Suppose a chemist has two clear liquids in unlabelled bottles. One evaporates quickly at room temperature; the other stays in the watch glass much longer. A good A level explanation does not stop at "one has a lower boiling point". It explains which particles attract, how strongly they attract, and therefore how much energy is needed to separate particles into the gas phase.

Boiling a simple molecular substance does not break the covalent bonds inside the molecules. It separates molecules from each other. The mark-earning chain is:

1. identify the intermolecular forces between molecules
2. compare the strength or number of those forces
3. link stronger forces to more energy needed to separate molecules
4. conclude a higher boiling temperature and lower volatility

Volatility means how easily a substance vaporises. A more volatile liquid has weaker intermolecular forces, so more molecules can escape from the liquid at a given temperature.

For this lesson, keep three force ideas available.

These are intermolecular attractions. They are not full covalent bonds, ionic bonds or metallic bonds. When an alkane boils, C-C and C-H bonds remain intact; the molecules move further apart.

Feynman diagnostic

Explain to a younger student why pentane has a higher boiling temperature than methane without using the phrase "stronger bonds" on its own. A secure answer says that pentane molecules have more electrons and a larger surface area, so London forces between pentane molecules are stronger. More energy is needed to separate the molecules during boiling. The diagnostic trap is saying that pentane's covalent bonds are harder to break: boiling separates molecules, not atoms within molecules.

Alkanes are non-polar simple molecular substances. Their main intermolecular forces are London forces. That means their boiling-temperature trends are explained mainly by electron number, polarisation and surface contact.

As the carbon chain length increases, the number of electrons increases. Larger electron clouds are more easily polarised, so temporary dipoles induce stronger dipoles in neighbouring molecules. Longer chains also have more surface contact between molecules. Both effects increase London forces, so boiling temperature increases.

Representative normal boiling temperatures show the pattern:

The trend is not "longer covalent chain means the chain is harder to snap". The molecules are not being snapped. The longer molecules experience stronger attractions to other molecules.

Branching changes the surface shape. A straight-chain alkane can make more contact with neighbouring molecules than a compact, branched isomer with the same molecular formula. Less contact gives weaker London forces, so a branched alkane has a lower boiling temperature than its straight-chain isomer.

Compare two C4H10 isomers:

Worked example

Pentane, 2-methylbutane and 2,2-dimethylpropane are structural isomers with formula C5H12. Predict the order of boiling temperature, highest first, and explain the order.

Route

They have the same molecular formula, so they have the same number of electrons. The main difference is shape. Pentane is the least branched and has the greatest surface contact between molecules, so it has the strongest London forces. 2-methylbutane is more branched, so contact is reduced. 2,2-dimethylpropane is the most compact, so it has the least contact and the weakest London forces.

Answer

Highest boiling temperature: pentane, then 2-methylbutane, then 2,2-dimethylpropane. The boiling temperature falls as branching increases because London forces between molecules become weaker.

Guided practice

Two hydrocarbons have similar molar masses. Molecule A is a straight-chain alkane; molecule B is more highly branched. Complete the explanation:

"Molecule B has a lower boiling temperature because..."

Model completion: "because its more compact shape gives less surface contact between molecules, so London forces are weaker and less energy is needed to separate the molecules."

Common-error contrast

"More branches means more bonds, so boiling temperature is higher" is not the relevant model. Branching can increase the number of named C-C connections in a displayed structure, but boiling depends on attractions between whole molecules. For isomeric alkanes, the compact shape matters more than counting bonds in the molecule.

Alcohols and alkanes can have similar numbers of electrons, but alcohols usually have much higher boiling temperatures. The important difference is the O-H group. An alcohol molecule can form hydrogen bonds with neighbouring alcohol molecules because hydrogen is bonded to oxygen and oxygen has lone pairs.

Compare propane and ethanol. Both have 26 electrons, but propane is non-polar and relies mainly on London forces. Ethanol has London forces as well, but it also has permanent dipole-dipole attractions and hydrogen bonding. Ethanol therefore has a much higher boiling temperature and is less volatile.

Approximate data make the comparison visible:

A Pearson-style explanation should refer to the forces in both substances. It is not enough to say "ethanol has hydrogen bonding"; the comparison needs the effect: stronger intermolecular forces mean more energy is needed to separate ethanol molecules, so ethanol has the higher boiling temperature and lower volatility.

Hydrogen halides add a useful diagnostic because one trend is regular and one point is anomalous. From HCl to HI, boiling temperature increases as the halogen atom gets larger and has more electrons. London forces become stronger down the group. HF does not fit that simple HCl-to-HI pattern because HF molecules form hydrogen bonds.

HF has a high boiling temperature compared with the other hydrogen halides because hydrogen bonding is much stronger than the dipole-dipole attractions and London-force pattern expected for a small molecule. The correct explanation is not "fluorine has the highest electronegativity, so HF must simply have the lowest boiling temperature because it is smallest." Size and polarisation explain HCl to HI; hydrogen bonding explains the HF anomaly.

Worked example

A student is given this order of boiling temperatures:

HCl < HBr < HI < HF

Explain the order in terms of intermolecular forces.

Model answer

HCl, HBr and HI all have London forces and permanent dipole-dipole attractions, but from HCl to HI the number of electrons and the polarisation of the electron cloud increase. London forces therefore become stronger, so boiling temperature increases from HCl to HI. HF has hydrogen bonding between molecules because H is bonded to highly electronegative F, so the intermolecular attractions are much stronger and HF has the highest boiling temperature.

Feynman diagnostic

Explain why "HI is more polar than HF, so HI should boil at a higher temperature" is a weak answer. The issue is that the premise is wrong and the model is incomplete. HF has the very polar H-F bond and forms hydrogen bonds; from HCl to HI the increase in boiling temperature is mainly caused by stronger London forces as electron number and polarisation increase.

Solubility is also a competition between particle attractions. To dissolve a substance, attractions between solute particles and attractions between solvent particles must be replaced by attractions between solute and solvent. A substance dissolves well when the new solute-solvent attractions are strong enough to compensate for the attractions that were disrupted.

Water is polar. Oxygen has a partial negative charge and the hydrogens have partial positive charges. This makes water a good solvent for many ionic compounds because water molecules can surround separated ions.

For a positive ion, such as Na+, the oxygen end of water points towards the ion. For a negative ion, such as Cl-, the hydrogen ends of water point towards the ion. These ion-dipole attractions are called hydration of the ions. At this stage, you only need the particle-level idea: water molecules surround and stabilise the ions. Enthalpy cycles for hydration and solution come later in the course.

Hydration explanation route

1. An ionic lattice contains positive and negative ions held by electrostatic attractions.
2. Water is polar.
3. Water molecules orientate around the ions: O end towards cations, H end towards anions.
4. Ion-dipole attractions form between water and the ions.
5. These hydrated ions can disperse through the water if the overall balance of interactions is favourable.

Water also dissolves simple alcohols well because water and alcohol molecules can hydrogen bond to each other. Methanol and ethanol are good examples. The O-H group in the alcohol can hydrogen bond with water molecules, so strong solute-solvent attractions form.

Do not overextend this to every molecule containing a polar bond. A polar molecule may still be a poor solvent match for water if it cannot form strong enough attractions with water to replace water-water hydrogen bonds.

Worked example

Explain why ethanol mixes with water.

Answer

Ethanol has an O-H group, so ethanol molecules can form hydrogen bonds with water molecules. The lone pairs on oxygen and the O-H bonds in water and ethanol allow strong intermolecular attractions between solute and solvent. These attractions help replace some ethanol-ethanol and water-water attractions, so ethanol is soluble in water.

The useful solvent-choice rule is often shortened to "like dissolves like", but the A level version needs particle reasoning. A good solvent forms similar or stronger intermolecular attractions with the solute than the attractions being disrupted in the separate solute and solvent.

Water is a poor solvent for many compounds that cannot hydrogen bond with it. Edexcel specifically includes polar molecules such as halogenoalkanes. A halogenoalkane has a polar C-X bond, so it is not "non-polar" in the simple sense. However, it cannot form strong hydrogen bonds with water because it has no H bonded to N, O or F, and water-water hydrogen bonding is difficult to replace with the weaker attractions available between water and the halogenoalkane.

This is a common assessment trap: polarity helps, but polarity alone does not guarantee solubility in water. The solvent decision asks what attractions can form between solvent and solute.

Non-aqueous solvents are often chosen for compounds with similar intermolecular forces to the solvent. For example, an alkane dissolves better in a non-polar hydrocarbon solvent than in water because both solute and solvent mainly use London forces. A halogenoalkane is often more compatible with an organic solvent that can provide London forces and dipole interactions than with water.

Worked example

Choose the better solvent, water or hexane, for dissolving a non-polar hydrocarbon oil. Justify your choice.

Model answer

Hexane is the better solvent. The oil and hexane are both molecular substances whose main intermolecular forces are London forces. The attractions formed between oil molecules and hexane molecules are similar to the attractions present in the separate substances, whereas water would need to disrupt strong water-water hydrogen bonds and would not form strong enough attractions with the non-polar oil.

Worked example

A student says chloromethane should dissolve well in water because the C-Cl bond is polar. Improve the answer.

Model answer

Chloromethane is polar, but it cannot form strong hydrogen bonds with water. Water molecules would have to disrupt water-water hydrogen bonds, and the attractions between chloromethane and water are not strong enough to compensate well. A non-aqueous solvent with similar intermolecular forces would usually be a better choice.

Pearson-aware recap

For boiling-temperature questions, name the substances being compared, identify the intermolecular forces in each, and link stronger forces to more energy needed to separate molecules. For solvent-choice questions, identify the solute particles, the solvent particles, and the new attractions that can form between them: hydration for some ions in water, hydrogen bonding for simple alcohols in water, and similar intermolecular forces for many non-aqueous solvent choices. The strongest answers avoid single-word labels such as "polar" and instead make the cause-and-effect chain visible.