1.Lesson overview
- 3.5 Shapes of molecules
- 3.6 Intermolecular forces, electronegativity and bond properties
- 3C Shapes of molecules
- Types of intermolecular force and hydrogen bonding (7.1–7.4)
- Intermolecular forces and physical properties (7.5)
- Choice of solvents (7.6)
- 1.3 Bonding (International AS)
- 1.8 Valence Electrons and Ionic Compounds
- 2.1 Types of Chemical Bonds
- 2.3 Structure of Ionic Solids
- 2.4 Structure of Metals and Alloys
- 2.5 Lewis Diagrams
- 2.6 Resonance and Formal Charge
- 2.7 VSEPR and Hybridization
- 3.1 Intermolecular and Interparticle Forces
- Use electron-pair repulsion to predict molecular and ionic shapes from the number of bonding pairs and lone pairs around a central atom.
- State the standard bond angles for common shapes and explain how lone pairs reduce bond angles.
- Use electronegativity and molecular geometry to distinguish polar bonds from polar molecules.
- Identify and rank London forces, permanent dipole–dipole attractions and hydrogen bonding.
- Explain differences in boiling point, volatility and solubility using intermolecular forces and molecular shape.
A molecule's formula does not determine its shape by itself. Electron pairs repel into a three-dimensional arrangement, and that shape determines whether individual bond dipoles cancel. The result controls intermolecular attractions, volatility, solubility and many real-world uses.
Use electron-pair repulsion to predict shapes and angles, account for the stronger repulsion of lone pairs, combine shape with bond polarity to decide molecular polarity, then compare London forces, permanent dipoles and hydrogen bonding.
2.Key language and ideas
- Electron-pair repulsion theory
- A model in which electron pairs around a central atom arrange themselves to minimise repulsion.
- Bond angle
- The angle between two bonds around a central atom.
- Lone pair
- A non-bonding pair of electrons that occupies more space around a central atom than a bonding pair.
- Bond dipole
- A separation of partial charge caused by unequal sharing of electrons in a polar covalent bond.
- Permanent dipole
- A molecule-wide uneven charge distribution resulting from bond polarity and non-cancelling shape.
- Hydrogen bond
- A strong intermolecular attraction involving bonded to , or and a lone pair on a neighbouring molecule.
3.Count electron domains before naming a shape
Around a central atom, bonding pairs and lone pairs repel. A double or triple bond counts as one electron domain for shape prediction because its electron density lies in one direction. First identify the total number of domains, arrange them as far apart as possible, then describe the shape made by the atoms only.
Two domains are linear, three trigonal planar, four tetrahedral, five trigonal bipyramidal and six octahedral. Electron-pair repulsion predicts ideal angles, while real angles reflect differences in pair-pair repulsion.
4.Polarity is a vector sum, not a bond label
Electron-pair repulsion sets an approximate molecular geometry, while bond electronegativities create individual bond dipoles. Overall molecular polarity depends on the vector sum of those dipoles, so polar bonds can cancel in a symmetric molecule. Intermolecular forces then help explain boiling points, solubility and viscosity, but comparisons should control for molecular size and shape. Hydrogen bonding requires an appropriate H–N, H–O or H–F donor and an acceptor, not merely the presence of hydrogen.
5.Know the standard shapes and their bond angles
Common molecular shapes must be recalled accurately: and are linear, is trigonal planar, is tetrahedral, is trigonal bipyramidal and is octahedral. State the angle only when it belongs to the named position in the shape.
For five domains, equatorial positions are apart and axial-equatorial angles are . For six domains, adjacent bonds are . These spatial distinctions matter when explaining repulsion or polarity.
| Electron domains | Molecular example | Shape and typical angle |
|---|---|---|
| 2 | Linear, | |
| 3 | Trigonal planar, | |
| 4, no lone pairs | Tetrahedral, | |
| 4, one lone pair | Trigonal pyramidal, about | |
| 4, two lone pairs | Non-linear, about |
6.Explain why lone pairs compress bond angles
Repulsion strength follows lone pair–lone pair–bond pair–bond pair. Lone pairs are held by one nucleus rather than shared between two, so their electron density is more concentrated near the central atom and occupies more space.
Thus methane has the ideal tetrahedral angle of , ammonia has about and water about . Do not say that lone pairs attract bonds closer together; they repel bonding pairs more strongly and force the bonding pairs nearer to one another.
7.Separate polar bonds from polar molecules
A bond is polar when atoms of different electronegativity share electrons unequally. A molecule is polar only if its bond dipoles do not cancel as vectors. Symmetrical has two polar bonds but no overall dipole; bent water has polar O–H bonds that reinforce to give a polar molecule.
Judge polarity after finding the shape. Consider identical outer atoms, different outer atoms, lone pairs and any geometry that makes dipoles cancel. This is more reliable than deciding from the formula or from the presence of one electronegative atom.
8.Rank intermolecular forces and predict physical properties
London forces arise from temporary fluctuations in electron density that induce dipoles in neighbouring particles. They act between all atoms and molecules and grow with electron number, polarizability and surface contact. Branching reduces surface contact, so more-branched isomers usually have lower boiling temperatures.
Permanent dipole–permanent dipole attractions occur between polar molecules. Hydrogen bonding is a particularly strong case when H is bonded to N, O or F and can interact with a lone pair on a neighbouring molecule. Stronger intermolecular forces raise boiling temperature and reduce volatility because more energy is needed to separate molecules.
9.Use shape to decide whether bond dipoles cancel
Molecular polarity is a vector question. A polar bond has an unequal sharing of electrons because the bonded atoms have different electronegativities, but the molecule is polar only if the individual bond dipoles do not cancel. Symmetrical linear, trigonal-planar or tetrahedral molecules can be non-polar even when every bond is polar.
Lone pairs matter twice: they occupy electron domains in VSEPR reasoning and they often produce a net dipole because the molecule becomes asymmetric. This is why CO₂ is non-polar whereas H₂O is polar, and why NH₃ has a net dipole whereas BF₃ does not. The resulting intermolecular forces then influence boiling point, solubility and viscosity.
10.Worked example 1: Predict shape and polarity of ammonia
Predict the shape, bond angle and polarity of .
- 1Nitrogen has three bonding pairs to H and one lone pair: four electron domains.
- 2The electron-domain arrangement is tetrahedral, but the atoms form a trigonal pyramidal shape.
- 3The lone pair repels bonding pairs more strongly, reducing the H–N–H angle from to about .
- 4The N–H bond dipoles do not cancel in this asymmetric shape, so ammonia is polar.
11.Worked example 2: Compare boiling temperatures of isomers
Explain why pentane has a higher boiling temperature than 2-methylbutane, although both have formula .
- 1Both are non-polar hydrocarbons, so London forces are the key intermolecular attraction.
- 2Pentane is less branched and has a larger surface area for close contact between molecules.
- 3This produces stronger London forces between pentane molecules.
- 4More energy is therefore needed to separate pentane molecules, giving the higher boiling temperature.
12.Worked example 3: Compare the polarity of CO₂ and H₂O, even though both contain polar C–O or O–H bonds.
Compare the polarity of CO₂ and H₂O, even though both contain polar C–O or O–H bonds.
- 1CO₂ is linear, so the two C–O bond dipoles point in opposite directions with equal magnitude and cancel.
- 2H₂O has two lone pairs on oxygen, giving a bent shape rather than a straight one.
- 3The O–H dipoles in bent H₂O do not cancel, so water has a permanent dipole.
- 4Water therefore has stronger permanent dipole interactions and hydrogen bonding, helping explain its unusually high boiling temperature.
13.Extended worked case: apply and evaluate
Both and have polar bonds. Explain why only water has a permanent molecular dipole.
- 1
is linear, so its two equal bond dipoles point in opposite directions and cancel.
- 2
is bent because oxygen has two lone pairs; its O–H dipoles do not cancel.
- 3
Water is polar and can form intermolecular hydrogen bonds, strongly affecting its physical properties.
Classify whole-molecule polarity from shape and bond dipoles together; 'contains polar bonds' is insufficient.
14.Relating structure to volatility and solubility
Comparing safe liquid samples or supplied boiling-point data lets you test predictions based on molecular shape and intermolecular forces. Use a controlled comparison rather than attributing every difference to one variable.
- 1PredictStage 1Classify samples by polarity and identify the strongest intermolecular force expected.
- 2CompareStage 2Use given boiling-point, evaporation-time or solubility data under the same conditions.
- 3ExplainStage 3Link the observed trend to the energy required to overcome attractions between particles.
15.Connections, patterns and applications
16.Exam method: from formula to molecular polarity
Never decide molecular polarity before you have considered the three-dimensional shape.
- 1Count domainsStage 1Count bonding regions and lone pairs around the central atom.
- 2Name the shapeStage 2Give the molecular shape, not only the electron-pair arrangement, and state a justified angle.
- 3Resolve dipolesStage 3Identify polar bonds and decide whether their vector effects cancel in that shape.
17.Common misconceptions
- Counting a double bond as two electron domains in repulsion theory.
- Calling carbon dioxide polar because the bonds are polar.
- Explaining water's boiling point with covalent O–H bonds rather than intermolecular hydrogen bonds.
18.Exam focus and retrieval
- Use non-linear rather than bent if the specification's terminology expects it, but explain the lone-pair effect.
- For boiling-point trends, first identify the strongest force and then compare electron number, shape or hydrogen bonding.
- Include values such as , and where the molecule warrants them.
- 1Why is non-polar but polar?
- 2Which pair repulsion is strongest?
- 3Why do larger alkanes generally have higher boiling temperatures?
19.Summary
- Electron domains repel to give predictable molecular shapes and bond angles.
- Lone pairs compress bond angles more than bonding pairs.
- Molecular polarity requires non-cancelling bond dipoles.
- London forces act between all particles; hydrogen bonding is especially strong and raises boiling temperature.
- Predict molecular shape and polarity with VSEPR theory, then use intermolecular forces to explain physical behaviour.
20.Curriculum alignment and applied reasoning
This extension turns the lesson into an exam-ready sequence: identify the evidence, apply the mechanism or calculation, then state a qualified conclusion. Core outcomes revisited here include: Use electron-pair repulsion to predict molecular and ionic shapes from the number of bonding pairs and lone pairs around a central atom.; State the standard bond angles for common shapes and explain how lone pairs reduce bond angles.; Use electronegativity and molecular geometry to distinguish polar bonds from polar molecules..
| Course | Mapped focus in this lesson |
|---|---|
| Cambridge International A Level Chemistry 9701 | 3.5 Shapes of molecules 3.6 Intermolecular forces, electronegativity and bond properties |
| Edexcel IAL Chemistry | 3C Shapes of molecules Types of intermolecular force and hydrogen bonding (7.1–7.4) Intermolecular forces and physical properties (7.5) Choice of solvents (7.6) |
| AQA International A-level Chemistry | 1.3 Bonding (International AS) |
| AP Chemistry | 1.8 Valence Electrons and Ionic Compounds 2.1 Types of Chemical Bonds 2.3 Structure of Ionic Solids 2.4 Structure of Metals and Alloys 2.5 Lewis Diagrams 2.6 Resonance and Formal Charge 2.7 VSEPR and Hybridization 3.1 Intermolecular and Interparticle Forces |
Scenario: Two molecules have polar bonds, but only one is polar overall. Use three-dimensional shape to explain this and predict which molecule is more likely to have stronger intermolecular attractions.
Worked reasoning: Bond dipoles cancel in a symmetrical geometry but give a net dipole in an asymmetrical one. A polar molecule can show permanent dipole–dipole attractions in addition to London forces; molecules with O–H, N–H or F–H may hydrogen bond. Relative size also affects London forces.
Exam-quality communication: Draw or name the molecular shape before concluding about polarity.
- Name the observation, quantity, structure or variable before interpreting it.
- Show the causal step or calculation route; do not jump from data to a conclusion.
- State a limitation, condition or comparison whenever the evidence cannot justify an absolute claim.