Lesson 2 of 6
Orbital overlap and molecular shape
Sigma/pi overlap, VSEPR and deductions for unfamiliar molecules.
A-Level 9476 (2026-2027)
I can explain covalent bonds using orbital overlap.Syllabus
Syllabus 9476, 2(c). Explain covalent bonds using orbital overlap.
I can explain required molecular shapes and angles by VSEPR.Syllabus
Syllabus 9476, 2(d). Explain required molecular shapes and angles by VSEPR.
I can predict analogous shapes and angles.Syllabus
Syllabus 9476, 2(e). Predict analogous shapes and angles.
End-on overlap gives sigma; sideways overlap gives pi
A double bond contains one sigma and one pi bond.
A sigma (σ) bond has electron density along the line joining the nuclei. It forms by end-on overlap: s-s in H2, s-p in a simple H-Cl description, or p-p. A pi (π) bond forms by sideways overlap of parallel p orbitals, producing electron density above and below that line. The two pi regions together constitute one bond.
The extra bond in ethene comes from sideways overlap
Two nuclei lie on the horizontal internuclear axis. A sigma bond occupies the region directly between them. Parallel vertical p orbitals on the two carbons overlap above and below that axis to form one pi bond.
| Connection | Sigma bonds | Pi bonds |
|---|---|---|
| Single bond | 1 | 0 |
| Double bond | 1 | 1 |
| Triple bond | 1 | 2 in mutually perpendicular planes |
| Whole ethene molecule | 5: four C-H and one C-C | 1: between the carbon atoms |
A carbon-carbon single bond can usually rotate without destroying its sigma overlap. Rotation about a double bond requires loss of the pi overlap and is restricted. Bond rotation is therefore not the same as rotating the entire molecule in space. Detailed d-orbital overlap is outside the overlap requirement here.
Check your understanding
How many sigma and pi bonds are in one molecule of propene, CH2=CHCH3?
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8 sigma and 1 pi
Count each bond once: six C-H and two C-C bonds each contain one sigma bond (8). The C=C also contains one pi bond.
Count electron regions, then describe the atoms
Multiple bonds count as one region. Lone pairs affect shape even though they are not atoms.
Make a guess. You are not marked.
Methane, CH4, has a bond angle of 109.5 degrees. Water also has four electron pairs around its central atom. What is the H-O-H angle?
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Smaller, about 104.5 degrees, as lone pairs repel more
Two of the four pairs on oxygen are lone pairs. Lone pairs repel more strongly than bonding pairs, so they push the O-H bonds closer together.
VSEPR places regions of electron density around a central atom as far apart as possible. A single, double or triple bond counts as one region for this purpose. A lone pair is another region. The electron-region arrangement includes lone pairs; the molecular shape names the positions of the atoms only.
| Molecule | Bonding regions / lone pairs | Molecular shape | Typical bond angle |
|---|---|---|---|
| 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 degrees between adjacent bonds; 180 between opposite bonds |
A lone pair is held by only one nucleus and occupies more space near the central atom than a bonding pair. The usual repulsion order is lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. The tetrahedral electron arrangement therefore gives progressively smaller bond angles from CH4 to NH3 to H2O. Exact angles also depend on the atoms and bonds: do not assign water's angle to every bent molecule.
Worked exampleDeduce an unfamiliar ion
Predict the shape of H3O+.
Try it first. Then show the worked stepsHide the worked steps
- Count valence electrons: oxygen contributes six; three hydrogens add three; the positive charge subtracts one. Total = eight.
- Three O-H shared pairs use six electrons, leaving one lone pair on O.
- Four electron regions give a tetrahedral arrangement; three bonded atoms and one lone pair give a trigonal pyramidal molecular shape.
Trigonal pyramidal. A basic VSEPR comparison predicts an angle near the tetrahedral value, modified by its lone pair; it does not justify copying the exact NH3 angle.
Worked exampleA double bond is one region
SO2 has two S-O bonding regions and one lone pair on S. What shape follows?
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- There are three electron regions, so their reference arrangement is trigonal planar.
- Only two of those regions contain bonded atoms.
- The molecule is bent, with an O-S-O angle slightly below the 120-degree reference.
Bent, not linear. Counting each double bond as two directions would give the wrong model.
Why can sulfur have more than eight outer electrons when oxygen, in the same group, cannot? 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 regions. A Period 2 atom such as N, O or F has only 2s and 2p orbitals, so it is limited to four regions and eight electrons. That is why PCl5 and SF6 exist, but NCl5 and OF6 do not. This is called expanding the octet.
| Example | Bonding regions / lone pairs | Molecular shape | Bond 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 |
| SF6 | 6 / 0 | Octahedral | 90 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 regions, 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 regions 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.
Worked exampleSix regions with two lone pairs
Predict the shape and bond angle of the ICl4- ion.
Try it first. Then show the worked stepsHide the worked steps
- Count electrons around iodine: seven valence electrons plus one for the negative charge = eight.
- Four I-Cl bonds use four of these electrons. The other four make two lone pairs.
- Four bonding pairs and two lone pairs give six regions, so the electron arrangement is octahedral.
- The two lone pairs sit opposite each other to keep as far apart as possible. The four Cl atoms lie in one plane.
Square planar, with Cl-I-Cl bond angles of 90 degrees. Iodine is in Period 5, so it can hold more than eight outer electrons.
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 regions in its second shell, so it cannot form five bonds.
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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