Dipole Moment

8 MCQs1 revision card9-step worked example
Source: NCERT Chemical Bonding and Molecular StructurePYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The question NEET likes to test on dipole moment is deceptively simple: given a molecule, does it have a net dipole moment, and if so, why? The trap is not in the formula — it is in the vector addition.

What is dipole moment? Dipole moment (μ) measures the polarity of a molecule. For a diatomic molecule, μ = q × d, where q is the magnitude of charge separation and d is the bond length (NCERT Class 11 Chemistry Chapter 4, page 10). The SI unit is C·m, but NEET problems overwhelmingly use the Debye (1 D = 3.336 × 10⁻³⁰ C·m).

The real test: vector cancellation in polyatomics. A polar bond does not guarantee a polar molecule. CO₂ has two polar C=O bonds, but its linear geometry makes the bond dipoles cancel to zero. BF₃ (trigonal planar, 120° symmetry) also has μ = 0 despite three polar B–F bonds. Meanwhile, H₂O has two polar O–H bonds at 104.5° — the vectors do not cancel, giving μ ≈ 1.85 D.

The high-frequency confusion: students see "polar bonds" and immediately conclude "polar molecule." NEET exploits this by offering symmetric molecules (CCl₄, BF₃, BeF₂) as distractors alongside genuinely polar ones (CHCl₃, NH₃, H₂O). The deciding factor is always molecular geometry, not bond polarity alone.

Key distinctions for NEET:

  • Zero dipole moment: linear symmetric (CO₂, BeF₂), trigonal planar symmetric (BF₃), tetrahedral symmetric (CCl₄, CH₄).
  • Non-zero dipole moment: bent (H₂O, SO₂), trigonal pyramidal (NH₃, NF₃), molecules with asymmetric substitution (CHCl₃).

Watch-out: When comparing dipole moments of similar molecules (e.g., NF₃ vs NH₃), remember that lone-pair dipole contribution matters. In NH₃, the lone pair reinforces the bond dipoles; in NF₃, it opposes them. Result: μ(NH₃) > μ(NF₃).

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The SI unit of dipole moment is:

MCQ 2Easy RecallPractice

Which of the following molecules has zero dipole moment?

MCQ 3Easy RecallPractice

1 Debye equals:

MCQ 4Direct ApplicationPractice

BF₃ has three polar B–F bonds, yet its dipole moment is zero. This is because:

MCQ 5Direct ApplicationPractice

Among H₂O, H₂S, and H₂Se, which has the highest dipole moment?

MCQ 6Direct ApplicationPractice

The dipole moment of NF₃ (0.23 D) is much less than that of NH₃ (1.47 D), despite both having trigonal pyramidal geometry. The best explanation is:

MCQ 7CalculationPractice

Two charges of +1.6 × 10⁻¹⁹ C and −1.6 × 10⁻¹⁹ C are separated by 1.0 × 10⁻¹⁰ m. The dipole moment in Debye is approximately:

MCQ 8Concept TrapPractice

Which of the following pairs correctly represents one molecule with zero dipole moment and one with non-zero dipole moment?

Quick recall before you leave

Worked Example

  1. 1

    Given

    - Bond length: d = 1.5 × 10⁻¹⁰ m - Partial charge: q = 0.20 × 1.6 × 10⁻¹⁹ C = 0.20e

  2. 2

    Required

    Dipole moment μ in Debye.

  3. 3

    Concept

    Dipole moment is the product of charge magnitude and separation distance. For a diatomic, no vector addition is needed — the bond dipole IS the molecular dipole (NCERT Class 11 Chemistry Chapter 4, page 10).

  4. 4

    Formula

    μ = q × d

  5. 5

    Substitution

    First, find q in coulombs: q = 0.20 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻²⁰ C Then: μ = 3.2 × 10⁻²⁰ C × 1.5 × 10⁻¹⁰ m

  6. 6

    Calculation

    μ = 3.2 × 1.5 × 10⁻²⁰⁻¹⁰ C·m = 4.8 × 10⁻³⁰ C·m Converting to Debye: μ = 4.8 × 10⁻³⁰ / 3.336 × 10⁻³⁰ = 1.44 D **Note on exact constants:** The elementary charge e = 1.6 × 10⁻¹⁹ C is used here as a given exact value (problem-defined). The conversion factor 1 D = 3.336 × 10⁻³⁰ C·m is a defined constant. Neither constrains the significant figures of the answer. The answer is reported to 2 significant figures, governed by the given values 0.20 and 1.5 (each 2 sig figs).

  7. 7

    Final answer

    μ ≈ 1.4 D (2 significant figures)

  8. 8

    Common trap

    A frequent error is forgetting to multiply the fractional charge (0.20e) by the value of e before using μ = q × d. Using 0.20 directly (without converting to coulombs) gives a nonsensical answer. Another trap: confusing the Debye conversion direction — you divide C·m by 3.336 × 10⁻³⁰ to get Debye, not multiply.

  9. 9

    Similar NEET-style question

    "The bond length of HCl is 1.27 × 10⁻¹⁰ m. If the dipole moment is 1.03 D, calculate the percentage ionic character of the bond." (Requires computing the theoretical μ for full charge separation, then taking the ratio.) ---

Before solving, remember these

Formula

Dipole moment

μ = q × d, where q is charge and d is bond length. SI: C·m, common: Debye (D) = 3.336 × 10⁻³⁰ C·m. Diatomic dipole moment depends on electronegativity difference and bond length.

-- NCERT Class 11 Chemistry, Ch. 4, p. 10

Formulas

Bond order from MO theory

Higher bond order: shorter, stronger bond. N₂: BO=3, O₂: BO=2, F₂: BO=1.

SymbolQuantitySI Unit
N_bbonding electrons-
N_aantibonding electrons-
BObond order-

Valid when

  • MO theory framework
  • Closed-shell molecule (or with appropriate treatment)

Dipole moment

Product of charge magnitude and bond length. SI: C·m. Common: Debye (1 D = 3.336e-30 C·m).

SymbolQuantitySI Unit
qchargeC
dbond lengthm
mudipole momentC*m or D

Valid when

  • Diatomic or vector-summed for polyatomic
  • Polar bond

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Hybridization counting lone pairs

Category: Inorganic Exception

Student counts only bonded atoms when assigning hybridization. Lone pairs count toward steric number too. Steric number = bond pairs + lone pairs → hybridization.

When it triggers

Molecule with central atom having lone pairs (e.g., NH₃: 3 bonds + 1 lp = 4 = sp³; H₂O: 2+2 = 4 = sp³).

How to avoid

Steric number formula: SN = (bond pairs) + (lone pairs). SN=2: sp; SN=3: sp²; SN=4: sp³; SN=5: sp³d; SN=6: sp³d². Lone pairs distort but still count.

Past Year Questions

14 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Given below are two statements : Statement I : A hypothetical diatomic molecule with bond order zero is quite stable. Statement II : As bond order increases, the bond length increases. In the light of the above statements, choose the most appropriate answer from the options given below :

1Statement I is false but Statement II is true
2Both Statement I and Statement II are true
3Both Statement I and Statement II are false
4Statement I is true but Statement II is false
NTA Answer: Option 3(final)
NEET 2025

Given below are two statements : Statement I : Like nitrogen that can form ammonia, arsenic can form arsine. Statement II : Antimony cannot form antimony pentoxide. In the light of the above statements, choose the most appropriate answer from the options given below :

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 4(final)
NEET 2022

Which statement regarding polymers is not correct?

1Thermosetting polymers are reusable
2Elastomers have polymer chains held together by weak intermolecular forces
3Fibers possess high tensile strength
4Thermoplastic polymers are capable of repeatedly softening and hardening on heating and cooling respectively
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 11 Chemistry Chapter 4, p.10

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