The shape of the lesson
By the end of the lesson, learners can
- Three shapes around carbon: Overlap the orbitals in the sigma and pi simulator and watch the shape around each carbon change from ethane to ethene to ethyne.
- Counting σ and π: Count every bond as one σ, then add one π for each double bond and two for each triple bond.
- 4 → sp3, 109.5°; 3 → sp2, 120°; 2 → sp, 180°.
- Every bond has one σ.
- Double +1 π; triple +2 π.
- π bond: planar, no rotation.
One picture every idea returns to
A carbon bonded to four atoms (single bonds only) is sp3, tetrahedral, 109.5°. A carbon in a double bond is bonded to three atoms: sp2, trigonal planar, 120°. A carbon in a triple bond (or with two double bonds) is bonded to two: sp, linear, 180°. Every single bond is a σ bond; a double bond is σ + π. The π bond forms by sideways overlap of p orbitals above and below the plane, so ethene is planar and there is no rotation about C=C, which is why cis-trans isomers exist.
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, Tetrahedral, flat or straight.. Every carbon atom in an organic molecule has one of three shapes, decided by how many atoms it is bonded to. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Three shapes around carbon | Overlap the orbitals in the sigma and pi simulator and watch the shape around each carbon change from ethane to ethene to ethyne. 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 |
| 17 to 30 min | Teach and check: Counting σ and π | Count every bond as one σ, then add one π for each double bond and two for each triple bond. Project tab 1 of the lab (Chemical bonding lab, shapes) and let a student drive it while the class predicts. | Frayer Model |
| 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, 13 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 Counting σ and π, 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.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
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 the C–H bonds; they are σ bonds too. | A double bond is one σ and one π, not two π. In propene only the two C=C carbons are planar; the CH 3 carbon is tetrahedral. |
Shapes of organic molecules, sigma and pi bonds: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
State the shape and bond angle around each carbon in CH3CH=CH2.
How many σ and π bonds are there in ethanal, CH3CHO?
Describe how the π bond in ethene forms and explain why ethene is planar.
How many σ and π bonds are there in CH2=CH–CN?
Explain why but-2-ene can form two isomers but butane cannot.
State the hybridisation of carbon in methane, ethene and ethyne.
Answer key and marking notes
- CH3 carbon: tetrahedral, 109.5°.
- CH carbon: trigonal planar, 120°.
- CH2 carbon: trigonal planar, 120°.
- Bonds: 4 C–H, 1 C–C, 1 C=O = 6 σ.
- 1 π (in C=O).
- Each carbon is sp2 hybridised; three σ bonds lie in one plane at 120°.
- Each carbon has an unhybridised p orbital at right angles to this plane; they overlap sideways to form the π bond.
- This overlap requires the p orbitals to be parallel, so all six atoms are held in one plane.
- Bonds: 3 C–H, C=C, C–C, C≡N = 6 σ.
- 1 π in C=C + 2 π in C≡N = 3 π.
- Rotation about the C=C is restricted because twisting would break the π bond, so groups are fixed on the same or opposite sides.
- Butane has only σ bonds, which rotate freely, so all arrangements interconvert.
sp3, sp2 and sp respectively.
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.
In ethane, each carbon atom forms four single bonds. What is the shape around each carbon, and the H-C-H bond angle?
- Atetrahedral, 109.5°
- Btrigonal planar, 120°
- Clinear, 180°
- Dsquare planar, 90°
How many sigma (σ) and pi (π) bonds are there in one molecule of ethene, CH2=CH2?
- A5 σ and 1 π
- B4 σ and 2 π
- C6 σ and 0 π
- D4 σ and 1 π
What is the shape around each carbon atom in ethene, and the approximate H-C-H bond angle?
- Atrigonal planar, 120°
- Btetrahedral, 109.5°
- Clinear, 180°
- Dpyramidal, 107°
How is the π bond in ethene formed?
- Aby sideways overlap of p orbitals on the two carbon atoms, giving electron density above and below the plane of the molecule
- Bby end-on overlap of orbitals directly between the two carbon nuclei
- Cby end-on overlap of two s orbitals
- Dby one carbon donating a lone pair to the other carbon
What is the O=C-C bond angle around the carbonyl carbon in propanone, CH3COCH3?
- A120°
- B109.5°
- C180°
- D107°
How many σ and π bonds are there in one molecule of propene, CH2=CHCH3?
- A8 σ and 1 π
- B7 σ and 1 π
- C7 σ and 2 π
- D9 σ and 0 π
Why is there no free rotation about the C=C bond in ethene, while there is free rotation about the C-C bond in ethane?
- Arotating about C=C would twist the p orbitals out of line and break the sideways overlap of the π bond
- Bthe C=C bond is shorter, so the hydrogen atoms collide during rotation
- Ca σ bond can never rotate, and C=C contains a σ bond
- Dethene has two π bonds that lock the carbons in place
Consider ethanal, CH3CHO. Which statement is correct?
- Ait has 6 σ and 1 π bonds; the angles are about 109.5° at the CH3 carbon and 120° at the CHO carbon
- Bit has 5 σ and 2 π bonds; all angles are about 120°
- Cit has 6 σ and 1 π bonds; all angles are about 120°, because the molecule contains a double bond
- Dit has 7 σ and 0 π bonds; all angles are about 109.5°
Quiz key
A carbon with four single bonds is sp3 hybridised: four bonding pairs and no lone pairs give a tetrahedral arrangement with angles of 109.5°. Trigonal planar, 120°, applies to an sp2 carbon with three regions of electron density, as in C=C or C=O. Linear, 180°, applies to an sp carbon with two regions, as in ethyne or a nitrile carbon. Square planar would put the pairs only 90° apart, which gives more repulsion than a tetrahedron, so carbon never adopts it.
There are four C-H single bonds (4 σ) and one C=C, which is made of one σ bond and one π bond. Total: 5 σ and 1 π. 4 σ and 2 π treats both parts of the double bond as π bonds; ethene has only one π bond, and the first bond between any two atoms is always σ. 6 σ and 0 π treats the double bond as two σ bonds, but two σ bonds cannot both lie along the same axis. 4 σ and 1 π forgets the σ component of C=C.
Each carbon in ethene is sp2 hybridised and has three regions of electron density (two C-H and the C=C counted as one region). These repel to 120° apart in a plane: trigonal planar. The whole molecule is flat. Tetrahedral, 109.5°, needs four regions, as in an alkane carbon. Linear needs only two regions, as in ethyne. Pyramidal, 107°, needs three bonding pairs and a lone pair, as in NH3; carbon in ethene has no lone pair.
Each sp2 carbon keeps one unhybridised p orbital at right angles to the molecular plane. These overlap sideways, giving two lobes of electron density above and below the C-C axis: the π bond. End-on overlap directly between the nuclei forms a σ bond, not a π bond. Overlap of two s orbitals is also end-on and gives a σ bond (as in H2). Neither carbon has a lone pair in ethene, so no dative bond is involved.
The carbonyl carbon has three regions of electron density: C=O (counted as one) and two C-C bonds. It is sp2 and trigonal planar, so the angles are about 120°. 109.5° is the angle around the CH3 carbons, which have four single bonds, not around the C=O carbon. 180° would need only two regions, as in a linear sp carbon. 107° is the angle for three bonding pairs and one lone pair, as in NH3; the carbonyl carbon has no lone pair.
Propene has six C-H bonds, one C-C single bond and one C=C. The single bonds give 6 + 1 = 7 σ, and the C=C adds one σ and one π: 8 σ and 1 π. (By the same count, but-2-ene, CH3CH=CHCH3, has 11 σ and 1 π.) 7 σ and 1 π forgets the σ part of the double bond. 7 σ and 2 π counts the double bond as two π bonds; a double bond always contains exactly one π. 9 σ and 0 π treats the C=C as two σ bonds.
The π bond depends on the two p orbitals staying parallel. Rotating one end would move them out of alignment, destroying the overlap and breaking the π bond, which needs a lot of energy. This restricted rotation keeps ethene planar and is the reason cis-trans isomers exist. A σ bond is symmetrical about the internuclear axis, so its overlap is unchanged on rotation; that is why ethane rotates freely, and why 'a σ bond can never rotate' is wrong. Bond length is not the reason: the H atoms on adjacent carbons are not close enough to block rotation. Ethene has one π bond, not two.
Ethanal has four C-H, one C-C and one C=O. The single bonds give 5 σ, and C=O adds one σ and one π: 6 σ and 1 π. The CH3 carbon has four single bonds (sp3, tetrahedral, 109.5°); the CHO carbon has three regions (sp2, trigonal planar, 120°). 5 σ and 2 π counts the C=O as two π bonds. A double bond only makes the atoms directly around it planar, so the CH3 carbon is still tetrahedral. 7 σ and 0 π treats C=O as two σ bonds, and ignores the planar carbonyl carbon.
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.