AP Chemistry • Compound Structure and Properties

VSEPR and Hybridization

Predicting three-dimensional molecular geometry from electron-pair repulsion and orbital mixing.

AP Chemistry illustration with a laboratory flask, atom, and microscope
Inside this lesson 10 sections 5 questions 6 figures

SECTION 1

Historical Context & Motivation

Lewis dot structures, introduced by Gilbert N. Lewis in 1916, provided a powerful way to represent bonding and lone pairs in molecules, but they revealed nothing about the three-dimensional arrangement of atoms in space. Chemists recognized early on that molecular shape profoundly influences physical properties such as polarity, boiling point, and reactivity. The question driving mid-twentieth-century structural chemistry was clear: given a Lewis structure, how can one predict a molecule's geometry without resorting to complex quantum-mechanical calculations? Two complementary models—VSEPR theory and hybridization—emerged to answer that question, each approaching it from a different angle: one purely geometric, the other rooted in orbital theory.

1916 • Lewis Dot Structures

Gilbert N. Lewis publishes his theory of shared electron-pair bonding, establishing the language of valence electrons and the octet rule that underpins both VSEPR and hybridization models.

1931 • Pauling's Hybridization Theory

Linus Pauling proposes that atomic orbitals on a central atom can mix—or hybridize—to form equivalent hybrid orbitals oriented in space, explaining the tetrahedral geometry of carbon in methane.

1940 • Sidgwick & Powell's Electron-Pair Model

Nevil Sidgwick and Herbert Powell observe that the arrangement of bonds around a central atom correlates with the total number of electron pairs, laying the groundwork for VSEPR.

1957 • Gillespie & Nyholm Formalize VSEPR

Ronald Gillespie and Ronald Nyholm refine the model, introducing the systematic rules for predicting geometry from electron-domain counts—the version taught worldwide today.

1960s–Today • X-Ray & Computational Validation

Advances in X-ray crystallography and computational chemistry validate VSEPR predictions for a vast number of molecules, while also revealing its limitations for transition-metal compounds.

Together, VSEPR and hybridization provide the conceptual toolkit that chemists—and AP Chemistry students—use to move from a flat Lewis structure to a prediction of bond angles, molecular shape, and the orbital description that rationalizes that shape. Understanding both models, and knowing when each is most useful, is essential for mastering topics on the AP exam ranging from intermolecular forces to organic reaction mechanisms.

SECTION 2

Core Principles & Definitions

Before diving into geometric predictions, it is important to distinguish several key terms. An electron domain (also called an electron group) is any region around a central atom where electrons are likely to be found: a single bond, a double bond, a triple bond, or a lone pair each count as one electron domain. The electron-domain geometry describes how all electron domains—bonding and nonbonding—are arranged around the central atom, while the molecular geometry describes the positions of only the atoms (excluding lone pairs). These two geometries coincide when no lone pairs are present, but diverge when they are, a distinction the AP exam frequently tests.

1

VSEPR Principle

Electron domains around a central atom repel one another and adopt the spatial arrangement that maximizes their separation, thereby minimizing repulsion energy.
2

Lone-Pair Repulsion Hierarchy

Repulsion strength follows the order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Lone pairs are 'fatter' because they spread closer to the nucleus and compress bond angles.
3

Hybridization as Orbital Mixing

Atomic orbitals on the same atom can combine (hybridize) to produce a set of degenerate hybrid orbitals oriented to match the electron-domain geometry predicted by VSEPR.
4

Conservation of Orbitals

The number of hybrid orbitals produced always equals the number of atomic orbitals that mix. Two atomic orbitals → two hybrid orbitals; three → three; and so on.
5

Sigma & Pi Framework

Hybrid orbitals form σ (sigma) bonds and hold lone pairs. Unhybridized p orbitals overlap side-by-side to form π (pi) bonds in double and triple bonds.
✦

KEY TAKEAWAY

Think of electron domains like inflated balloons tied together at a central knot: each balloon pushes the others away, and the resulting arrangement depends on how many balloons there are—two balloons line up straight (linear), three spread into a triangle (trigonal planar), and four point to the corners of a tetrahedron. VSEPR tells you the balloon arrangement; hybridization tells you which atomic orbitals were mixed to create the balloons in the first place.

SECTION 3

VSEPR Geometries — Visual Overview

The following diagram presents the five fundamental electron-domain geometries that arise from two through six electron domains, along with representative molecular geometries when lone pairs are present. Each electron-domain geometry serves as the 'parent' from which specific molecular shapes descend by replacing bonding pairs with lone pairs.

VSEPR Electron-Domain GeometriesLinear2 ED · 180°sp · BeCl₂Trigonal Planar3 ED · 120°sp² · BF₃Tetrahedral4 ED · 109.5°sp³ · CH₄Trig. Bipyramidal5 ED · 90°/120°sp³d · PCl₅▼ Effect of Lone Pairs on Tetrahedral Parent Geometry ▼Trigonal Pyramidal3 BP + 1 LP · ≈107°NH₃Bent2 BP + 2 LP · ≈104.5°H₂OSeesaw4 BP + 1 LPSF₄Gold = central atom · Colored = bonding atoms · Dashed = atom behind plane · Ellipses = lone pairs
The top row shows the five parent electron-domain geometries for 2–5 electron domains. The bottom row illustrates how replacing bonding pairs with lone pairs in the tetrahedral and trigonal bipyramidal parents produces distinct molecular geometries (trigonal pyramidal, bent, and seesaw). Note that bond angles decrease as lone pairs are added because of the greater repulsive 'spread' of nonbonding electron density.

Several patterns are worth committing to memory. First, the ideal bond angles—180°, 120°, 109.5°, 90°/120°, and 90°—arise solely from the number of electron domains and assume all domains are equivalent. Second, lone pairs compress bond angles below these ideals because lone-pair electron density sits closer to the nucleus and exerts a broader repulsive 'footprint' than bonding pairs. Third, the molecular geometry name is always determined by the arrangement of atoms, not electron domains—which is why NH3 is described as trigonal pyramidal despite having a tetrahedral electron-domain geometry.

SECTION 4

Hybridization — Orbital Mixing Framework

While VSEPR predicts the geometry that electron domains adopt, hybridization provides the quantum-mechanical rationale: the central atom's atomic orbitals combine to form a new set of equivalent hybrid orbitals whose orientations match the VSEPR-predicted geometry. The key idea is that an isolated carbon atom has non-equivalent 2s and 2p orbitals, yet in methane all four C–H bonds are experimentally identical. Pauling resolved this contradiction by showing that one 2s and three 2p orbitals can mathematically combine into four equivalent sp³ hybrid orbitals pointing toward the vertices of a tetrahedron.

Mapping Electron Domains to Hybridization

The rule is straightforward: count the total number of electron domains (σ bonds + lone pairs) on the central atom—this equals the number of atomic orbitals that must hybridize. A double or triple bond counts as a single electron domain because only the σ component uses a hybrid orbital; the additional π bonds use unhybridized p orbitals.

Electron DomainsHybridizationAtomic Orbitals MixedGeometryUnhybridized p Orbitals
2spone s + one pLinear (180°)2
3sp²one s + two pTrigonal Planar (120°)1
4sp³one s + three pTetrahedral (109.5°)0
5sp³done s + three p + one dTrigonal Bipyramidal0
6sp³d²one s + three p + two dOctahedral0

Relationship between electron-domain count, hybridization, and geometry

Sigma and Pi Bond Accounting

Every covalent bond contains exactly one σ bond. Additional bonding interactions in double and triple bonds are π bonds formed by lateral overlap of unhybridized p orbitals. Thus, a double bond = 1 σ + 1 π and a triple bond = 1 σ + 2 π. In ethene (C2H4), each carbon is sp² hybridized with three σ bonds (two C–H and one C–C) and one π bond from the remaining unhybridized p orbital on each carbon. In ethyne (C2H2), each carbon is sp hybridized with two σ bonds and two π bonds from two unhybridized p orbitals.

💡

AP Exam Tip

When asked to determine hybridization on the AP Chemistry exam, use the shortcut: count the number of atoms bonded to the central atom plus the number of lone pairs on the central atom. That total equals the number of hybrid orbitals, which directly gives the hybridization. Do not count multiple bonds separately—a double or triple bond still contributes only one to the count.

SECTION 5

Comprehensive Geometry Classification

The full range of molecular geometries testable on the AP Chemistry exam can be organized by the number of electron domains and the number of lone pairs. The following table provides a complete reference, including ideal bond angles and the modifications caused by lone-pair compression.

EDBPLPED GeometryMolecular GeometryBond AngleExample
220LinearLinear180°CO₂
330Trig. PlanarTrigonal Planar120°BF₃
321Trig. PlanarBent< 120°SO₂
440TetrahedralTetrahedral109.5°CH₄
431TetrahedralTrigonal Pyramidal≈107°NH₃
422TetrahedralBent≈104.5°H₂O
550Trig. BipyramidalTrigonal Bipyramidal90°, 120°PCl₅
541Trig. BipyramidalSeesaw< 90°, < 120°SF₄
532Trig. BipyramidalT-shaped< 90°ClF₃
523Trig. BipyramidalLinear180°XeF₂
660OctahedralOctahedral90°SF₆
651OctahedralSquare Pyramidal< 90°BrF₅
642OctahedralSquare Planar90°XeF₄

Complete VSEPR geometry table: ED = electron domains, BP = bonding pairs, LP = lone pairs

Hybridization and Orbital DiagramsEnergy-level representation showing orbital mixingsp³ (e.g., CH₄)Atomic Orbitals↑↓2s↑2p↑→Hybrid↑↑↑↑4 × sp³sp² (e.g., C in C₂H₄)Atomic Orbitals↑↓2s↑2p↑↑→Hybrid↑↑↑3 × sp²↑p (π)sp (e.g., C in C₂H₂)Atomic Orbitals↑↓2s↑2p↑↑→Hybrid↑↑2 × sp↑↑2p (2 × π)Legends orbitalp orbital (hybridized)p orbital (unhybridized, π)Hybrid orbital↑ = electron (up spin)↑↓ = electron pair
Orbital box diagrams for sp³, sp², and sp hybridization of carbon. In each case, the number of atomic orbitals mixed equals the number of hybrid orbitals produced. Unhybridized p orbitals (green boxes) remain available for π bonding.

A critical observation from this diagram is the relationship between multiple bonding and hybridization. As the hybridization changes from sp³ to sp² to sp, the number of unhybridized p orbitals increases from 0 to 1 to 2, enabling progressively more π bonds. This explains why sp²-hybridized atoms can form one double bond, while sp-hybridized atoms can form either two double bonds (as in CO2) or one triple bond (as in C2H2). In the trigonal bipyramidal and octahedral cases, d orbitals join the mix, which is why expanded octets occur only in atoms from period 3 and below, where d orbitals are energetically accessible.

SECTION 6

Worked Example — Analyzing XeF₄

Xenon tetrafluoride (XeF4) is an excellent example because it involves an expanded octet. Let us determine its Lewis structure, electron-domain and molecular geometry, bond angles, polarity, and hybridization.

Predicting Geometry and Hybridization for XeF₄

  1. 1

    Step 1 — Draw the Lewis Structure

    Count total valence electrons: Xe contributes 8, and each F contributes 7, giving 8 + 4(7) = 36 valence electrons. Xenon is the central atom. Place four Xe–F single bonds (8 electrons), then distribute remaining electrons to satisfy octets on F (24 electrons as 12 lone pairs on fluorines), leaving 4 electrons = 2 lone pairs on Xe. The Lewis structure shows Xe bonded to four F atoms with two lone pairs on Xe.
    36 total valence e⁻; 4 bonding pairs + 2 lone pairs on Xe
  2. 2

    Step 2 — Count Electron Domains on Central Atom

    Xenon has 4 bonding pairs (one to each F) and 2 lone pairs, for a total of 6 electron domains.
    6 electron domains
  3. 3

    Step 3 — Determine Electron-Domain Geometry

    Six electron domains adopt an octahedral electron-domain geometry to maximize separation (all angles 90°).
    Octahedral electron-domain geometry
  4. 4

    Step 4 — Determine Molecular Geometry

    With 4 bonding pairs and 2 lone pairs in an octahedral arrangement, the lone pairs position themselves opposite each other (trans, at 180°) to minimize lone pair–lone pair repulsion. The four fluorine atoms then occupy the equatorial plane, giving a square planar molecular geometry with 90° F–Xe–F bond angles.
    Square planar molecular geometry, 90° bond angles
  5. 5

    Step 5 — Assign Hybridization

    Six electron domains require six hybrid orbitals: one s + three p + two d = sp³d². Four of these hybrid orbitals overlap with F 2p orbitals to form σ bonds, and two hold lone pairs.
    sp³d² hybridization
  6. 6

    Step 6 — Assess Polarity

    Although each Xe–F bond is polar (ΔEN = 1.4), the square planar geometry is perfectly symmetric. Every bond dipole has an equal and opposite counterpart, so the individual dipole moments cancel. XeF₄ is a nonpolar molecule.
    Nonpolar — symmetric cancellation of bond dipoles

SECTION 7

Strengths, Limitations & Common Pitfalls

VSEPR and hybridization are remarkably effective for main-group compounds, but like all models they have boundaries. The table below summarizes where these models shine and where they falter, which is important context for both understanding chemistry deeply and for AP free-response questions that ask students to evaluate the validity of a model.

StrengthLimitation
Accurately predicts geometry for most main-group molecules using simple electron-pair countingFails for many transition-metal complexes where crystal field theory or ligand field theory is needed
Correctly explains how lone pairs compress bond angles relative to ideal valuesCannot predict exact bond angles—only approximate trends (e.g., NH₃ is ≈107°, not exactly 109.5°)
Hybridization links orbital theory to observable molecular geometry in an intuitive wayHybridization is a mathematical construct, not directly observable; MO theory provides a more rigorous picture
Straightforward σ/π accounting enables prediction of bond order and rotational rigidityStruggles with delocalized π systems; resonance must be invoked alongside hybridization
No math or computation required—purely qualitative reasoningDoes not predict bond energies, magnetic properties, or spectroscopic data

VSEPR and hybridization: strengths vs. limitations

⚠

Common Pitfall

Students frequently confuse electron-domain geometry with molecular geometry. Remember: the electron-domain geometry accounts for all electron domains (bonding + lone pairs), but the molecular geometry describes only the positions of atoms. Always name both when asked on the AP exam, and be explicit about which one you are reporting.
✦

MODELS IN CONTEXT

Think of VSEPR as a reliable GPS that gets you to the right neighborhood—it tells you the molecular shape with high confidence—but molecular orbital theory is the street-level map that shows you exactly where the electron density lives. For the AP Chemistry exam, VSEPR and hybridization are the primary tools you'll deploy; MO theory appears only for diatomics. Master the simpler model first, then appreciate where it hands off to the more advanced one.

SECTION 8

Connection to Molecular Orbital Theory

Hybridization is classified as part of valence bond (VB) theory, which treats bonds as localized overlaps between orbitals on adjacent atoms. An alternative and more complete framework, molecular orbital (MO) theory, constructs orbitals that are delocalized over the entire molecule by combining all atomic orbitals of appropriate symmetry. MO theory naturally explains phenomena that VB theory handles awkwardly—paramagnetism of O2, for instance—without needing resonance structures. On the AP exam, MO theory is applied mainly to homonuclear and heteronuclear diatomics (O2, N2, NO, etc.), but understanding its relationship to VB theory deepens your grasp of bonding models overall.

FeatureValence Bond / HybridizationMolecular Orbital Theory
Orbital scopeLocalized between two atomsDelocalized over entire molecule
Bond orderCounted by σ/π analysis½(bonding e⁻ − antibonding e⁻)
MagnetismCannot predict (predicts O₂ as diamagnetic—incorrect)Correctly predicts O₂ as paramagnetic
ResonanceRequires multiple Lewis structuresDelocalization emerges naturally
AP exam scopeAll main-group polyatomicsHomonuclear & heteronuclear diatomics

Valence Bond theory vs. Molecular Orbital theory

For AP Chemistry, the take-home message is that VSEPR and hybridization remain your go-to toolkit for predicting and explaining the geometry and bonding of polyatomic main-group species. When a question involves diatomic bond order, magnetic behavior, or relative bond energies among diatomics, switch to MO theory. The two frameworks are not contradictory—they are complementary levels of approximation applied to different classes of problems.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

The ion ICl4− has a square planar molecular geometry. What is the electron-domain geometry around the central iodine atom?
  • A. Tetrahedral
  • B. Trigonal bipyramidal
  • C. Octahedral
  • D. Square planar
Show answer (C)
The correct answer is C. A square planar molecular geometry arises when two lone pairs occupy axial positions in an octahedral arrangement of six electron domains (4 bonding pairs + 2 lone pairs). Answer A (tetrahedral) has only four electron domains and cannot produce a square planar shape. Answer B (trigonal bipyramidal) has five electron domains; with two lone pairs in an equatorial arrangement, it would produce a T-shape, not square planar. Answer D (square planar) is incorrect because 'square planar' describes the molecular geometry, not the electron-domain geometry—the electron-domain geometry must account for the lone pairs and is therefore octahedral.

Question 2: BASIC CALCULATION

What is the hybridization of the central nitrogen atom in NO3− (nitrate ion)?
  • A. sp
  • B. sp²
  • C. sp³
  • D. sp³d
Show answer (B)
The correct answer is B. In NO₃⁻, nitrogen is bonded to three oxygen atoms (via resonance structures showing one double bond and two single bonds that are equivalent by resonance), with no lone pairs on nitrogen. This gives 3 electron domains, which requires sp² hybridization. Answer A (sp) would correspond to 2 electron domains (linear geometry). Answer C (sp³) would require 4 electron domains, which would mean a lone pair on nitrogen; however, the formal charge analysis and Lewis structure show that nitrogen has no lone pair in nitrate. Answer D (sp³d) requires 5 electron domains and an expanded octet, which nitrogen cannot achieve as a period-2 element.

Question 3: INTERMEDIATE

Consider the molecule SOCl2 (thionyl chloride), in which S is the central atom. Which of the following correctly describes its molecular geometry and polarity?
  • A. Trigonal planar; nonpolar
  • B. Trigonal pyramidal; polar
  • C. Bent; polar
  • D. Tetrahedral; nonpolar
Show answer (B)
The correct answer is B. Sulfur in SOCl₂ has 4 electron domains: one S=O double bond, two S–Cl single bonds, and one lone pair. (Total valence electrons: 6 + 6 + 2×7 = 26; after placing bonds and lone pairs on terminal atoms, one lone pair remains on S.) Four electron domains give a tetrahedral electron-domain geometry; with one lone pair, the molecular geometry is trigonal pyramidal. The asymmetric arrangement of different atoms (O and Cl have different electronegativities) and the lone pair ensure the dipole moments do not cancel, making the molecule polar. Answer A (trigonal planar) ignores the lone pair on S. Answer C (bent) would require two lone pairs and only two bonding pairs, which doesn't match. Answer D (tetrahedral) would require four bonding pairs with no lone pairs.

Question 4: APPLIED

5 pts
Acetic acid (CH3COOH) contains two carbon atoms in different bonding environments. (a) Draw the Lewis structure for acetic acid. (1 pt) (b) Identify the hybridization of each carbon atom. (1 pt) (c) Predict the approximate bond angles around each carbon atom. (1 pt) (d) Identify the total number of σ bonds and π bonds in the molecule. (1 pt) (e) Explain why the C–C bond in acetic acid is a single bond and can rotate freely, while the C=O bond is a double bond with restricted rotation. Relate your answer to the orbital overlap model. (1 pt)
Show answer
Rubric breakdown: (a) 1 point — Lewis structure: The methyl carbon (C₁) is bonded to three H atoms and one C. The carboxyl carbon (C₂) is bonded to C₁, a doubly-bonded O, and an –OH group. All octets satisfied (hydrogen has duets); total of 24 valence electrons. (b) 1 point — C₁ (CH₃ carbon) has 4 electron domains (4 single bonds, no lone pairs) → sp³ hybridized. C₂ (COOH carbon) has 3 electron domains (one C–C σ bond, one C=O, one C–O σ bond; double bond counts as one domain) → sp² hybridized. (c) 1 point — Around C₁ (sp³): bond angles ≈ 109.5° (tetrahedral). Around C₂ (sp²): bond angles ≈ 120° (trigonal planar). (d) 1 point — σ bonds: C₁–H × 3, C₁–C₂, C₂=O (σ component), C₂–O(H), O–H = 7 σ bonds total. π bonds: C₂=O contains 1 π bond. Total: 7 σ + 1 π. (e) 1 point — The C–C single bond consists of end-to-end (σ) overlap of an sp³ hybrid orbital on C₁ with an sp² hybrid orbital on C₂. Because σ bonds have cylindrical symmetry about the internuclear axis, rotation about this bond does not break the overlap and is therefore free. The C=O double bond consists of a σ bond (head-on overlap of an sp² hybrid on C₂ with an orbital on O) plus a π bond (side-by-side overlap of the unhybridized p orbital on C₂ with a p orbital on O). Rotation about the C=O axis would destroy the lateral p-orbital overlap needed for the π bond, so rotation is restricted.

Question 5: CRITICAL THINKING

4 pts
A student measures the bond angles in three molecules and records the following data: Molecule | Measured Bond Angle CH₄ | 109.5° NH₃ | 107.3° H₂O | 104.5° All three molecules have four electron domains around the central atom. (a) Explain the observed trend in bond angles from CH₄ to NH₃ to H₂O using VSEPR theory. (2 pts) (b) A classmate claims that because H₂O has a smaller bond angle than NH₃, the oxygen atom in water must use sp² hybridization rather than sp³. Evaluate this claim. (1 pt) (c) Predict the approximate bond angle in NF₃ relative to NH₃ and justify your prediction. (1 pt)
Show answer
Rubric breakdown: (a) 2 points (1 for identifying lone-pair repulsion trend, 1 for explaining the mechanism): — CH₄ has 4 bonding pairs and 0 lone pairs; the four equivalent bond pairs adopt ideal tetrahedral angles of 109.5°. — NH₃ has 3 bonding pairs and 1 lone pair. Lone pairs occupy more angular space near the nucleus than bonding pairs because they are not shared with another atom and spread more broadly. The lone pair exerts greater repulsion on the bonding pairs, compressing the H–N–H angles below 109.5° to ≈107.3°. — H₂O has 2 bonding pairs and 2 lone pairs. Two lone pairs exert even more compressive force on the bonding pairs, reducing the H–O–H angle further to ≈104.5°. — The trend is: as the number of lone pairs increases from 0 to 1 to 2, bond angles decrease because lone pair–bond pair repulsion is stronger than bond pair–bond pair repulsion. (b) 1 point — The classmate's claim is incorrect. The hybridization of oxygen in H₂O is sp³, not sp², because there are 4 electron domains (2 bonding pairs + 2 lone pairs), and 4 electron domains require 4 hybrid orbitals. The reduction in bond angle from the ideal 109.5° does not change the hybridization; it simply reflects the greater repulsive effect of lone pairs relative to bonding pairs. Hybridization is determined by the total number of electron domains, not by the exact bond angle. (c) 1 point — NF₃ is expected to have a bond angle slightly less than that of NH₃ (i.e., < 107.3°). Fluorine is more electronegative than hydrogen, so bonding electron density in N–F bonds is pulled farther from the nitrogen nucleus compared to N–H bonds. This makes the N–F bonding pairs 'thinner' (they occupy less angular space near N) and relatively less repulsive toward each other, allowing the lone pair on nitrogen to compress the F–N–F angles more than in NH₃. Experimentally, the F–N–F angle is about 102°.

SUMMARY

Summary

Summary

VSEPR theory predicts that electron domains around a central atom arrange themselves to minimize repulsion, producing characteristic electron-domain geometries (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral). When one or more domains are lone pairs, the molecular geometry differs from the electron-domain geometry, and bond angles are compressed below their ideal values because lone pair–bond pair repulsion exceeds bond pair–bond pair repulsion.

Hybridization provides the orbital-level rationale: atomic orbitals on the central atom mix to form hybrid orbitals (sp, sp², sp³, sp³d, sp³d²) whose number equals the total number of electron domains. Hybrid orbitals form σ bonds and hold lone pairs, while unhybridized p orbitals overlap laterally to form π bonds in double and triple bonds. Together, these two models allow you to move from a Lewis structure to a complete three-dimensional molecular description—including geometry, bond angles, polarity, and orbital description—using straightforward counting and reasoning.

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