Intermolecular forces: AQA A-level Chemistry

Distinguish the three forces between molecules.

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Intermolecular forces

5 practice questions

Identifying all intermolecular forces

Learning goalDistinguish the three forces between molecules.

🔓 CRACK IT

Keep the location of an attraction clear. Covalent bonds hold atoms together within a molecule. Intermolecular forces attract one molecule to another.

Start with induced dipole-dipole attractions, which act between all atoms and molecules. Polar molecules also have permanent dipole-dipole attractions. Suitable molecules can form hydrogen bonds as well.

List all the forces present before comparing substances. Naming hydrogen bonding does not mean the other applicable forces disappear.

London dispersion is another name for induced dipole-dipole interactions. Use the specific force name when explaining an attraction.

ForceWhen presentAttraction
Induced dipole–dipole (van der Waals)Between all atoms and moleculesOpposite instantaneous and induced partial charges
Permanent dipole–dipoleAlso between polar moleculesOpposite permanent partial charges
Hydrogen bondingH bonded to N/O/F and a neighbouring N/O/F lone pairδ+ H bonded to N/O/F and a lone pair on N/O/F
EXAM TRAPA molecule with oxygen but no O–H/N–H/F–H may form hydrogen bonds to δ+ H atoms in water while being unable to hydrogen-bond to itself.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 39
  1. State the strongest intermolecular force in HCl. [1]
  2. State the strongest intermolecular force in CH4. [1]
  3. State the strongest intermolecular force in NH3. [1]
  4. State the strongest intermolecular force in CH3OCH3. [1]
  5. Explain why all four have van der Waals forces. [2]

Total: 6 marks

Show answer & marking points
  1. Permanent dipole–dipole [1].
  2. Induced dipole–dipole (van der Waals) [1].
  3. Hydrogen bonding [1].
  4. Permanent dipole–dipole [1].
  5. Continuously moving electrons produce instantaneous dipoles [1], which induce dipoles in neighbouring molecules and cause attraction [1].
✓ DONE WHENYou can identify every applicable force rather than naming only the strongest.

Induced dipole–dipole (van der Waals) forces

Learning goalExplain electron-number and branching effects.

🔓 CRACK IT

Electrons are always moving. For an instant, they can be distributed unevenly, giving one side of a particle a small negative charge and the other a small positive charge.

  1. This instantaneous dipole distorts the electron cloud of a neighbouring particle.
  2. That distortion creates an induced dipole in the neighbour.
  3. Opposite partial charges attract. New dipoles continually form as the electrons move.

A larger electron cloud is usually more easily distorted, or more polarisable. This tends to give stronger van der Waals attractions and a higher boiling point.

For isomers with the same electron count, more branching usually reduces surface contact between molecules. The overall attractions are then weaker, so less energy is needed to separate them.

FactorConsequence
More electrons / larger electron cloudGreater polarisability, stronger temporary dipoles
Greater molecular surface contactMore neighbouring electron clouds can interact
Worked example

Halogen boiling-point trend

Explain Cl2 < Br2 < I2 in boiling point.

Worked answer

Electron number and polarisability increase down the group. van der Waals forces strengthen, requiring more energy to separate molecules.

Boiling point increases.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 40

The table gives Group 0 data.

ElementNumber of electronsBoiling point / °C
Ne10−246
Ar18−186
Kr36−153
Xe54−108
  1. Name the force between the atoms. [1]
  2. Explain the trend from Ne to Xe. [4]
  3. Predict how Rn’s boiling point compares with Xe’s and justify your answer. [3]

Total: 8 marks

Show answer & marking points
  1. Induced dipole–dipole (van der Waals) forces [1].
  2. More electrons [1] give a more polarisable cloud [1], stronger temporary dipoles and attractions [1], requiring more energy to separate atoms [1].
  3. Above −108 °C [1]; more electrons / greater polarisability [1] give stronger van der Waals forces [1].
Exam-style question 41

The three C5H12 isomers have the same electron count but different shapes.

CompoundBoiling point / °C
Pentane36.1
2-Methylbutane27.8
2,2-Dimethylpropane9.5
  1. Place them in increasing volatility. [1]
  2. Explain why boiling points differ despite identical molecular formulae. [4]

Total: 5 marks

Show answer & marking points
  1. Pentane < 2-methylbutane < 2,2-dimethylpropane [1].
  2. Same electron count / similar potential for van der Waals forces [1]. More branching makes molecules compact and reduces surface contact [1]; attractions become weaker [1], so less energy is needed and boiling point falls [1].
✓ DONE WHENYou can explain trends through polarisability or surface contact and energy.

Permanent dipole–dipole attractions

Learning goalIdentify attractions between neighbouring polar molecules.

🔓 CRACK IT

A polar molecule has a permanent uneven distribution of charge. The δ+ end of one molecule attracts the δ− end of a neighbouring molecule.

These permanent dipole-dipole attractions act as well as induced dipole-dipole attractions. For otherwise comparable molecules, the extra attractions can increase the energy needed to separate them.

HCl has an overall dipole because it has one polar bond. SO₂ is bent, so its bond dipoles do not cancel. Use the shape as well as bond polarity when deciding whether this force is present.

Do not rank boiling points from polarity alone. A large non-polar molecule can have stronger overall attractions than a small polar molecule.

Permanent dipole–dipole force
Electrostatic attraction between the δ+ end of one polar molecule and the δ− end of a neighbouring molecule.
Worked example

HCl versus F2

HCl and F₂ have similar relative molecular masses. Explain why HCl has the higher boiling point.

Worked answer

Both have van der Waals forces. HCl is polar and has additional permanent dipole–dipole attractions, increasing the energy needed to separate molecules.

HCl has stronger overall attractions and the higher boiling point.
EXAM TRAPDo not rank boiling points from dipole size alone. A larger non-polar molecule can have stronger overall forces than a smaller polar molecule.

🧪 TEST IT

Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 42

A student claims that the molecule with the largest permanent dipole must have the highest boiling point.

MoleculeDipole moment / DBoiling point / °C
CH3F1.85−78
CH3Cl1.87−24
CH3Br1.824
CH3I1.6242
  1. Explain why all four have permanent dipoles. [2]
  2. Explain the boiling-point trend from CH3F to CH3I. [4]
  3. Evaluate the claim using the data. [2]

Total: 8 marks

Show answer & marking points
  1. C–X is polar [1]; unlike outer atoms make the tetrahedral molecule asymmetric, so dipoles do not cancel [1].
  2. More electrons [1], greater polarisability [1], stronger van der Waals forces [1], greater energy needed to separate molecules [1].
  3. False: CH3Cl has the largest dipole but CH3I boils highest [1]. van der Waals forces can dominate the total attraction [1].
Exam-style question 43

All C–Cl bonds are polar.

MoleculePermanent dipole / DBoiling point / °C
CH40−162
CH3Cl1.87−24
CH2Cl21.6040
CHCl31.0461
CCl4077
  1. Identify the two non-polar molecules. [2]
  2. Explain why CCl4 has no permanent dipole despite polar C–Cl bonds. [2]
  3. Explain why boiling point rises from CH3Cl to CCl4 as permanent dipole decreases. [4]
  4. Evaluate “CCl4 has weaker intermolecular forces than CH3Cl because CCl4 has no permanent dipole.” [2]

Total: 10 marks

Show answer & marking points
  1. CH4 [1] and CCl4 [1].
  2. Symmetrical tetrahedral arrangement [1] makes four equal C–Cl dipoles cancel [1].
  3. Replacing H by Cl increases electron number [1] and polarisability [1], strengthening van der Waals forces [1] and increasing the energy needed to boil [1].
  4. False: both have van der Waals forces [1]; CCl4’s more polarisable cloud gives stronger overall attractions and a higher boiling point [1].
✓ DONE WHENYou can explain permanent-dipole effects while also considering van der Waals forces.

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