Lesson 2 of 3
Shapes and forces
Use overlap, repulsion and polarity.
A-Level 8873, revised syllabus (2026-2027)
I can connect overlap to sigma and pi bonds.Syllabus
Syllabus 8873, 2(c). Connect overlap to sigma and pi bonds.
I can explain the required shapes with VSEPR.Syllabus
Syllabus 8873, 2(d). Explain the required shapes with VSEPR.
I can predict analogous shapes and angles.Syllabus
Syllabus 8873, 2(e). Predict analogous shapes and angles.
I can infer bond polarity.Syllabus
Syllabus 8873, 2(f). Infer bond polarity.
I can infer molecular polarity.Syllabus
Syllabus 8873, 2(g). Infer molecular polarity.
I can explain intermolecular attractions.Syllabus
Syllabus 8873, 2(h). Explain intermolecular attractions.
I can explain gas liquefaction.Syllabus
Syllabus 8873, 2(i). Explain gas liquefaction.
I can connect hydrogen bonding to properties.Syllabus
Syllabus 8873, 2(j). Connect hydrogen bonding to properties.
Orbital overlap builds bonds; repulsion sets shape
Separate the type of overlap from the arrangement of electron domains.
A sigma bond forms by head-on overlap along the line joining the nuclei. It can use s-s overlap (H2), s-p overlap (HCl) or p-p overlap. A pi bond forms by sideways overlap of parallel p orbitals, with electron density above and below the internuclear axis. A single bond is sigma; a double bond is one sigma plus one pi; a triple bond is one sigma plus two pi.
Head-on and sideways overlap
The sigma overlap lies on the internuclear axis. Parallel p orbitals overlap sideways in two regions for one pi bond, leaving the axis as a nodal region for that pi orbital.
VSEPR places electron domains as far apart as possible to reduce repulsion. Around a central atom, a single, double or triple bond counts as one bonding domain. Lone pairs also repel, but are omitted from the name of the molecular shape. Lone-pair/bond-pair repulsion is greater than bond-pair/bond-pair repulsion, compressing some angles.
| Species | Bonding domains; lone pairs | Molecular shape | Angles |
|---|---|---|---|
| BF3 | 3; 0 | Trigonal planar | 120 degrees |
| CO2 | 2; 0 | Linear | 180 degrees |
| CH4 | 4; 0 | Tetrahedral | 109.5 degrees |
| NH3 | 3; 1 | Trigonal pyramidal | About 107 degrees |
| H2O | 2; 2 | Bent | About 104.5 degrees |
| SF6 | 6; 0 | Octahedral | 90 and 180 degrees |
The table follows the same repulsion argument throughout: two bonding domains point in opposite directions (linear), three spread in one plane (trigonal planar), and four point towards the corners of a tetrahedron. Six equivalent domains point along three perpendicular axes (octahedral). NH3 and H2O both start from four electron domains. Replace one tetrahedral bonding direction by a lone pair to obtain a pyramidal atom arrangement; replace two to obtain a bent arrangement. Lone pairs occupy more angular space near the central atom, so they compress the remaining bond angles.
Worked exampleTransfer the method to an unfamiliar ion
Predict the shape and bond angle of NH4+.
Try it first. Then show the worked stepsHide the worked steps
- Its four shared pairs are four bonding domains around nitrogen.
- There is no lone pair on nitrogen.
- Four domains adopt a tetrahedral arrangement to minimise repulsion.
NH4+ is tetrahedral, about 109.5 degrees. Its parent NH3 is pyramidal because NH3 has a lone pair.
Worked exampleFind the lone pairs before naming the shape
Deduce the shape of SCl2, taking sulfur as the central atom.
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- Count 6 + 2(7) = 20 outer electrons. Draw two S-Cl single bonds, using four electrons.
- Complete each terminal Cl octet with three lone pairs: twelve more electrons. The four electrons left belong on S as two lone pairs.
- Only count domains around S: two S-Cl bonding domains plus two lone pairs, giving four. The Cl lone pairs do not add to the domain count around S.
- The electron-domain arrangement is tetrahedral. Naming only the atoms gives a bent molecule, analogous to H2O. Lone-pair repulsion compresses the Cl-S-Cl angle below 109.5 degrees.
SCl2 is bent and polar because its equal S-Cl dipoles do not cancel. VSEPR gives an angle smaller than 109.5 degrees; it does not justify copying the exact 104.5-degree water value to every bent molecule.
Sulfur in SF6 has twelve outer electrons, but oxygen in the same group never goes past eight. An element in Period 3 or below has empty d orbitals in its outer shell (3d for Period 3). They are close enough in energy to be used, so the central atom can hold five or six electron domains. A Period 2 atom such as N, O or F has only 2s and 2p orbitals, so it is limited to four domains and eight electrons. That is why PCl5 and SF6 exist, but NCl5 and OF6 do not. This is called expanding the octet.
| Example | Bonding domains; lone pairs | Molecular shape | Angles |
|---|---|---|---|
| PCl5 | 5; 0 | Trigonal bipyramidal | 90, 120 and 180 degrees |
| SF4 | 4; 1 | See-saw | Less than 90 and less than 120 degrees |
| ClF3 | 3; 2 | T-shaped | Slightly less than 90 degrees |
| XeF2 | 2; 3 | Linear | 180 degrees |
| BrF5 | 5; 1 | Square pyramidal | Slightly less than 90 degrees |
| XeF4 | 4; 2 | Square planar | 90 degrees |
Where the lone pairs go decides the shape. With five domains, the three equatorial positions are 120 degrees apart, while the two axial positions are at 90 degrees to three others. A lone pair takes an equatorial position, where it has fewer close neighbours at 90 degrees. So one lone pair gives a see-saw, two give a T-shape and three give a straight line. With six domains every position is the same, so one lone pair gives a square pyramid. A second lone pair goes opposite the first, as far from it as possible, which leaves a square plane.
Check your understanding
Phosphorus forms PCl3 and PCl5. Nitrogen, in the same group, forms NCl3 but not NCl5. Why?
Show the answer
P is in Period 3 and can use empty 3d orbitals; N in Period 2 has no 2d orbitals.
Phosphorus can hold five bonding pairs, ten electrons, by using its 3d orbitals. Nitrogen is limited to four electron pairs in its second shell, so it cannot form five bonds.
Ethene contains five sigma bonds in total (four C-H and one C-C) plus one C-C pi bond. Sideways overlap requires the two p orbitals to remain aligned, so the double bond cannot rotate freely without disrupting that overlap. This explains later cis-trans isomerism.
Exam-style question
Predict the shape of a ClF3 molecule and its F-Cl-F bond angle. Explain your answer. [3]
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A polar bond does not guarantee a polar molecule
Combine bond direction, shape and the possible attractions between particles.
Make a guess. You are not marked.
H2S molecules have more electrons than H2O molecules. Which has the higher boiling point?
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H2O, because its molecules form hydrogen bonds
Water molecules form hydrogen bonds, because H is bonded to the small, very electronegative O. These are far stronger than the extra dispersion forces in H2S.
Electronegativity describes an atom's attraction for a bonding electron pair. In H-Cl, Cl attracts the pair more strongly: H is partially positive and Cl partially negative. These are partial charges within a covalent bond, not full H+ and Cl- ions. The molecular dipole is the combined effect of all bond dipoles in three dimensions.
| Molecule | Reason | Overall polarity |
|---|---|---|
| CO2 | Equal C=O bond dipoles point in opposite directions in a linear molecule. | Non-polar |
| BF3 | Three equal B-F dipoles cancel in its trigonal plane. | Non-polar |
| SF6 | Six equal S-F dipoles cancel in an octahedron. | Non-polar |
| H2O | The bent shape prevents the O-H dipoles cancelling. | Polar |
| NH3 | Pyramidal arrangement gives a net dipole. | Polar |
| CHCl3 | The tetrahedral molecule has unlike surrounding atoms, so its dipoles do not cancel. | Polar |
Instantaneous dipole-induced dipole attractions occur in all particles: a momentary uneven electron distribution induces a dipole in a neighbouring particle, and the nearby opposite partial charges attract electrostatically. They are the principal intermolecular attraction in Br2 and the attraction between noble-gas atoms. Larger, more easily polarised electron clouds usually give stronger attractions. CHCl3 also has permanent dipole-permanent dipole attraction between oppositely charged ends of neighbouring molecules.
Hydrogen bonding is an electrostatic attraction between the partially positive H covalently bonded to N, O or F and a lone pair on N, O or F in a neighbouring particle. Water provides O-H donors and oxygen lone pairs; ammonia provides N-H donors and a nitrogen lone pair. Merely containing hydrogen is insufficient: methane cannot hydrogen-bond to itself.
The same possible attractions exist in a gas and its liquid, but gas particles are usually far apart. Cooling reduces kinetic energy and compression brings particles closer, allowing attractions to hold particles in a liquid. A larger attraction generally makes liquefaction easier. Pressure alone does not guarantee liquefaction at every temperature; this is a qualitative explanation, not an ideal-gas calculation.
Hydrogen bonding makes water's boiling point relatively high for such a small molecule because substantial energy is needed to separate molecules. In ordinary ice, each water molecule hydrogen-bonds to four neighbours in an open, approximately tetrahedral network: its two O-H groups donate two hydrogen bonds and its two oxygen lone pairs accept two. On melting, some of that open arrangement collapses, so liquid water is denser than ice and ice floats. The covalent O-H bonds remain intact.
Check your understanding
CH4 and H2O are both small molecules. Why does water boil at a much higher temperature?
Show the answer
Water molecules form hydrogen bonds; methane has only weak instantaneous dipole attractions.
More energy is needed to separate water molecules, held by hydrogen bonds. Boiling does not break the covalent O-H bonds.