The shape of the lesson
By the end of the lesson, learners can
- Two ways to overlap: In the simulator, bring two orbitals together end-on, then turn two p orbitals parallel and bring them alongside each other.
- Hybrid orbitals: A carbon atom has one 2s and three 2p orbitals.
- Why π bonds matter: The shape of the π bond has consequences you will use throughout organic chemistry.
- σ: end-on overlap, density on the axis.
- π: sideways p overlap, density above and below.
- sp3 109.5°, sp2 120°, sp 180°.
- π is weaker and blocks rotation.
One picture every idea returns to
A σ (sigma) bond forms by direct, end-on overlap of orbitals, putting the electron density on the line between the nuclei. A π (pi) bond forms by sideways overlap of two p orbitals, putting the density above and below that line. A single bond is one σ bond; a double bond is one σ + one π; a triple bond is one σ + two π. Carbon mixes its 2s and 2p orbitals into hybrid orbitals: sp3 (four, tetrahedral, 109.5°), sp2 (three, trigonal planar, 120°, one p left for a π bond) and sp (two, linear, 180°, two p left). The π bond is weaker than the σ bond because sideways overlap is smaller.
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, End-on, or sideways.. A covalent bond is overlapping orbitals, and the way they overlap matters. 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: Two ways to overlap | In the simulator, bring two orbitals together end-on, then turn two p orbitals parallel and bring them alongside each other. Project tab 6 of the lab (Chemical bonding lab, sigma and pi) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Hybrid orbitals | A carbon atom has one 2s and three 2p orbitals. | Pairs |
| 21 to 30 min | Teach and check: Why π bonds matter | The shape of the π bond has consequences you will use throughout organic chemistry. | Corners |
| 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 Why π bonds matter, 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.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
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 |
|---|---|
| Count σ and π bonds in a whole molecule by counting every bond as one σ, then adding one π for each double bond and two for each triple bond. | Ethene, H 2 C=CH 2, has 5 σ and 1 π, not 4 and 2. Forgetting the C–H bonds is the most common slip. |
Sigma and pi bonds, hybridisation: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Describe, with reference to orbitals, how the double bond in ethene is formed.
State the number of σ bonds and π bonds in (a) ethane, C2H6, and (b) carbon dioxide, O=C=O.
State the hybridisation of carbon and the bond angle around it in (a) methane, (b) ethene and (c) ethyne.
The bond energy of C=C is about 610 kJ mol−1 and of C–C about 350 kJ mol−1. Explain why the C=C value is less than twice the C–C value.
Explain why there is no free rotation about a C=C double bond.
Hydrogen cyanide has the structure H–C≡N. State the total number of σ and π bonds, the hybridisation of carbon and the H–C–N bond angle.
Answer key and marking notes
- Each carbon is sp2 hybridised; one sp2 orbital from each carbon overlaps end-on to form a σ bond.
- Each carbon has one unhybridised p orbital, perpendicular to the plane of the molecule.
- These p orbitals overlap sideways to form a π bond, with electron density above and below the plane.
- (a) Ethane: six C–H and one C–C, all single, so 7 σ and 0 π.
- (b) CO2: two double bonds, so 2 σ and 2 π.
- (a) sp3, 109.5°.
- (b) sp2, 120°.
- (c) sp, 180°.
- A double bond is one σ bond plus one π bond.
- The π bond forms by sideways overlap of p orbitals, which is less effective than end-on overlap, so it is weaker than a σ bond (about 260 kJ mol−1 against 350).
- Rotation would twist the two p orbitals out of alignment.
- That would destroy the sideways overlap and break the π bond, which needs a lot of energy.
- 2 σ bonds (C–H and the σ part of C≡N) and 2 π bonds.
- Carbon is sp hybridised, so the molecule is linear: 180°.
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 is a σ (sigma) bond formed?
- Aby end-on (head-on) overlap of orbitals, giving electron density concentrated along the line between the two nuclei
- Bby sideways overlap of two p orbitals, giving electron density above and below the line between the nuclei
- Cby the transfer of an electron from one orbital to another
- Donly by overlap of two s orbitals
Why is a π bond weaker than a σ bond between the same two atoms?
- Asideways overlap of p orbitals is less effective, so the bonding electron density is further from the nuclei and less strongly attracted
- Ba π bond contains only one electron
- Ca π bond is formed from s orbitals, which are smaller
- Dthe electrons in a π bond lie exactly between the nuclei, so they repel each other more
How many σ bonds and π bonds are there in one molecule of ethene, C2H4?
- A5 σ and 1 π
- B4 σ and 1 π
- C4 σ and 2 π
- D6 σ and 0 π
What is the hybridisation of each carbon atom in ethene, and what is the H–C–H bond angle?
- Asp2, about 120°
- Bsp3, about 109.5°
- Csp, 180°
- Dsp2, about 109.5°
The diagram shows two p orbitals overlapping between two atoms. What type of bond is being formed?
- Aone π bond, with electron density above and below the internuclear axis
- Bone σ bond, because the orbitals overlap between the nuclei
- Ctwo π bonds, one above and one below the axis
- Done σ bond formed from two s orbitals
How many σ and π bonds are there in one molecule of ethyne, H–C≡C–H, and what is the H–C–C bond angle?
- A3 σ and 2 π, 180°
- B2 σ and 3 π, 180°
- C3 σ and 2 π, 120°
- D5 σ and 0 π, 109.5°
Propene is CH2=CH–CH3. What is the hybridisation of carbon atoms 1, 2 and 3, reading from left to right?
- Asp2, sp2, sp3
- Bsp2, sp2, sp2
- Csp, sp, sp3
- Dsp3, sp3, sp3
Rotation about the C=C bond in an alkene does not occur at room temperature, but rotation about a C–C single bond does. What is the reason?
- Arotation would twist the parallel p orbitals out of line, breaking the π bond
- Bthe σ bond in C=C is too short to rotate
- Cthe double bond contains two σ bonds locked on either side of the axis
- Dthe hydrogen atoms on the two carbons repel each other too strongly
Quiz key
A σ bond forms when orbitals overlap end-on, so the shared electron density lies along the internuclear axis; it can be formed from s–s, s–p, p–p (end-on) or hybrid orbitals. Sideways overlap of p orbitals with density above and below the axis describes a π bond. Transfer of an electron produces ions, not a covalent bond. s–s overlap (as in H2) is only one way of making a σ bond; the C–H bonds in methane use hybrid orbitals of carbon.
In a π bond the p orbitals overlap side by side, so the overlap is smaller and the electron density sits above and below the axis, further from the two nuclei; it is attracted less strongly and is easier to break, which is why alkenes react by breaking the π bond. Like any covalent bond, a π bond is one shared pair of two electrons. π bonds use p orbitals, never s orbitals, because s orbitals are spherical and cannot overlap sideways. The electrons lying on the axis between the nuclei describes a σ bond.
Ethene has four C–H single bonds (4 σ) and one C=C double bond, which is one σ plus one π, so the totals are 5 σ and 1 π. 4 σ and 1 π forgets that the double bond also contains a σ bond. 4 σ and 2 π treats both parts of the double bond as π bonds. 6 σ and 0 π counts the double bond as two σ bonds, but two σ bonds cannot both lie along the same axis.
Each carbon uses one s and two p orbitals to make three sp2 hybrid orbitals for its three σ bonds, which point to the corners of a triangle at about 120°; the remaining p orbital forms the π bond. sp3 and 109.5° describe a carbon with four σ bonds, as in ethane. sp and 180° describe a carbon with two σ bonds, as in ethyne. The last option has the right hybridisation but the tetrahedral angle, which does not match three electron pairs in a plane.
The p orbitals are parallel to each other and perpendicular to the axis, so they overlap sideways, giving a single π bond whose electron density is in two regions, above and below the axis. A σ bond needs end-on overlap with density on the axis itself, and there is no overlap on the axis in the diagram. The upper and lower regions together make up one π bond holding one pair of electrons, not two bonds. The dumbbell shapes are p orbitals; s orbitals are spherical.
There are two C–H σ bonds, and the triple bond is one σ plus two π bonds (formed from two pairs of p orbitals at right angles), so the totals are 3 σ and 2 π. Each carbon forms two σ bonds using sp hybrid orbitals, which point in opposite directions, so the molecule is linear at 180°. 2 σ and 3 π treats the whole triple bond as π bonds. 120° is the angle for sp2 carbon, as in ethene. 5 σ and 0 π with 109.5° treats every bond as a single σ bond around tetrahedral carbon.
Carbons 1 and 2 each form three σ bonds and take part in the C=C π bond, so they are sp2 (about 120°). Carbon 3 forms four single σ bonds and is sp3 (about 109.5°). Making carbon 3 sp2 ignores the fact that the double bond affects only the two carbons it joins. sp would need a triple bond or two double bonds on those carbons. All sp3 would mean no π bond at all, as in propane.
The π bond exists only while the two p orbitals are parallel and overlap sideways. Turning one carbon relative to the other would reduce this overlap to zero and break the π bond, which needs much more energy than is available at room temperature; this restricted rotation is why E/Z isomers exist. The σ part has cylindrical symmetry about the axis and would allow rotation whatever its length. A double bond is one σ plus one π, not two σ bonds. Repulsion between hydrogen atoms is small and is not the cause; ethane rotates freely even though it has more hydrogens.
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.