Edexcel International A Level Chemistry: Intermolecular forces (YCH11)

This guide teaches Topic 7: Intermolecular Forces (outcomes 7.1 to 7.6) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017. The topic sits in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols, an IAS compulsory unit, so everything here is AS content.

Use it with the revision notes and the practice questions. Polarity comes from Unit 1; revise ionic and covalent bonding and shapes of molecules if needed. Course hub: Edexcel International A Level Chemistry. Checklist: printable checklist. Find your gaps with the free 10-minute diagnostics.

What this topic covers

Outcome	What you must be able to do
7.1	Describe London forces, permanent dipole-permanent dipole interactions and hydrogen bonds
7.2	Explain the interactions that give hydrogen bonding in H₂O, liquid NH₃ and liquid HF
7.3	Explain water’s high melting and boiling temperatures and why ice is less dense than water
7.4	Predict whether hydrogen bonding is present in molecules like those in 7.2
7.5	Explain boiling temperature trends: alkane chain length, branching, alcohols versus alkanes, HF to HI
7.6	Explain solvent choice: hydration of ions, alcohols in water, poor solubility of halogenoalkanes, non-aqueous solvents

7.1 The three intermolecular forces

These forces hold separate molecules to one another. Covalent bonds inside each molecule are far stronger, and none of them break when a simple molecular substance melts or boils.

London forces (instantaneous dipole-induced dipole)

Electrons in a molecule are always moving. At any instant the electron cloud may be slightly uneven, so one end is briefly δ− and the other δ+. This instantaneous dipole distorts the electron cloud of a neighbouring molecule and creates an induced dipole in it. The two dipoles attract. The pattern keeps shifting, but there is always a net attraction.

London forces act between all molecules, polar or not. They get stronger when:

• the molecule has more electrons, because a larger electron cloud is more easily distorted (more polarisable);
• the molecules have more surface contact, because more of each molecule is close enough to interact.

Permanent dipole-permanent dipole interactions

A molecule with a permanent dipole (polar bonds arranged so they do not cancel) has a δ+ end and a δ− end at all times. The δ+ end of one molecule attracts the δ− end of the next. Examples include HCl, CH₃Cl and propanone. These molecules also have London forces, which are often the bigger contribution.

Hydrogen bonds

A hydrogen bond forms when a hydrogen atom covalently bonded to N, O or F is attracted to a lone pair on an N, O or F atom in another molecule. The X-H bond is very polar and the tiny, unshielded H atom carries a concentrated δ+ charge, so it can approach the lone pair closely. The three atoms (X-H···Y) lie close to a straight line, at about 180°, because that keeps the two electron-rich atoms furthest apart. Hydrogen bonds are the strongest of the three, but far weaker than a covalent bond.

Worked example 1. Name the types of intermolecular force in each of: Br₂, ICl, CH₃OH.

• Br₂: identical atoms, no permanent dipole. Only London forces.
• ICl: Cl is more electronegative than I, so the molecule has a permanent dipole. Permanent dipole-permanent dipole interactions, plus London forces.
• CH₃OH: has an O-H bond and lone pairs on O. Hydrogen bonds, plus London forces and permanent dipole interactions.

7.2 Hydrogen bonding in water, ammonia and hydrogen fluoride

Each molecule has H bonded to N, O or F and lone pairs, so it can both donate (through H) and accept (through a lone pair) hydrogen bonds.

Molecule	H atoms on N/O/F	Lone pairs on N/O/F	Hydrogen bonds per molecule
H₂O	2	2	Up to 4 (two as donor, two as acceptor)
NH₃	3	1	Limited by the single lone pair
HF	1	3	Limited by the single H atom

In liquid HF the molecules link into zig-zag chains, F-H···F-H···F. In liquid NH₃, the single lone pair leaves most H atoms unused. Water is balanced, so in ice every molecule takes part in four hydrogen bonds.

This explains the order of boiling temperatures: H₂O (373 K) > HF (293 K) > NH₃ (240 K). An individual F-H···F bond is stronger than an O-H···O bond because F is more electronegative, but water forms twice as many hydrogen bonds per molecule, so overall more energy is needed to separate water molecules.

7.3 The anomalous properties of water

High melting and boiling temperatures

Compare water with the other Group 16 hydrides, which have the same shape.

Hydride	Electrons	Boiling temperature / K
H₂O	10	373
H₂S	18	213
H₂Se	36	232
H₂Te	54	271

From H₂S to H₂Te, boiling temperature rises as the number of electrons rises, because London forces get stronger. On this trend water, with the fewest electrons, should boil below 213 K, yet it boils 160 K above H₂S. The extra energy is needed to break hydrogen bonds, which H₂S, H₂Se and H₂Te cannot form (S, Se and Te are not electronegative enough). The same reasoning explains its high melting temperature.

Why ice floats

Ice locks every H₂O into four hydrogen bonds pointing to the corners of a tetrahedron. The near-linear, fairly long hydrogen bonds hold the molecules in a rigid, open framework with much empty space. When ice melts, some hydrogen bonds break and the framework partly collapses. Molecules can then pack more closely, so liquid water is denser than ice.

Worked example 2. At 0 °C, ice has density 0.917 g cm⁻³ and liquid water has density 1.00 g cm⁻³. Find the volume of one mole (18.0 g) of each, and the extra volume per molecule in ice (L = 6.02 × 10²³ mol⁻¹).

• Ice: 18.0 / 0.917 = 19.6 cm³
• Liquid: 18.0 / 1.00 = 18.0 cm³
• Extra per molecule: (19.63 − 18.00) / 6.02 × 10²³ = 2.71 × 10⁻²⁴ cm³.

This extra volume is empty space in the framework.

7.4 Predicting hydrogen bonding

Test for hydrogen bonding between molecules of one substance:

1. Is there an H atom bonded directly to N, O or F?
2. Is there a lone pair on an N, O or F atom to accept it?

Both must be yes. A C-H bond never counts, even if F or O is elsewhere in the molecule.

Worked example 3. Which of these form hydrogen bonds between their own molecules: hydrazine (H₂N-NH₂), methoxymethane (CH₃OCH₃), fluoromethane (CH₃F), hydrogen peroxide (HO-OH)?

• Hydrazine: N-H bonds and N lone pairs. Yes.
• Methoxymethane: the O has lone pairs, but every H is bonded to C. No. (It can still accept a hydrogen bond from a water molecule, which matters for solubility.)
• Fluoromethane: the H atoms are on C, not on F. No. It has permanent dipole-permanent dipole and London forces.
• Hydrogen peroxide: O-H bonds and lone pairs on O. Yes.

7.5 Explaining physical properties

Alkanes: chain length

Alkane	Electrons	Boiling temperature / K
Butane	34	273
Pentane	42	309
Hexane	50	342
Heptane	58	372
Octane	66	399

Alkanes are non-polar, so London forces are the only intermolecular forces. Each extra CH₂ adds 8 electrons and more surface area, so London forces get stronger and boiling temperature rises. The rise per CH₂ shrinks along the series (36 K from butane to pentane, 27 K from heptane to octane).

Alkanes: branching

Hexane (342 K), 2-methylpentane (333 K) and 2,2-dimethylbutane (323 K) are isomers with 50 electrons each, so the difference cannot come from electron number. A straight chain can lie alongside its neighbours with a lot of surface in contact. Branched molecules are more compact and closer to spherical, so they touch over a smaller area. Fewer points of contact means weaker London forces, so less energy is needed to separate the molecules.

Alcohols versus alkanes

Compare propan-1-ol (CH₃CH₂CH₂OH, 34 electrons, 370 K) with butane (34 electrons, 273 K). Equal electron numbers mean similar London forces. Propan-1-ol also has an O-H group, so its molecules form hydrogen bonds, which need extra energy to break. It boils 97 K higher, so it is much less volatile.

Hydrogen halides HF to HI

Worked example 4. Explain this data: HF 293 K, HCl 188 K, HBr 206 K, HI 238 K.

• HCl to HI: electrons rise from 18 to 54, so London forces get stronger. The bonds get less polar down the group, so permanent dipole interactions get weaker, but the increase in London forces is larger, so boiling temperature rises.
• HF: by the HCl to HI trend it should boil lowest, but it boils 105 K above HCl. HF molecules form hydrogen bonds (H on F, three lone pairs on F), which need much more energy to overcome.

7.6 Choosing a solvent

A substance dissolves well when the new solute-solvent attractions are comparable to those broken in the solute and the solvent.

Water dissolving ionic compounds: hydration

Water molecules are polar. The δ− oxygen atoms cluster around cations, and the δ+ hydrogen atoms point towards anions. These ion-dipole attractions are called hydration, and forming them releases energy. If enough energy is released to make up for breaking the ionic lattice and separating some water molecules, the solid dissolves. Some ionic compounds, such as calcium carbonate, stay insoluble because hydration does not compensate.

Water dissolving simple alcohols

Methanol, ethanol and propan-1-ol mix with water in all proportions. The O-H group forms hydrogen bonds with water molecules, replacing the hydrogen bonds broken in each pure liquid. As the hydrocarbon chain gets longer, more of the molecule cannot hydrogen bond, so solubility falls.

Why water is a poor solvent for halogenoalkanes

1-chlorobutane is polar but cannot form hydrogen bonds with water. To dissolve it, water would have to break hydrogen bonds and form only weaker permanent dipole and London attractions in their place. That exchange is not favourable, so the liquids form separate layers. Polarity alone does not make a substance water-soluble.

Non-aqueous solvents

A non-polar or weakly polar solute dissolves best in a solvent with similar intermolecular forces. Iodine and waxes dissolve in hexane or cyclohexane: London forces in the solute are replaced by similar ones with the solvent. Halogenoalkanes dissolve in ethanol or other halogenoalkanes (“like dissolves like”).

Worked example 5. Choose water or hexane as the better solvent for: potassium bromide, ethanol, naphthalene (a non-polar hydrocarbon).

• Potassium bromide: water, which hydrates the ions.
• Ethanol: water, through hydrogen bonds.
• Naphthalene: hexane. Only London forces in both; it cannot hydrogen bond with water.

Suggested practicals

The specification suggests investigating solubility of simple molecules in different solvents, measuring the enthalpy change of vaporisation of water, and measuring temperature changes when substances dissolve.

Worked example 6. A 150 W immersion heater keeps a beaker of water boiling on a balance. Over 600 s the mass falls by 38.5 g. Find the enthalpy change of vaporisation.

• Energy supplied = 150 × 600 = 90 000 J = 90.0 kJ
• Moles evaporated = 38.5 / 18.0 = 2.14 mol
• ΔvapH = 90.0 / 2.14 = +42.1 kJ mol⁻¹

Heat lost to the surroundings makes this method overestimate the true value. The energy breaks hydrogen bonds between molecules, not O-H bonds.

Common errors

• Claiming that boiling a simple molecular liquid splits its covalent bonds.
• Counting a C-H bond as able to hydrogen bond.
• Comparing alcohols and alkanes without matching electron numbers first.
• Explaining branching with “fewer electrons”. Isomers have identical electron counts; contact area is the point.
• Claiming HCl boils lower than HI because it has weaker dipoles. HCl has the stronger dipole; London forces decide the trend.
• Saying a halogenoalkane is insoluble because it is non-polar. It is polar; it cannot hydrogen bond with water.

Next: the practice questions.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11), Specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 7: Intermolecular Forces, Unit 2.