AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/2/2-7)
Unit 2 · Topic 2.7
2.7 VSEPR and Hybridization
VSEPR theory predicts a molecule's shape from the idea that electron domains around a central atom repel each other and spread as far apart as possible. From the shape you can find bond angles, decide whether the molecule is polar, name the central atom's hybridization and count sigma and pi bonds.
Key terms
- VSEPR theory
- electron domain
- molecular geometry
- hybridization
- sigma and pi bonds
- molecular polarity
Electron domains and VSEPR
An electron domain is any region of electron density around the central atom: a lone pair, a single bond, a double bond or a triple bond. A multiple bond counts as one domain.
VSEPR (valence shell electron pair repulsion) theory says domains repel each other, because electrons are all negative, so they arrange themselves as far apart as possible. The arrangement of all domains sets the electron geometry. The molecular geometry, the shape you name, describes only where the atoms are.
Lone pairs take up more room than bonding pairs, so they squeeze bond angles a little. Methane (no lone pairs) has 109.5° angles; ammonia (1 lone pair) about 107°; water (2 lone pairs) about 104.5°.
The shapes
For 5 and 6 domains you only need the shape and angles. Hybridization involving d orbitals isn't tested.
| Domains | Lone pairs | Molecular shape | Ideal angles | Hybridization | Example |
|---|---|---|---|---|---|
| 2 | 0 | linear | 180° | sp | CO₂ |
| 3 | 0 | trigonal planar | 120° | sp² | BF₃ |
| 3 | 1 | bent | less than 120° | sp² | SO₂ |
| 4 | 0 | tetrahedral | 109.5° | sp³ | CH₄ |
| 4 | 1 | trigonal pyramidal | less than 109.5° | sp³ | NH₃ |
| 4 | 2 | bent | less than 109.5° | sp³ | H₂O |
| 5 | 0 | trigonal bipyramidal | 90°, 120°, 180° | not tested | PCl₅ |
| 5 | 1 | seesaw | about 90° and 120° | not tested | SF₄ |
| 5 | 2 | T-shaped | about 90° | not tested | ClF₃ |
| 5 | 3 | linear | 180° | not tested | XeF₂ |
| 6 | 0 | octahedral | 90° | not tested | SF₆ |
| 6 | 1 | square pyramidal | about 90° | not tested | BrF₅ |
| 6 | 2 | square planar | 90° | not tested | XeF₄ |
Polarity of a whole molecule
A molecule is polar if its bond dipoles don't cancel, leaving one end partly negative. Two things matter: polar bonds and shape.
If identical polar bonds are arranged symmetrically with no lone pairs on the central atom (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral), the dipoles cancel and the molecule is nonpolar: CO₂, BF₃, CCl₄. Square planar XeF₄ and linear XeF₂ are also nonpolar, because their lone pairs sit symmetrically opposite each other.
Bent, trigonal pyramidal, seesaw, T-shaped and square pyramidal molecules are polar when their bonds are polar: H₂O, NH₃, SO₂, SF₄, BrF₅. A tetrahedral molecule with different outer atoms, like CH₂Cl₂, is also polar, because its bond dipoles aren't all equal and don't cancel.
Hybridization, sigma and pi bonds
Hybridization names how an atom's orbitals are arranged: count its electron domains. 2 domains is sp (180°), 3 is sp² (120°) and 4 is sp³ (109.5°). You don't need to know how hybrid orbitals are derived.
Every single, double or triple bond contains exactly one sigma (σ) bond, formed by head-on orbital overlap. A single bond is just 1 σ. A double bond is 1 σ + 1 π; a triple bond is 1 σ + 2 π. A pi (π) bond forms from side-by-side overlap, which is weaker than head-on overlap, so σ bonds are stronger than π bonds.
A π bond locks the atoms so they can't rotate around the bond. That's why cis- and trans-1,2-dichloroethene are different compounds: in the cis isomer both Cl atoms are on the same side of the C=C bond (polar), and in the trans isomer they're on opposite sides (nonpolar).
You won't be tested on
Molecular orbital diagrams, and bonding, nonbonding and antibonding orbitals, aren't assessed. Stick with Lewis diagrams, VSEPR and the sp, sp², sp³ labels.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Shape, angle and polarity
Predict the molecular geometry, approximate bond angle, hybridization and polarity of NH₃.
Show the solutionHide the solution
- Step 1: Lewis diagram: N has 5 valence electrons and each H has 1, total 8. N forms three N–H bonds and keeps one lone pair.
- Step 2: 4 electron domains (3 bonds + 1 lone pair) gives a tetrahedral electron geometry and sp³ hybridization.
- Step 3: Naming only the atoms, the shape is trigonal pyramidal. The lone pair pushes the bonds together slightly, so the angle is a bit less than 109.5° (about 107°).
- Step 4: N–H bonds are polar (N is more electronegative), and the pyramid isn't symmetric, so the dipoles don't cancel.
Answer: Trigonal pyramidal, about 107°, sp³, polar
- Example 2
Counting σ and π bonds and naming hybridization
Acrylonitrile has the structure H₂C=CH–C≡N. How many σ and π bonds does it contain? Give the hybridization of each carbon atom.
Show the solutionHide the solution
- Step 1: List the bonds: three C–H single bonds (3 σ), one C=C double bond (1 σ + 1 π), one C–C single bond (1 σ) and one C≡N triple bond (1 σ + 2 π).
- Step 2: Total σ = 3 + 1 + 1 + 1 = 6. Total π = 1 + 2 = 3.
- Step 3: The CH₂ carbon has 3 domains (two C–H and the C=C), so it is sp². The CH carbon also has 3 domains (C=C, C–H, C–C), so it is sp².
- Step 4: The carbon in C≡N has 2 domains (the C–C and the triple bond), so it is sp, with a 180° angle.
Answer: 6 σ bonds and 3 π bonds; the first two carbons are sp² and the nitrile carbon is sp.
- Example 3
Lone pairs change everything (classic trap)
CO₂ and SO₂ both have two oxygen atoms bonded to a central atom. Explain why CO₂ is nonpolar but SO₂ is polar.
Show the solutionHide the solution
- Step 1: CO₂: C has 2 double bonds and no lone pairs, so 2 domains and a linear shape. The two C=O dipoles point in opposite directions and cancel.
- Step 2: SO₂: the molecule has 18 valence electrons, and its Lewis diagram leaves S with one lone pair, so S has 3 domains. The shape is bent (just under 120°).
- Step 3: In a bent molecule, the two S–O dipoles don't point in opposite directions, so they add up to a net dipole.
- Step 4: Checking only the formula (AB₂) without drawing the Lewis diagram misses the lone pair.
Answer: CO₂ is linear, so its bond dipoles cancel; SO₂ has a lone pair on S, making it bent, so its bond dipoles don't cancel and it is polar.
Common mistakes
- Naming the shape from electron geometry instead of atom positions. NH₃ has tetrahedral electron geometry but a trigonal pyramidal shape.
- Counting a double or triple bond as more than one electron domain.
- Calling a molecule polar just because it has polar bonds. Check whether the shape lets the dipoles cancel.
- Saying a double bond is two σ bonds. It's one σ and one π.
On the exam
- Many free-response questions ask you to draw a Lewis diagram and then state a shape, a bond angle or hybridization. Justify the shape by stating the number of bonding and nonbonding domains on the central atom.
- Expect questions linking shape to polarity, and polarity to intermolecular forces in Unit 3.
Connected topics
Videos
Check yourself
4 questions on 2.7 VSEPR and Hybridization. Pick an answer to see if you got it, and why.
Acrylonitrile, used to make plastics and synthetic fibers, has the condensed structure CH₂=CH–C≡N.
The first carbon is bonded to two hydrogen atoms and double-bonded to the second carbon. The second carbon is bonded to one hydrogen atom and to the third carbon, which is triple-bonded to nitrogen. Nitrogen has one lone pair.
Described molecular structure
How many sigma (σ) bonds and pi (π) bonds are in one molecule of acrylonitrile?
What is the hybridization of each carbon atom, listed in order from the CH₂ carbon to the carbon bonded to nitrogen?
Which of the following is closest to the C–C–N bond angle in acrylonitrile?
Which of the following molecules is polar?
0 of 4 answered