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

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:

1. 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.
2. Otherwise the substance is covalent molecular. Identify which IMFs are present in order of strength: hydrogen bonding > dipole-dipole > dispersion only.
3. 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:

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:

Examples in context

Example 1. LNG storage at the Hastings terminal. AGL Energy's Crib Point LNG proposal and the existing Hastings facility in Victoria store liquefied natural gas at $-162^{\circ}\text{C}$. Methane $\text{CH}_4$ is held together in the liquid only by very weak dispersion forces, so the boiling point is extremely low. Compare ethane ($-89^{\circ}\text{C}$), propane ($-42^{\circ}\text{C}$) and butane ($-1^{\circ}\text{C}$): each additional $\text{CH}_2$ group adds about $25$ electrons, increasing the instantaneous dipole, and raising the boiling point by roughly $30^{\circ}\text{C}$. This molecular-mass trend is exploited in cryogenic distillation columns at the Longford gas plant, which separate $\text{CH}_4$, $\text{C}_2 \text{H}_6$ and $\text{C}_3 \text{H}_8$ by their distinct boiling points.

Example 2. Graphite electrodes at the Tomago aluminium smelter. The Tomago smelter in NSW uses massive carbon anodes consumed during electrolysis of $\text{Al}_2 \text{O}_3$. Graphite is a covalent layered substance: each carbon forms three $sp^2$ covalent bonds in flat hexagonal sheets, with the fourth valence electron delocalised between sheets. The strong in-plane covalent bonds give graphite a sublimation point above $3600^{\circ}\text{C}$, while the weak interlayer dispersion forces let sheets slide (lubricant) and let electrons conduct along the sheets. By contrast, diamond's full $sp^3$ network has no free electrons, so it does not conduct, even though it has very similar bond strength. The choice of graphite (not diamond) for anodes follows directly from this electron-mobility difference.

Try this

Q1. Explain why methane boils at $-162^{\circ}\text{C}$ but water boils at $100^{\circ}\text{C}$, despite similar molar masses. [3 marks]

• Cue. Methane: weak dispersion only. Water: hydrogen bonding (O-H to lone pair on O), much stronger; more energy needed to separate molecules.

Q2. Rank the boiling points of $\text{HF}$, $\text{HCl}$, $\text{HBr}$, $\text{HI}$ and explain the trend. [3 marks]

• Cue. Expected order by molar mass: HI > HBr > HCl > HF. Actual: HF (highest, hydrogen bonding), then HI > HBr > HCl by dispersion. Trap: HF breaks the trend due to H-bonding.

Q3. Compare graphite and diamond. (a) Describe the bonding in each. (b) State and explain one physical property that differs. (c) Predict which is harder and which conducts electricity. [2+2+2 marks]

• Cue. (a) Both covalent network; diamond $sp^3$ tetrahedral, graphite $sp^2$ layered with delocalised pi. (b) Conductivity: graphite conducts, diamond does not; or hardness: diamond very hard, graphite slippery. (c) Diamond harder; graphite conducts.