Dipole Moment in Chemical Bonding — Worksheet

20 questions covering concepts, geometry, vectors, and calculations. Answers are hidden; use the buttons or print to reveal.
Topic: Molecular Polarity & Dipole Moment (μ)
Level: High school / Intro college
Time: ~40–60 min
Reference data & formulas
  • Dipole moment (vector): \(\vec{\mu} = q\,\vec{r}\)
  • 1 Debye (D) = 3.33564 × 10^{-30} C·m
  • Elementary charge, e = 1.602 × 10^{-19} C
  • Percent ionic character ≈ (\mu_\text{obs}/\mu_\text{calc, 100% ionic}) × 100%
  • Resultant of two equal bond moments at angle θ: \(\mu_\text{net} = 2\,\mu_b\,\cos(\tfrac{\theta}{2})\)
Tip: On print, all answers auto-expand. For digital use, click “Show answer”.

Section A — Multiple Choice (6)

Q1
MCQ
The practical unit of dipole moment most commonly used in chemistry is:
  1. Coulomb
  2. Newton·meter
  3. Debye
  4. Tesla
Answer: C — Debye (D). (SI unit is C·m, but D is convenient.)
Q2
MCQ
Which molecule has an overall dipole moment of ~0 D?
  1. SO₂ (bent)
  2. NH₃ (trigonal pyramidal)
  3. H₂O (bent)
  4. BF₃ (trigonal planar)
Answer: D — BF₃. Symmetric trigonal planar geometry cancels bond dipoles.
Q3
MCQ
Correct qualitative order of dipole moments for hydrogen halides is:
  1. HCl > HF > HBr > HI
  2. HF > HCl > HBr > HI
  3. HI > HBr > HCl > HF
  4. HF > HBr > HCl > HI
Answer: B — Electronegativity difference dominates over the increase in bond length across the series.
Q4
MCQ
Which has the larger dipole moment?
  1. NF₃
  2. NH₃
  3. Both equal
  4. Insufficient information
Answer: B — In NF₃, the N–F bond dipoles partly oppose the lone-pair dipole; NH₃ has a larger net μ.
Q5
MCQ
Which isomer has μ ≈ 0?
  1. trans-1,2-dichloroethene
  2. cis-1,2-dichloroethene
  3. 1,1-dichloroethene
  4. None of the above
Answer: A — In the trans isomer, equal C–Cl bond moments cancel by symmetry.
Q6
MCQ
For ABn with all peripheral atoms identical, which geometries guarantee μ = 0?
  1. Linear AB₂
  2. Trigonal planar AB₃
  3. Tetrahedral AB₄
  4. All of the above
Answer: D — Perfectly symmetric geometries cancel vectorially.

Section B — Short Answer (7)

Q7
Short
Define dipole moment. Write its vector formula and two common units.
Answer: Separation of positive and negative charge in a system: \(\vec{\mu} = q\,\vec{r}\). Units: SI → C·m; practical → Debye (D).
Q8
Short
CO₂ is linear while SO₂ is bent. Explain why CO₂ has μ = 0 but SO₂ has μ > 0.
Answer: In CO₂, equal C=O bond dipoles are colinear and opposite → cancel. In SO₂, the bond angle (~120°) prevents complete cancellation → net μ along the bisector.
Q9
Short
Why is μ(NH₃) > μ(NF₃)? Refer to lone-pair and bond-dipole directions.
Answer: In NH₃, N→H bond dipoles add with the lone-pair dipole (toward N). In NF₃, highly polar N–F bonds point opposite to the lone-pair dipole, reducing the net μ.
Q10
Short
From the set {BF₃, PF₅, SF₆, XeF₄, IF₅, SF₄}, identify all polar molecules.
Answer: IF₅ (square pyramidal) and SF₄ (seesaw) are polar. BF₃, PF₅, SF₆, XeF₄ are symmetric → μ ≈ 0.
Q11
Short
Indicate the direction of the molecular dipole in acetone (CH₃–CO–CH₃). Mark δ⁺/δ⁻.
Answer: Carbonyl C is δ⁺, O is δ⁻. The dipole vector points from C (δ⁺) toward O (δ⁻), i.e., along C=O toward oxygen.
Q12
Short
How does carbon hybridization affect the dipole of a C–X bond (X = halogen)? Compare sp, sp², sp³.
Answer: More s-character → carbon more electronegative → smaller C–X polarity. Thus μ(sp) < μ(sp²) < μ(sp³) for the same X.
Q13
Short
Why is cis-1,2-dichloroethene polar but the trans isomer nearly nonpolar?
Answer: In the cis isomer, C–Cl bond moments reinforce; in the trans isomer, equal magnitudes oppose and largely cancel by symmetry.

Section C — Problems / Calculations (7)

Q14
Calc
Percent ionic character of HCl. Given: μobs = 1.08 D; r = 127.5 pm. Compute μcalc for 100% ionic and % ionic.
Solution: μcalc = (e·r)/D = (1.602×10⁻¹⁹ C × 1.275×10⁻¹⁰ m)/(3.33564×10⁻³⁰) ≈ 6.12 D. % ionic ≈ (1.08/6.12)×100 ≈ 17.6%.
Q15
Calc
Water model: each O–H bond has μb = 1.5 D; H–O–H angle = 104.5°. Estimate μ(H₂O).
Solution: μ = 2 μb cos(θ/2) = 2×1.5×cos(52.25°) ≈ 3.0×0.612 ≈ 1.84 D.
Q16
Calc
For diatomic XY: q = 0.32 e, r = 200 pm. Compute μ in Debye.
Solution: q = 0.32×1.602×10⁻¹⁹ = 5.126×10⁻²⁰ C. μ = q·r = 5.126×10⁻²⁰ × 2.00×10⁻¹⁰ = 1.025×10⁻²⁹ C·m ≈ (1.025×10⁻²⁹)/(3.33564×10⁻³⁰) = 3.07 D.
Q17
Calc
Approximate μ(SO₂) from bond vectors: assume each S=O bond has μb = 2.5 D and ∠OSO = 119°. (Planar, moments symmetric.)
Solution: μ ≈ 2 μb cos(θ/2) = 2×2.5×cos(59.5°) ≈ 5×0.508 ≈ 2.54 D.
Q18
Calc
Molecule AB has μ = 2.00 D and bond length r = 180 pm. Estimate the effective charge separation as a fraction of e.
Solution: q = μ/r = (2×3.33564×10⁻³⁰)/(1.80×10⁻¹⁰) = 3.71×10⁻²⁰ C ≈ (3.71×10⁻²⁰)/(1.602×10⁻¹⁹) = 0.231 e.
Q19
Calc
CO has μobs = 0.112 D and bond length 112.8 pm. What is its percent ionic character (magnitude)?
Solution: μcalc(100% ionic) = (e·r)/D = (1.602×10⁻¹⁹×1.128×10⁻¹⁰)/(3.33564×10⁻³⁰) ≈ 5.41 D. % ionic ≈ (0.112/5.41)×100 ≈ 2.07%.
Q20
Calc
A linear triatomic A–B–A has equal bond moments of 1.6 D each, colinear at 180°. Find μ.
Answer: 0 D — equal and opposite bond dipoles cancel in a symmetric linear molecule.

Teacher Notes

Emphasize vector nature of μ, role of geometry (VSEPR), and that large electronegativity difference alone does not guarantee large μ; bond length and cancellation matter. Encourage students to draw dipole arrows on Lewis structures before guessing polarity.