Difficulty: Intermediate | Prerequisites: Lewis structures, electronegativity (see Part 1 notes)
VSEPR theory predicts molecular shapes by counting electron domains (bonding pairs and lone pairs) around a central atom and arranging them to minimise repulsion. Hybridization (sp, sp2, sp3) follows directly from the number of electron domains. Whether a molecule is polar overall depends on both the polarity of its individual bonds and the symmetry of its geometry.
VSEPR (Valence Shell Electron Pair Repulsion)
A model that predicts molecular geometry based on the idea that electron domains around a central atom arrange themselves as far apart as possible to minimise repulsion.
Electron domain (region of electron density)
A bonding pair, a lone pair, or a multiple bond (counts as one domain) around the central atom. In simple terms, each "blob" of electrons, whether it is a single bond, double bond, triple bond, or lone pair, counts as one domain.
Lone pair (unshared pair)
A pair of valence electrons on the central atom that is not involved in bonding. Lone pairs take up more space than bonding pairs and compress bond angles.
Bond angle
The angle between two adjacent bonds on the same central atom. Determined by electron domain geometry, then adjusted if lone pairs are present.
Molecular geometry (molecular shape)
The three-dimensional arrangement of atoms only (ignoring lone pairs) in a molecule. Electron geometry includes lone pairs; molecular geometry does not.
Hybridization
The mixing of atomic orbitals (s, p, and sometimes d) to form new, equivalent hybrid orbitals used in bonding. In simple terms, atoms remix their orbitals to match the number of directions they need to bond in.
sp hybridization
Two hybrid orbitals formed from one s and one p orbital. Produces a linear arrangement (180 degrees). Found in molecules with two electron domains, such as CO2 and C2H2.
sp2 hybridization
Three hybrid orbitals from one s and two p orbitals. Produces a trigonal planar arrangement (120 degrees). Found in molecules with three electron domains, such as BF3, C2H4, and formaldehyde.
sp3 hybridization
Four hybrid orbitals from one s and three p orbitals. Produces a tetrahedral arrangement (109.5 degrees). Found in molecules with four electron domains, such as CH4, NH3, and H2O.
Sigma (s) bond
A bond formed by head-on overlap of orbitals along the internuclear axis. Every single bond is a sigma bond, and every double or triple bond contains exactly one sigma bond.
Pi (p) bond
A bond formed by side-by-side overlap of p orbitals above and below the internuclear axis. The second bond in a double bond and the second and third bonds in a triple bond are pi bonds.
Dipole moment
A measure of the overall polarity of a molecule. Depends on both the magnitude of bond dipoles and whether they cancel by symmetry.
Resonance structures
Two or more Lewis structures for the same molecule that differ only in the placement of electrons (not atoms). The true structure is an average (hybrid) of all resonance forms.
The core method
Draw the Lewis structure.
Count electron domains on the central atom (each single bond, double bond, triple bond, or lone pair = one domain).
Determine electron domain geometry (how all domains arrange).
Determine molecular geometry (the shape defined by atoms only, ignoring lone pairs).
Geometry reference
2 domains: linear, 180 degrees (e.g. BeCl2, CO2)
3 domains: trigonal planar, 120 degrees (e.g. BF3, SO3). If one lone pair, molecular shape is bent.
4 domains: tetrahedral, 109.5 degrees (e.g. CH4, SiH4). One lone pair gives trigonal pyramidal (approximately 107 degrees, e.g. NH3, PH3). Two lone pairs give bent (approximately 104.5 degrees, e.g. H2O, H2S).
Lone pairs compress bond angles
The progressive decrease in bond angles from CH4 (109.5) to NH3 (approximately 107) to H2O (approximately 104.5) is explained by the increasing number of lone pairs on the central atom.
Lone pairs occupy more space than bonding pairs because they are held closer to the central nucleus. They push bonding pairs closer together.
Key geometry examples from the quiz
SO3: trigonal planar (three bonding domains, no lone pairs on S in the dominant Lewis structure).
NF3: trigonal pyramidal (three bonding pairs + one lone pair). This is why NF3 is polar while BF3 (trigonal planar, no lone pair) is nonpolar.
BH3: trigonal planar, and B has fewer than an octet (only 6 electrons). BH3 has the largest bond angle among H2O, NH3, BH3, CH4, SiH4 because it is 120 degrees.
H2O: bent, with two lone pairs on oxygen.
C2H4 (ethylene): each carbon has three electron domains (one double bond + two single bonds), so the geometry around each C is trigonal planar. H-C-H angles are approximately 120 degrees.
The glycinium cation example
H-O-C bond angle: oxygen has four electron domains (two bonding + two lone pairs), so the geometry around O is tetrahedral but the molecular shape is bent, giving approximately 105 degrees.
The two C atoms and two O atoms lie in the same plane because the carboxyl carbon is sp2 (trigonal planar).
The N atom has four electron domains (three N-H bonds + one lone pair), making it tetrahedral/trigonal pyramidal, not planar.
Bent geometry with lone pairs on the central atom (Lewis diagram question)
A Lewis structure showing X bonded to two Z atoms, with two lone pairs on X, represents a bent molecule. The two lone pairs are what create the bent shape from a tetrahedral electron geometry.
Species that are not planar
PCl3: three bonding pairs + one lone pair = trigonal pyramidal, not planar.
CO3 2-, NO3-, BF3 are all trigonal planar (no lone pairs on the central atom).
ClF3 is T-shaped (five electron domains, two lone pairs), which is technically planar but for a different reason.
Rule: number of electron domains = number of hybrid orbitals
2 electron domains = sp (linear, 180 degrees)
3 electron domains = sp2 (trigonal planar, 120 degrees)
4 electron domains = sp3 (tetrahedral, 109.5 degrees)
Key examples
C2H2 (ethyne): each C has two electron domains (one triple bond + one single bond). Hybridization = sp. Linear.
C2H4 (ethylene): each C has three electron domains (one double bond + two single bonds). Hybridization = sp2. Trigonal planar.
C2H6 (ethane): each C has four electron domains (four single bonds). Hybridization = sp3. Tetrahedral.
CH4: four bonding pairs around C = sp3.
H2O: four electron domains on O (two bonding + two lone) = sp3.
HCONH2 (formamide): the carbon has three electron domains (one double bond to O, one single to H, one single to N) = sp2. The H-C-N bond angle is approximately 120 degrees.
Hybridization changes during reactions
C2H2 + Cl2 gives C2H2Cl2: the triple bond in C2H2 partially breaks. Carbon goes from sp (two domains) to sp2 (three domains) as one pi bond opens and Cl atoms add.
Benzene (C6H6) + 3 H2: each carbon goes from sp2 (three domains in the aromatic ring) to sp3 (four single bonds in cyclohexane).
C2H4 + O3 forming a cyclic ozonide: each carbon goes from sp2 to sp3 as the double bond opens and new single bonds form.
Both CH4 and H2O are sp3
This is a common exam question. Despite their different molecular shapes (tetrahedral vs bent), both central atoms have four electron domains and therefore sp3 hybridization.
Counting rules
Every single bond = 1 sigma bond.
Every double bond = 1 sigma + 1 pi.
Every triple bond = 1 sigma + 2 pi.
Examples
N2 (triple bond): 1 sigma + 2 pi bonds. This is why N2 has the largest bond dissociation energy and strongest intramolecular forces among BeCl2, SO2, N2, O2, and F2.
O2 (double bond): 1 sigma + 1 pi. Bond order = 2.
F2 (single bond): 1 sigma, 0 pi. Bond order = 1.
C2H2: 3 sigma bonds (one C-C + two C-H) and 2 pi bonds (from the triple bond). Total: 3 sigma, 2 pi.
CO2: 2 sigma bonds + 2 pi bonds (each C=O is 1 sigma + 1 pi).
HCN: 2 sigma bonds (H-C + C-N) + 2 pi bonds (from the triple bond C to N).
Pi bonding and planarity
Pi bonds prevent rotation around the bond axis, which is why C2H4 is planar.
Molecules with only single bonds (like C2H6) have free rotation and are not planar.
Species without pi bonds
CH4 has only single bonds, so all bonds are sigma. No pi bonding occurs.
A molecule is polar if it has polar bonds AND those bond dipoles do not cancel each other out. Symmetry is the deciding factor.
Nonpolar molecules with polar bonds (bond dipoles cancel)
CCl4: four polar C-Cl bonds arranged tetrahedrally. The dipoles cancel perfectly. Zero net dipole moment.
CO2: two polar C=O bonds in a linear arrangement. The dipoles point in opposite directions and cancel.
CS2: same logic as CO2, linear and symmetric.
BF3: three polar B-F bonds in a trigonal planar arrangement. Dipoles cancel.
SF6: six polar S-F bonds in an octahedral arrangement. Dipoles cancel.
PF5: five polar P-F bonds in a trigonal bipyramidal arrangement. Dipoles cancel. Zero dipole moment.
Polar molecules (bond dipoles do not cancel)
H2O: bent shape means the two O-H dipoles do not cancel. Strong dipole moment.
NH3: trigonal pyramidal, so the three N-H dipoles do not cancel.
NF3: trigonal pyramidal (lone pair on N), so despite the trigonal planar electron domain geometry, the molecular shape is asymmetric and the molecule is polar. This is the key difference from BF3, which is planar with no lone pair.
HF: only two atoms with a large electronegativity difference. Largest dipole moment among CO, CO2, O2, HF, and F2.
SO2: bent shape (three electron domains on S, including one lone pair), polar.
PCl3: trigonal pyramidal due to a lone pair on P, so it has a net dipole. Among CS2, PCl3, SF6, and CF4, only PCl3 has bond dipoles that do not cancel.
BrF: two atoms with different electronegativities, so it is polar. Has a larger dipole moment than Cl-F because Br is less electronegative than Cl, creating a bigger difference with F.
Largest dipole moment questions
Among CS2, PCl3, SF6, and CF4: PCl3 has the greatest dipole moment because the Cl atoms draw electron density away from P and the bond dipoles do not cancel (trigonal pyramidal).
Among CO2, H2O, CH4, C2H4, PH3: H2O has the largest dipole moment due to its bent geometry and the large electronegativity difference between O and H.
Students often confuse electron geometry with molecular geometry. The electron geometry includes lone pairs in the arrangement; the molecular geometry describes only where the atoms sit. A tetrahedral electron geometry can give tetrahedral, trigonal pyramidal, or bent molecular geometry depending on how many lone pairs are present.
Thinking a molecule with polar bonds must be polar. Symmetry can cause polar bond dipoles to cancel completely (CCl4, CO2, SF6).
Forgetting that a double or triple bond counts as one electron domain. Some students count each bond in a double bond separately, leading to incorrect geometry predictions.
Mixing up BF3 (trigonal planar, nonpolar) and NF3 (trigonal pyramidal, polar). The lone pair on nitrogen is what breaks the symmetry.
Assuming hybridization changes only when new atoms are added. Hybridization changes whenever the number of electron domains changes, even if the same atoms are present in a different bonding arrangement.
VSEPR geometry questions appear in both MCQ and FRQ sections. The exam loves asking for bond angles given a Lewis structure.
Hybridization tied to geometry is a frequent MCQ topic. Know the direct link: 2 domains = sp, 3 = sp2, 4 = sp3.
"Which molecule is polar / has the largest dipole moment" is one of the most common question types. Always check both bond polarity and molecular symmetry.
Sigma and pi bond counting comes up regularly, especially for organic-adjacent molecules (ethylene, acetylene, benzene).
Resonance and its effect on bond length and bond order is tested, particularly for SO2, O3, NO3-, CO3 2-, and benzene.
Expect questions about hybridization changes during a reaction (before and after).
True or False: A molecule with four electron domains around the central atom always has a tetrahedral molecular geometry. (False, it depends on how many are lone pairs.)
Fill in the blank: sp2 hybridization gives ______ electron domains arranged at ______ degrees. (3 domains, 120 degrees)
True or False: CO2 is a polar molecule. (False, the bond dipoles cancel due to linear symmetry.)
Fill in the blank: A double bond consists of ______ sigma bond(s) and ______ pi bond(s). (1 sigma, 1 pi)
True or False: NF3 is nonpolar because it is trigonal planar like BF3. (False, NF3 is trigonal pyramidal due to its lone pair, so it is polar.)
Q: What is the hybridization of the carbon atoms in ethyne (H-C triple bond C-H), and what is the bond angle?
A: sp hybridization. Each carbon has two electron domains (one triple bond and one single bond to H). The bond angle is 180 degrees (linear).
Q: The bond angles in CH4, NH3, and H2O decrease in that order. Explain why using VSEPR theory.
A: All three central atoms have four electron domains (sp3). CH4 has zero lone pairs (ideal tetrahedral, 109.5 degrees). NH3 has one lone pair, which compresses the bond angles to approximately 107 degrees. H2O has two lone pairs, which compress the bond angles further to approximately 104.5 degrees. Lone pairs take up more space than bonding pairs.
Q: Explain why BF3 is nonpolar but NF3 is polar, even though both have three fluorine atoms bonded to the central atom.
A: BF3 has three bonding domains and no lone pairs, making it trigonal planar. The three B-F bond dipoles are arranged symmetrically at 120 degrees and cancel. NF3 has three bonding domains and one lone pair, making it trigonal pyramidal. The asymmetric shape means the N-F bond dipoles do not cancel, giving a net dipole moment.
Q: In the reaction C2H4 + O3 forming an ozonide, how does the hybridization of the carbon atoms change?
A: Before the reaction, each carbon in C2H4 has three electron domains (sp2). After the reaction, each carbon has four electron domains (sp3), because the double bond opens and new single bonds form to oxygen atoms.
Q: Which of the following molecules has resonance structures and a bent geometry: CO2, O3, CH4, or BeF2?
A: O3. It has two resonance structures (the double bond can be drawn on either side), and the central oxygen has three electron domains (two bonding + one lone pair), giving a bent molecular geometry.
Molecular polarity connects directly to intermolecular forces (Unit 3). Polar molecules experience dipole-dipole forces; molecules with N-H, O-H, or F-H bonds can hydrogen-bond. Nonpolar molecules rely on London dispersion forces only. The geometry you determine here tells you which IMFs apply.
Hybridization and sigma/pi bond counting connect to organic chemistry concepts in AP Chemistry, particularly when you encounter reactions that break pi bonds and form new sigma bonds.
Resonance structures connect to formal charge analysis and to understanding why certain molecules have equal bond lengths (like benzene's six equivalent C-C bonds or the three equivalent N-O bonds in nitrate).
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