The shape of the lesson
By the end of the lesson, learners can
- Flickering dipoles: Switch the simulator to the id-id view and watch the electron clouds wobble.
- Polar molecules: Switch to the polarity view and look at the molecule as a whole.
- Comparing boiling points: To compare two simple molecular substances, name every type of intermolecular force present in each, then decide which is strongest overall.
- id-id: all molecules; stronger with more electrons and more contact.
- pd-pd: polar molecules only, in addition to id-id.
- Polar molecule: polar bonds that do not cancel by symmetry.
- Boiling overcomes intermolecular forces, not covalent bonds.
One picture every idea returns to
Intermolecular forces act between molecules and are much weaker than covalent bonds. Instantaneous dipole-induced dipole (id-id) forces act between all molecules: a momentary uneven electron distribution in one molecule induces a dipole in its neighbour. They grow with the number of electrons and with the contact area between molecules. Permanent dipole-permanent dipole (pd-pd) forces act as well between polar molecules, whose δ+ end of one attracts the δ− end of the next. A molecule with polar bonds is only polar if the bond dipoles do not cancel by symmetry. The strongest intermolecular force, hydrogen bonding, has its own lesson.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Forces between molecules.. Boiling water does not break a single O–H bond. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Flickering dipoles | Switch the simulator to the id-id view and watch the electron clouds wobble. Project tab 7 of the lab (Chemical bonding lab, intermolecular forces) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Polar molecules | Switch to the polarity view and look at the molecule as a whole. Project tab 5 of the lab (Chemical bonding lab, molecular polarity) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Comparing boiling points | To compare two simple molecular substances, name every type of intermolecular force present in each, then decide which is strongest overall. | Connect-Extend-Challenge |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Comparing boiling points, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Never say that boiling a simple molecular substance “breaks covalent bonds”. | It overcomes intermolecular forces. And do not assume the more polar molecule always boils higher: id-id forces grow so quickly with electron number that they often dominate, as HBr against HCl shows. |
Intermolecular forces: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Explain how instantaneous dipole-induced dipole forces arise between two argon atoms.
The boiling points of the noble gases increase from helium to xenon. Explain this trend.
Tetrachloromethane, CCl4, contains polar C–Cl bonds but is a non-polar molecule. Explain why.
Butane (Mr 58, b.p. −1 °C) and propanone (Mr 58, b.p. 56 °C) have similar numbers of electrons. Explain the difference in boiling point.
Place these in order of increasing boiling point and give a reason: 2,2-dimethylpropane, pentane, hexane.
A student writes: “When iodine sublimes, the I–I bonds break.” Correct this statement.
Answer key and marking notes
- Electrons move constantly, so at any instant the electron distribution in one atom may be uneven, giving an instantaneous dipole.
- This dipole distorts the electron cloud of a neighbouring atom, inducing a dipole in it.
- The δ+ end of one attracts the δ− end of the other.
- The number of electrons per atom increases down the group.
- Larger electron clouds are more easily polarised, so the id-id forces between atoms are stronger and more energy is needed to separate them.
- The molecule is tetrahedral and symmetrical.
- The four equal C–Cl bond dipoles point to the corners of the tetrahedron and cancel, so there is no overall dipole.
- Both have id-id forces of similar strength, because they have similar numbers of electrons.
- Propanone contains a polar C=O bond and is a polar molecule, so it also has permanent dipole-permanent dipole forces.
- More energy is needed to overcome the intermolecular forces in propanone.
- 2,2-dimethylpropane < pentane < hexane.
- Hexane has the most electrons; of the two C5H12 isomers, the unbranched pentane has the larger contact area, so stronger id-id forces.
- Sublimation overcomes the weak id-id forces between I2 molecules.
- The covalent I–I bonds within the molecules are not broken; iodine vapour is still I2.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
How do instantaneous dipole-induced dipole (id-id) forces arise between non-polar molecules?
- Aat any instant the electron cloud of a molecule may be unevenly spread, creating a temporary dipole that induces a dipole in a neighbouring molecule
- Beach molecule has a permanent dipole because its bonds are polar
- Ca hydrogen atom on one molecule is attracted to a lone pair on the next
- Delectrons are transferred between neighbouring molecules, forming ions
The boiling points of the halogens increase from fluorine to iodine. What is the reason?
- Athe molecules have more electrons, so the instantaneous dipole-induced dipole forces are stronger
- Bthe covalent bond within each molecule becomes stronger down the group
- Cthe molecules become more polar down the group
- Dhydrogen bonding becomes stronger as the molecules get larger
Which molecule has polar bonds but is non-polar overall?
- ACCl4
- BCHCl3
- CHCl
- DNH3
Pentane, CH3CH2CH2CH2CH3, boils at 36 °C, but its isomer 2,2-dimethylpropane, C(CH3)4, boils at 10 °C. Why?
- Athe long chain of pentane has a larger surface area of contact between molecules, so the id-id forces are stronger
- Bpentane has more electrons than 2,2-dimethylpropane
- Cpentane has permanent dipoles but 2,2-dimethylpropane does not
- Dthe covalent bonds in pentane are stronger
Propanone, CH3COCH3 (Mr 58), boils at 56 °C, while butane, C4H10 (Mr 58), boils at −1 °C. What is the main reason for the difference?
- Apropanone has a polar C=O bond and a permanent dipole, so it has permanent dipole-permanent dipole forces as well as id-id forces
- Bpropanone molecules form hydrogen bonds with each other
- Cbutane has no intermolecular forces at all
- Dpropanone has a much larger relative molecular mass than butane
Which molecules have a permanent dipole? CO2, SO2, BF3, PCl3
- ASO2 and PCl3 only
- Ball four, since they all contain polar bonds
- CCO2 and SO2 only
- DSO2 only
The boiling points of HCl, HBr and HI are −85 °C, −67 °C and −35 °C. HCl has the most polar bond, yet it has the lowest boiling point. What does this show?
- Ain these molecules id-id forces, which increase with the number of electrons, contribute more than permanent dipole-permanent dipole forces
- Bpermanent dipole-permanent dipole forces are always stronger than id-id forces
- Cthe H–I bond is the strongest, so HI is hardest to boil
- DHCl forms hydrogen bonds, which lower its boiling point
Which types of intermolecular force are present between molecules in liquid chloromethane, CH3Cl?
- Ainstantaneous dipole-induced dipole forces and permanent dipole-permanent dipole forces
- Bpermanent dipole-permanent dipole forces only
- Cinstantaneous dipole-induced dipole forces only
- Dhydrogen bonds and permanent dipole-permanent dipole forces
Quiz key
Electrons are always moving, so at one instant there may be more electron density on one side of a molecule, giving a temporary dipole. This induces an opposite dipole in a neighbour and the two attract; the dipoles keep changing but the attraction persists on average. Permanent dipoles cause permanent dipole-permanent dipole forces, which only polar molecules have. The attraction of a hydrogen atom to a lone pair is hydrogen bonding. No electrons are transferred between molecules; that would produce ionic bonding.
From F2 (18 electrons) to I2 (106 electrons) the electron clouds become larger and more easily distorted, so larger temporary dipoles form and the id-id forces between molecules get stronger, needing more energy to separate. Boiling does not break the X–X covalent bond, and those bonds actually get weaker from Cl2 to I2. Each halogen molecule has two identical atoms, so it is non-polar. There is no hydrogen in these molecules, so hydrogen bonding is impossible.
Each C–Cl bond in CCl4 is polar (C 2.5, Cl 3.0), but the four bonds point to the corners of a regular tetrahedron, so the dipoles cancel and the molecule has no overall dipole. In CHCl3 one Cl is replaced by H, so the dipoles no longer cancel and the molecule is polar. HCl has only one polar bond, so it must be polar. NH3 is pyramidal, so its three N–H dipoles add up to give an overall dipole.
Both molecules have 42 electrons, but the straight chain of pentane can lie alongside its neighbours with a large area of contact, so more id-id attractions form. The nearly spherical 2,2-dimethylpropane touches its neighbours over only a small area, so its id-id forces are weaker and it boils at a lower temperature. Both have exactly the same number of electrons, because they are isomers. Both are hydrocarbons with almost non-polar C–H bonds, so neither has significant permanent dipoles. Boiling does not break covalent bonds.
With the same Mr and similar numbers of electrons (32 and 34), the id-id forces are similar. Propanone also has a polar C=O group (δ+ on C, δ− on O), so its molecules attract each other by permanent dipole-permanent dipole forces too, and more energy is needed to boil it. Propanone has no H atom bonded to O, N or F, so it cannot hydrogen bond with itself. Butane does have intermolecular forces (id-id); without them it could not be liquefied. The Mr values are equal, as the question states.
SO2 is bent (a lone pair on S) and PCl3 is pyramidal (a lone pair on P), so their bond dipoles do not cancel and both are polar. CO2 is linear and BF3 is trigonal planar, so their identical polar bonds cancel exactly and they are non-polar. Saying all four assumes that polar bonds always give a polar molecule, ignoring shape. Including CO2 treats it as bent like SO2, but carbon has no lone pair. SO2 only misses the pyramidal shape of PCl3.
The polarity of the bond falls from HCl to HI, so permanent dipole forces get weaker, yet the boiling point rises. The rise must come from id-id forces, which increase with the number of electrons (18, 36, 54). So for these molecules the id-id contribution is the larger one. If dipole forces were always stronger, HCl would boil highest. The H–I bond is actually the weakest of the three, and boiling does not break it anyway. Chlorine is not N, O or F, so HCl does not form hydrogen bonds, and hydrogen bonds would raise, not lower, the boiling point.
Every molecule has electrons, so id-id forces are always present. The C–Cl bond is polar and the three C–H bonds do not cancel it, so chloromethane has a permanent dipole and also has permanent dipole-permanent dipole forces. Choosing dipole forces only forgets that id-id forces exist between all molecules. id-id only ignores the polar C–Cl bond. There are no hydrogen bonds because the hydrogen atoms are bonded to carbon, not to N, O or F.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.