← Unit 1: How can the diversity of materials be explained?
How can the versatility of non-metals be explained?
the nature of intermolecular forces (dispersion, dipole-dipole and hydrogen bonding) and the relationship of structure to physical properties of covalent molecular, covalent network and covalent layered (graphite) substances, including the allotropes of carbon
A focused VCE Chemistry Unit 1 answer on intermolecular forces. Covers dispersion, dipole-dipole and hydrogen bonding, ranking and predicting boiling points, and the structure and properties of covalent molecular, covalent network and covalent layered (graphite, graphene) substances and the allotropes of carbon.
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What this dot point is asking
VCAA wants you to name and explain the three intermolecular forces (dispersion, dipole-dipole, hydrogen bonding), to rank substances by boiling point using these forces, and to explain the structure and properties of the three classes of covalent material: covalent molecular (water, sugar, iodine), covalent network (diamond, silicon dioxide) and covalent layered (graphite), including the allotropes of carbon (diamond, graphite, graphene, fullerenes).
The answer
Intermolecular forces (IMFs)
Intermolecular forces are attractions between molecules, not within them. Breaking an IMF (boiling or melting a molecular substance) takes far less energy than breaking the covalent bonds inside a molecule.
The three IMFs to know, weakest to strongest in general:
1. Dispersion forces (London forces). Present between all molecules (and atoms). They arise from instantaneous, fluctuating dipoles in the electron cloud of one molecule inducing a dipole in a neighbour. Strength increases with the number of electrons (and surface area). Iodine (I2) is a solid at room temperature because of strong dispersion forces between large electron clouds; chlorine (Cl2) is a gas because of weaker dispersion.
2. Dipole-dipole attractions. Present only between polar molecules. The partial positive end of one molecule attracts the partial negative end of another. Stronger than dispersion (for molecules of similar size), but weaker than hydrogen bonds.
3. Hydrogen bonding. A special, particularly strong dipole-dipole interaction. Requires all three of:
- An H atom covalently bonded to N, O or F (highly electronegative, small).
- A lone pair on N, O or F on a neighbouring molecule to receive it.
- Direct line-up (the H, plus the donor and acceptor atoms, are roughly collinear).
The classic hydrogen-bonded substances: water, ammonia, hydrogen fluoride, alcohols, carboxylic acids, amides, proteins, DNA.
Hydrogen bonds are typically 5 to 10 times stronger than ordinary dipole-dipole, but still much weaker than covalent bonds.
Predicting boiling points
Apply this order of decisions:
- If a substance is an ionic compound, metallic or a covalent network solid, the bonding inside is ionic, metallic or covalent (not IMF). Very high melting/boiling points.
- Otherwise the substance is covalent molecular. Identify which IMFs are present in order of strength: hydrogen bonding > dipole-dipole > dispersion only.
- Within a single IMF type (especially dispersion-only substances), boiling point increases with the number of electrons (or molar mass) and with greater surface contact (straight-chain isomers boil higher than branched).
A quick comparison:
| Substance | Dominant IMF | bp (deg C) |
|---|---|---|
| CH4 | Dispersion | -161 |
| HCl | Dipole-dipole | -85 |
| H2O | Hydrogen bond | 100 |
| HF | Hydrogen bond | 20 |
| NH3 | Hydrogen bond | -33 |
| CO2 | Dispersion only (non-polar) | -78 (sublimes) |
| I2 | Dispersion (many electrons) | 184 |
Three classes of covalent material
Covalent molecular substances. Discrete molecules held together by IMFs in the solid and liquid states. Examples: water (H2O), sucrose (C12H22O11), iodine (I2), CO2 (dry ice). Properties:
- Low to moderate melting/boiling points (only weak IMFs to break).
- Generally soft as solids.
- Do not conduct electricity in any state (no charged particles free to move).
- Solubility depends on polarity: polar dissolves in polar (water), non-polar dissolves in non-polar (hexane).
Covalent network (covalent lattice) substances. Every atom in the solid is bonded covalently to its neighbours in a 3D giant lattice. There are no discrete molecules. Examples: diamond, silicon dioxide (quartz), silicon carbide. Properties:
- Very high melting/boiling points (covalent bonds must break to melt).
- Extremely hard.
- Do not conduct electricity in pure diamond/quartz (no delocalised electrons).
- Insoluble in everything.
Covalent layered substances. Atoms within a layer are covalently bonded, but layers themselves are held together by weak dispersion forces. The standout example is graphite.
Allotropes of carbon
Allotropes are different structural forms of the same element. Carbon has several worth knowing:
| Allotrope | Structure | Bonding within structure | Properties |
|---|---|---|---|
| Diamond | 3D tetrahedral network | Each C bonds to 4 other C via single covalent bonds. All 4 valence electrons localised in bonds. | Extremely hard, very high mp (sublimes ~3550 deg C), no electrical conductivity, transparent. |
| Graphite | 2D hexagonal sheets stacked with weak dispersion between sheets | Each C bonds to 3 other C in a flat hexagonal layer; the 4th valence electron is delocalised in each layer. | Soft and slippery (sheets slide), conducts along sheets, high mp. |
| Graphene | A single layer of graphite (one sheet) | Same hexagonal bonding within the sheet | Extremely strong (tensile), excellent electrical and thermal conductor along the sheet, transparent and flexible. |
| Fullerenes (e.g. buckminsterfullerene C60) | Closed cages of 60 C atoms in a soccer-ball pattern of pentagons and hexagons | Each C bonds to 3 others; some delocalisation | Discrete molecular substance with relatively low mp; soluble in organic solvents; potential drug-delivery and electronic applications. |
| Carbon nanotubes | A graphene sheet rolled into a cylinder | Each C bonds to 3 others, with delocalisation along the tube | Very high tensile strength along axis, conducts along the tube. |
Worked example
Rank the following by boiling point and justify: ethane (C2H6), ethanol (C2H5OH), and butane (C4H10).
- Ethane: non-polar, dispersion only, small. Lowest bp (-89 deg C).
- Butane: non-polar, dispersion only, but more electrons and more surface area than ethane. Higher than ethane (-1 deg C).
- Ethanol: polar with an O-H group, capable of hydrogen bonding. Despite having fewer electrons than butane, the hydrogen bonds raise the bp far above what dispersion would give. Highest bp (78 deg C).
So the order is C2H6 < C4H10 < C2H5OH. A common student error is to rank purely by molar mass and put butane above ethanol; hydrogen bonding wins.
Common traps
Confusing intermolecular and intramolecular forces. Boiling water breaks IMFs (hydrogen bonds), not the O-H covalent bonds. Water vapour is still H2O.
Calling all polar molecules hydrogen bonded. HCl is polar but does not hydrogen bond (Cl is too large and not electronegative enough; only N, O, F qualify).
Forgetting that all molecules have dispersion forces. Non-polar molecules have only dispersion. Polar molecules have dispersion plus dipole-dipole.
Saying graphite has no covalent bonds because the layers slide. Within each layer the bonding is strong covalent. The layers slide because the layer-to-layer force is weak dispersion.
Calling diamond a metal because it is hard. Diamond is a covalent network, not metallic. It does not conduct.
Quoting H2O bp as evidence that water is "an exception to molecular substances". Water is a normal covalent molecular substance; its high bp is fully explained by hydrogen bonding, not an exception to the model.
In one sentence
Intermolecular forces (dispersion, dipole-dipole, hydrogen bonding) hold covalent molecular substances together with weak attractions that set their melting and boiling points, while covalent network substances like diamond and covalent layered substances like graphite involve continuous covalent bonding through the structure and explain the very different properties of the allotropes of carbon.
Past exam questions, worked
Real questions from past VCAA papers on this dot point, with our answer explainer.
2024 VCE4 marksExplain why the boiling point of water (100 degrees C) is much higher than that of hydrogen sulfide (-60 degrees C), even though both molecules have the same shape and similar molar mass. Identify the intermolecular forces present in each.Show worked answer →
A 4-mark answer needs both substances analysed, the IMFs named, and the H-bonding criterion stated.
H2O: bent shape, two O-H bonds. O is highly electronegative (3.44) and small. H2O molecules have hydrogen bonding (a strong dipole-dipole attraction between an H bonded to N, O or F on one molecule and a lone pair on an N, O or F of another), plus weaker dipole-dipole and dispersion forces.
H2S: bent shape, two S-H bonds. S is not electronegative enough (2.58) to support hydrogen bonding. H2S has dipole-dipole attractions plus dispersion forces, no hydrogen bonds.
Boiling point comparison: hydrogen bonds are about 5 to 10 times stronger than typical dipole-dipole forces. Boiling H2O requires breaking the hydrogen-bond network; boiling H2S requires breaking only dipole-dipole and dispersion. That accounts for the 160 degree difference despite the molecules being structurally similar.
Markers reward stating the three criteria for hydrogen bonding (H bonded to N, O or F; lone pair on N, O or F to receive it; only those three are electronegative and small enough).
2025 VCE3 marksDiamond and graphite are both allotropes of carbon. Diamond is one of the hardest known substances but does not conduct electricity, while graphite is soft and slippery and conducts electricity along the layers. Explain these properties in terms of structure and bonding.Show worked answer →
A 3-mark answer needs both structures and the link to each property.
Diamond: each C atom forms 4 single covalent bonds to 4 neighbours in a 3D tetrahedral lattice (covalent network). All 4 valence electrons are in covalent bonds, so there are no delocalised electrons to carry charge. The 3D network of strong covalent bonds means no plane along which the structure can slide, so diamond is extremely hard and does not conduct.
Graphite: each C atom forms 3 single covalent bonds to 3 neighbours, arranged in flat hexagonal sheets (covalent layered). The 4th valence electron is delocalised within each layer, so graphite conducts electricity along the layers. The layers are held together by weak dispersion forces only, so they slide past one another easily, making graphite soft and slippery (used as a dry lubricant and in pencils).
Both diamond and graphite have very high melting points because melting either requires breaking many covalent bonds.