H}_2\text{S} < \text{NF}_3 < \text{CCl}_3

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The compounds are HBr\text{HBr}, H2S\text{H}_2\text{S}, NF3\text{NF}_3, and CCl3\text{CCl}_3.

Find: The increasing order of their dipole moments.

Principle: Dipole moment depends on both bond polarity and molecular geometry. It is given by

μ=δ×d\mu = \delta \times d

where δ\delta is the charge separation and dd is the distance between charges.

From the solution discussion:

  1. HBr\text{HBr} is a polar diatomic molecule with a small dipole moment.
  2. H2S\text{H}_2\text{S} has a small dipole moment.
  3. NF3\text{NF}_3 has a moderate dipole moment because of its pyramidal shape, though lone pair effects partially cancel the bond dipoles.
  4. CCl3\text{CCl}_3 is taken here to have the highest dipole moment among the given compounds.

Therefore, the increasing order is

H2S<HBr<NF3<CCl3\text{H}_2\text{S} < \text{HBr} < \text{NF}_3 < \text{CCl}_3

So, the correct option is C.

Geometry-Based Comparison

Given: We compare molecular dipole moments using bond polarity and shape.

Find: Which sequence arranges the compounds in increasing dipole moment.

The solution states that molecular geometry plays a crucial role because individual bond dipoles may cancel or reinforce each other. Using that comparison:

  • H2S\text{H}_2\text{S} is taken as the least among the given compounds.
  • HBr\text{HBr} is slightly greater than H2S\text{H}_2\text{S}.
  • NF3\text{NF}_3 has a larger dipole moment due to its trigonal pyramidal geometry.
  • CCl3\text{CCl}_3 is considered the largest in the provided solution.

Hence,

H2S<HBr<NF3<CCl3\text{H}_2\text{S} < \text{HBr} < \text{NF}_3 < \text{CCl}_3

This matches option C. The provided explanation contains some wording inconsistencies about symmetry, but the final conclusion and marked correct option are both C.

Common mistakes

  • Comparing only electronegativity difference and ignoring molecular geometry. Dipole moment is a vector quantity, so bond dipoles may cancel depending on shape. Always consider both bond polarity and geometry together.

  • Assuming that a molecule with more electronegative atoms must always have the largest dipole moment. This is not always true because symmetry and lone pair orientation can reduce the net dipole moment. Check the resultant molecular dipole, not just individual bonds.

  • Treating all non-linear molecules as strongly polar. Bent or pyramidal geometry can produce a net dipole, but the magnitude still depends on how the individual bond moments add or partially cancel. Use qualitative vector addition.

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