The shape of the lesson
By the end of the lesson, learners can
- Count the pairs: Load each molecule in the 3D simulator and rotate it.
- Worked examples: Apply the four steps to ions and molecules with double bonds.
- Lone pairs push harder: A lone pair is attracted by only one nucleus, so it sits closer to the central atom and spreads out more.
- Pairs: outer electrons + one per bond, adjusted for charge, halved.
- Repulsion: lone–lone > lone–bond > bond–bond.
- Double bond: one region of electron density.
- Angles: 109.5° → 107° → 104.5° as lone pairs are added.
One picture every idea returns to
In VSEPR theory the electron pairs around a central atom repel one another and arrange themselves as far apart as possible, which fixes the shape. Count the bonding pairs (a double or triple bond counts as one region) and the lone pairs. Lone pairs are held closer to the nucleus and repel more than bonding pairs: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Each lone pair closes the remaining bond angles by roughly 2.5°, so CH4 is 109.5°, NH3 107° and H2O 104.5°. The shape is named from the positions of the atoms, not the lone pairs.
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, Pairs that repel.. Electron pairs around a central atom push each other as far apart as they can. 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: Count the pairs | Load each molecule in the 3D simulator and rotate it. Project tab 1 of the lab (Chemical bonding lab, VSEPR shapes) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Worked examples | Apply the four steps to ions and molecules with double bonds. | Pairs |
| 21 to 30 min | Teach and check: Lone pairs push harder | A lone pair is attracted by only one nucleus, so it sits closer to the central atom and spreads out more. | 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, 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 Lone pairs push harder, 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 |
|---|---|
| When explaining a bond angle, always name both the number of bonding pairs and the number of lone pairs, and state that lone pairs repel more than bonding pairs. “Water is bent because of its lone pairs” is not enough for the marks. | Also, name the shape from the atoms: NH 3 is pyramidal, not tetrahedral, even though its four pairs point towards the corners of a tetrahedron. |
Shapes of molecules: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
State the shape of an ammonia molecule and its bond angle, and explain your answer.
Predict the shape and bond angle of the PH4+ ion.
Phosphorus pentachloride, PCl5, has five bonding pairs around phosphorus. Name its shape and state its two bond angles.
The bond angles in CH4, NH3 and H2O are 109.5°, 107° and 104.5°. Explain this trend.
Predict the shape of the sulfur dioxide molecule, which has two S=O double bonds and one lone pair on sulfur, and estimate the O–S–O angle.
State the shape of (a) beryllium chloride, BeCl2, and (b) the tetrachloroaluminate ion, AlCl4−, with bond angles.
Answer key and marking notes
- Nitrogen has three bonding pairs and one lone pair.
- The pairs repel to be as far apart as possible, and the lone pair repels more strongly than the bonding pairs.
- Shape: pyramidal; bond angle about 107°.
- P has 5 outer electrons, + 4 from H, − 1 for the charge = 8, so four bonding pairs and no lone pairs.
- Tetrahedral, 109.5°.
- Trigonal bipyramidal.
- 120° between the three equatorial bonds, 90° between axial and equatorial bonds.
- Each central atom has four electron pairs, arranged roughly tetrahedrally.
- CH4 has no lone pairs, NH3 one and H2O two.
- Lone pairs repel more strongly than bonding pairs, so each extra lone pair pushes the bonds closer and reduces the angle.
- Three regions of electron density (two double bonds and one lone pair) arrange trigonal planar.
- With one lone pair, the molecule is non-linear (bent), with an angle slightly less than 120° (about 119°).
- (a) Two bonding pairs, no lone pairs: linear, 180°.
- (b) Al has 3 outer electrons, + 4 from Cl, + 1 for the charge = 8, four bonding pairs: tetrahedral, 109.5°.
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.
According to VSEPR theory, which shows the repulsions between electron pairs in decreasing order of strength?
- Alone pair–lone pair > lone pair–bond pair > bond pair–bond pair
- Bbond pair–bond pair > lone pair–bond pair > lone pair–lone pair
- Clone pair–bond pair > lone pair–lone pair > bond pair–bond pair
- Dall electron pairs repel equally, whether bonding or lone pairs
What are the shape of an ammonia molecule, NH3, and its H–N–H bond angle?
- Apyramidal, 107°
- Btrigonal planar, 120°
- Ctetrahedral, 109.5°
- Dnon-linear, 104.5°
Why is the H–O–H bond angle in water (104.5°) smaller than the H–N–H angle in ammonia (107°)?
- Aoxygen has two lone pairs, which repel the bonding pairs more than the single lone pair on nitrogen
- Boxygen is more electronegative, so it pulls the hydrogen atoms closer together
- Cwater has fewer electron pairs around the central atom than ammonia
- Dthe O–H bonds are shorter, so the hydrogen atoms are pushed together
What is the shape of a boron trifluoride molecule, BF3?
- Atrigonal planar, 120°
- Bpyramidal, 107°
- CT-shaped, 90°
- Dtetrahedral, 109.5°
What are the shape and bond angle of sulfur hexafluoride, SF6?
- Aoctahedral, 90°
- Bhexagonal planar, 60°
- Ctrigonal bipyramidal, 90° and 120°
- Dtetrahedral, 109.5°
The diagram shows a molecule of phosphorus pentachloride, PCl5. Which bond angles are found in this molecule?
- A90° and 120° (and 180° between the two axial bonds)
- Bonly 72°, since five bonds share 360°
- Conly 109.5°
- Donly 90°
What are the shape and approximate H–N–H bond angle of the amide ion, NH2−?
- Anon-linear, about 104.5°
- Blinear, 180°
- Cpyramidal, about 107°
- Dtrigonal planar, 120°
Carbon dioxide, O=C=O, is linear but sulfur dioxide, O=S=O, is non-linear. What is the reason?
- Athe sulfur atom has a lone pair as well as two double bonds, while the carbon atom has no lone pair
- Bsulfur atoms are larger than carbon atoms, so the bonds bend
- Cthe S=O bonds are polar but the C=O bonds are not
- Dthe lone pairs on the oxygen atoms push the bonds out of line in SO2 but not in CO2
Quiz key
A lone pair is attracted by only one nucleus, so it is more compact and closer to the central atom than a bonding pair, and it repels other pairs more strongly. So lone pair–lone pair repulsion is greatest, then lone pair–bond pair, then bond pair–bond pair. The second option is the exact reverse. The third option puts the mixed pair above two lone pairs, which is inconsistent with lone pairs being the stronger repellers. If all pairs repelled equally, NH3 and H2O would have 109.5° angles like CH4, but the angles are 107° and 104.5°.
Four electron pairs around N point towards the corners of a tetrahedron, but one is a lone pair, so the atoms form a pyramid; the stronger lone pair repulsion squeezes the bonds from 109.5° to about 107°. Trigonal planar at 120° ignores the lone pair and is the shape of BF3. Tetrahedral at 109.5° is the arrangement of the four pairs, but the shape names only the atoms, and the angle is reduced. Non-linear at 104.5° is water, which has two lone pairs.
Both O and N have four electron pairs arranged roughly tetrahedrally, but O has two lone pairs and two bonding pairs, while N has one lone pair and three bonding pairs. The extra lone pair on O gives more lone pair repulsion, closing the bond angle by about 2.5° per lone pair (109.5 to 107 to 104.5). Electronegativity does not pull the hydrogen atoms together; it pulls the bonding electrons towards O. Both molecules have four pairs around the central atom. Bond length is not what fixes the angle in VSEPR; the number of lone pairs is.
Boron forms three bonding pairs and has no lone pair, so the three pairs spread out as far as possible in a plane at 120°. Pyramidal at 107° would need a lone pair on the central atom, as in NH3. T-shaped needs two lone pairs as well as three bonding pairs (as in ClF3). Tetrahedral needs four electron pairs, but boron has only three, which is why BF3 is electron-deficient.
Six bonding pairs around S point to the corners of an octahedron: four in a square plane and one above and one below, with every adjacent F–S–F angle 90°. A flat hexagon at 60° would keep the pairs much closer together than necessary. Trigonal bipyramidal with 90° and 120° is the shape for five pairs, as in PCl5. Tetrahedral is the shape for four pairs; sulfur can expand its octet to six pairs.
Five bonding pairs and no lone pairs give a trigonal bipyramid. The three equatorial Cl atoms lie in a plane at 120° to each other, and the two axial Cl atoms are at 90° to that plane and 180° to each other. 72° assumes all five bonds lie flat in one plane, which would put the pairs too close together. 109.5° is the tetrahedral angle for four pairs. Only 90° describes an octahedral molecule such as SF6.
N has 5 outer electrons, plus 1 for the charge, making 6; two are used in the N–H bonds (with one from each H), giving 2 bonding pairs and 2 lone pairs. Four pairs with two lone pairs gives a non-linear shape like water, about 104.5°. Linear ignores the lone pairs and counts only the two bonds. Pyramidal at 107° is for three bonding pairs and one lone pair, as in NH3. Trigonal planar would need three pairs and no lone pairs, and forgets the extra electron from the charge.
In CO2 carbon uses all four outer electrons in two double bonds, so there are two regions of electron density and they point in opposite directions (180°). In SO2 sulfur has six outer electrons: four are used in the two double bonds and two form a lone pair, giving three regions, so the molecule is bent (about 119°). Atom size does not decide the shape. Both C=O and S=O bonds are polar. Both molecules have lone pairs on their oxygen atoms; only lone pairs on the central atom affect the shape.
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.