IB Chemistry HLTopic 2 — Models of Bonding & StructurePaper 1 & 2Core idea~11 min read
Intermolecular Forces
Water boils at 100°C. Methane, a molecule of almost the same size, boils at −162°C. Neither difference has anything to do with the covalent bonds inside the molecules — it is entirely about the far weaker attractions between them. Those attractions are the most useful thing in this whole topic.
📘 What you need to know
Intermolecular forces act between molecules. Intramolecular forces (covalent bonds) act within them.
Intermolecular forces are far weaker than covalent bonds, and they are what you break when something melts or boils.
Four types: London (dispersion) forces, dipole–dipole, dipole–induced dipole, and hydrogen bonding.
The first three together are called van der Waals forces.
London forces exist between all molecules and get stronger with more electrons and more surface area.
Hydrogen bonding needs H covalently bonded to O, N or F, plus a lone pair on an O, N or F nearby.
Strength order: London < dipole–dipole < hydrogen bonding.
Inside a molecule, and between molecules
This distinction is the source of more lost marks than almost anything else in the course, so get it straight now.
Two very different kinds of attraction
A covalent O–H bond is worth about 463 kJ mol⁻¹. The hydrogen bond between molecules is worth roughly 20. That factor of twenty is why boiling is easy and decomposition is not.
If an exam question asks why a covalent substance has a low boiling point and you write “weak covalent bonds”, you will get nothing. The covalent bonds are strong. It is the forces between the molecules that are weak.
London (dispersion) forces
These exist between every atom and every molecule, including ones with no permanent charge at all. Here is how something with no dipole manages to attract something else with no dipole.
Electrons move. At any given instant they might happen to be bunched slightly to one side of a molecule, giving it a fleeting temporary dipole. That momentary charge repels the electrons in a neighbouring molecule, pushing them away and creating an induced dipole lined up to attract it. The two flicker in and out of existence constantly, but on average they pull the molecules together.
How a molecule with no dipole still attracts
Use the full name in exams. “Induced dipole” or “instantaneous dipole” on their own are not the term the mark scheme wants — write London (dispersion) forces.
What makes them stronger
Two factors, and both come up constantly.
Number of electrons. More electrons means a bigger, floppier electron cloud that distorts more easily, so the temporary dipoles are larger.
Surface area. Long straight molecules touch each other along their whole length. Branched, ball-shaped molecules only meet at a few points, so there is less contact and weaker attraction.
Boiling points of the noble gases
Xenon has 27 times as many electrons as helium and boils 161°C higher, purely because its electron cloud is so much easier to distort.
The surface area effect in one comparison. Pentane and 2,2-dimethylpropane are isomers — identical formula, identical number of electrons. Pentane is a long chain and boils at 36°C. 2,2-dimethylpropane is a compact ball and boils at 10°C. Same electrons, less contact, weaker forces.
Dipole–dipole attractions
If a molecule is polar, it has a permanent δ+ end and a permanent δ− end. Line a few of them up and the positive end of one is attracted to the negative end of the next. These are in addition to London forces, not instead of them.
A fair comparison. Butane and propanone both have 34 electrons, so their London forces are about the same. But propanone is polar and butane is not. Butane boils at −0.5°C; propanone boils at 56°C. That 56°C gap is the dipole–dipole attraction doing its work.
Hydrogen bonding
This is a dipole–dipole attraction taken to an extreme. When hydrogen bonds to oxygen, nitrogen or fluorine, the electronegativity difference is so large and the hydrogen atom so small that the hydrogen ends up with an unusually concentrated δ+. It then attracts a lone pair on an O, N or F in the next molecule very strongly.
🧩 You need both of these, or it is not a hydrogen bond
A hydrogen atom covalently bonded to O, N or F.
A lone pair on an O, N or F atom in a neighbouring molecule for that hydrogen to point at.
Count how many each molecule can form: water has 2 hydrogens and 2 lone pairs, so 2. Ammonia has 3 hydrogens but only 1 lone pair, so 1.
🧠 Why water is the odd one out
Water can form two hydrogen bonds per molecule, so every molecule is locked into a network. That is why it boils at 100°C when H2S, a bigger molecule with more electrons, boils at −60°C. Sulfur is not electronegative enough for hydrogen bonding, so H2S only has dipole–dipole and dispersion.
All four, side by side
Force
Acts between
Rough strength
Example
London (dispersion)
all molecules and atoms
1–50 kJ mol⁻¹
Cl2, CH4, Xe
Dipole–induced dipole
a polar and a nonpolar molecule
weak
HCl with Cl2
Dipole–dipole
two polar molecules
moderate
propanone, HCl
Hydrogen bonding
H–O, H–N or H–F and a lone pair
strongest of the four
H2O, NH3, HF, ethanol
“Van der Waals forces” is an umbrella term covering the first three rows only. Hydrogen bonding is not a van der Waals force, and it is not a covalent bond either — it sits in a category of its own.
Worked examples
WE 1
Explain why the boiling points increase from F2 to Cl2 to Br2 to I2 [3]
Mark 1: identify the only force present
All four are nonpolar, so the only force is London (dispersion).
Mark 2: what changes down the group
The number of electrons increases: F₂ 18 → I₂ 106.
Mark 3: link to strength
Bigger electron clouds distort more easily, so larger temporary dipoles and stronger attractions.
More electrons, stronger London forces, more energy neededsay “more electrons”, not “bigger molecules” — size is a consequence, not the cause
WE 2
Butane (Mr 58) boils at −0.5°C, propanone (Mr 58) at 56°C. Explain the difference.
Step 1: rule out the obvious explanation
Same Mr and the same number of electrons, so London forces are similar. Size is not the answer.
Step 2: check for polarity
Butane is a hydrocarbon → nonpolar. Propanone has a polar C=O and is not symmetrical → permanent dipole.
Step 3: name the extra force
Propanone has dipole–dipole attractions on top of its dispersion forces.
Propanone: stronger total forces, so a higher boiling point“on top of” matters — it does not replace dispersion, it adds to it
WE 3
State how many hydrogen bonds one HF molecule can form, and explain
Step 1: count the donor hydrogens
HF has 1 hydrogen bonded to fluorine.
Step 2: count the lone pairs available
F has 7 valence electrons, one in the bond, so 3 lone pairs.
Step 3: the smaller number is the limit
A hydrogen bond needs one of each, so the single hydrogen is the bottleneck.
1 hydrogen bond per molecule on averagethis is why HF boils lower than water despite the stronger individual bond
💡 Exam tips
Always name the force. “Stronger forces” is worth nothing. “Stronger London dispersion forces” is worth a mark.
Compare fairly. If two molecules have the same number of electrons, the difference cannot be dispersion — look for polarity or hydrogen bonding.
Write London (dispersion) forces in full at least once in your answer.
For hydrogen bonding, always mention both requirements: H bonded to O/N/F and a lone pair to accept it.
Dispersion forces are present in everything, including polar molecules. Never say a polar molecule “has dipole–dipole instead”.
When asked about melting or boiling, talk about intermolecular forces, never bond strength.
⚠ Common mix-ups
Saying covalent bonds break on boiling. They do not. Only intermolecular forces are overcome.
Calling a hydrogen bond a covalent bond. It is an intermolecular attraction, roughly twenty times weaker.
Including hydrogen bonding in van der Waals forces. That term covers the other three only.
Explaining trends with “bigger molecule”. Say more electrons, or more surface area — those are the actual causes.
Thinking any molecule containing hydrogen can hydrogen bond. CH4 cannot: carbon is not electronegative enough.
Forgetting the lone pair. A molecule with an H–O bond still needs a partner with a lone pair to bond to.
Up next: Physical Properties of Covalent Substances — putting these four forces to work to predict melting points, solubility and conductivity.
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