VSEPR Theory and Molecular Geometry

🔬 VSEPR Theory and Molecular Geometry

Part 1 of 7 — Introduction to VSEPR

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 1

• Understanding the core concepts covered in Part 1
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

What Is an Electron Domain?

An electron domain (also called an electron group or region of electron density) is any of the following around a central atom:

Critical Rule

A double bond counts as ONE electron domain. A triple bond also counts as ONE electron domain. Only the number of regions of electron density matters, not the total number of electrons.

🔑 Key Concept: A double or triple bond counts as one electron domain — only the number of distinct regions of electron density matters, not the bond order or total electron count.

Examples:

• $CO_{2}$: C has 2 double bonds → 2 electron domains
• $H_{2}O$: O has 2 single bonds + 2 lone pairs → 4 electron domains
• $NH_{3}$: N has 3 single bonds + 1 lone pair → 4 electron domains
• HCN: C has 1 single bond + 1 triple bond → 2 electron domains

Identify the number of electron domains around the central atom.

The Steric Number

The steric number is the total number of electron domains around the central atom. It is calculated as:

$\boxed{\text{Steric Number} = \text{(number of atoms bonded to central atom)} + \text{(number of lone pairs on central atom)}}$

The steric number determines the electron domain geometry — the arrangement of ALL electron groups (both bonding and lone pairs) in 3D space.

💡 Tip: The steric number equals the number of "things" attached to the central atom — count each bond (regardless of type) and each lone pair as one.

Example Calculation

For $H_{2}O$:

• Bonded atoms = 2 (two H atoms)
• Lone pairs on O = 2
• Steric number = 2 + 2 = 4
• Electron domain geometry = Tetrahedral

🔑 Key Concept: The steric number maps directly to the electron domain geometry — memorize the five base shapes (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral).

Note: The molecular geometry (shape based only on atom positions) may differ from the electron domain geometry when lone pairs are present. We'll explore this distinction next.

Determine the steric number for each central atom.

Two Types of Geometry

This is one of the most important distinctions in VSEPR theory:

Electron Domain Geometry

• Describes the arrangement of all electron domains (bonding + lone pairs)
• Determined solely by the steric number
• Think of it as the "invisible scaffolding"

Molecular Geometry

• Describes the arrangement of only the atoms (ignoring lone pairs)
• This is the actual shape of the molecule
• It's what we observe experimentally

When Are They Different?

🔑 Key Concept: Electron domain geometry ≠ molecular geometry when lone pairs are present. The electron domain geometry includes lone pairs; the molecular geometry shows only atom positions.

They are the same when there are no lone pairs on the central atom.

They are different when lone pairs are present — because lone pairs take up space in the electron domain geometry but are invisible in the molecular shape.

Example: $CH_{4}$ vs. $NH_{3}$ vs. $H_{2}O$

All three have the same electron domain geometry (tetrahedral), but the molecular geometry changes as lone pairs replace bonding pairs.

💡 Tip: To find the molecular geometry, start with the electron domain geometry and "remove" the lone pairs — what remains is the molecular shape.

Test your understanding of the difference between electron domain and molecular geometry.

Select the correct answers for each scenario.

📐 Linear, Trigonal Planar, and Tetrahedral Geometries

Part 2 of 7 — The Core Geometries

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 2

• Understanding the core concepts covered in Part 2
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

Trigonal Planar Geometry

When a central atom has 3 electron domains (steric number = 3), they spread out equally in a flat plane, 120° apart.

$\boxed{\text{Bond angle} = 120°}$

Characteristics

• Shape: flat triangle with the central atom at the center
• Bond angle: exactly 120°
• All atoms lie in the same plane

Examples

Important: Boron is Special

Boron (B) commonly forms only 3 bonds and has no lone pairs, making it naturally trigonal planar. $BF_{3}$ has only 6 electrons around B — it is an electron-deficient compound (an exception to the octet rule).

Test your knowledge of these geometries.

Tetrahedral Geometry

When a central atom has 4 electron domains (steric number = 4), they arrange in a three-dimensional shape called a tetrahedron.

$\boxed{\text{Bond angle} = 109.5°}$

Characteristics

• Shape: 3D triangular pyramid with 4 vertices
• Bond angle: approximately 109.5° (the "tetrahedral angle")
• NOT flat — atoms extend above and below a central plane

Why 109.5°?

The angle 109.5° is the angle that maximizes the distance between 4 points on a sphere. It's derived from the geometry of a regular tetrahedron:

$\boxed{\cos(109.5°) = -\frac{1}{3}}$

Examples

The tetrahedral geometry is extremely common in organic chemistry — every $sp^{3}$-hybridized carbon is tetrahedral.

Enter the ideal bond angle for each geometry.

Summary of Core Geometries

The Pattern

🔑 Key Concept: As the steric number increases, the bond angle decreases (180° → 120° → 109.5°). More electron domains must share the space around the central atom.

As the steric number increases:

• The bond angle decreases (180° → 120° → 109.5°)
• The geometry becomes more three-dimensional
• The electron domains spread into more directions

Hybridization Connection

💡 Tip: Each base geometry maps to a specific hybridization — sp (linear), $sp^{2}$ (trigonal planar), $sp^{3}$ (tetrahedral). Knowing one tells you the other.

Each geometry corresponds to a specific hybridization of the central atom:

• sp → linear (2 hybrid orbitals)
• $sp^{2}$ → trigonal planar (3 hybrid orbitals)
• $sp^{3}$ → tetrahedral (4 hybrid orbitals)

This connection between VSEPR geometry and orbital hybridization is fundamental to understanding bonding in organic and inorganic chemistry.

Identify the electron domain geometry and bond angle for each molecule.

Comprehensive check on linear, trigonal planar, and tetrahedral geometries.

🔷 Trigonal Bipyramidal and Octahedral Geometries

Part 3 of 7 — 5 and 6 Electron Domains

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 3

• Understanding the core concepts covered in Part 3
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

Trigonal Bipyramidal Geometry

When a central atom has 5 electron domains, they arrange in a trigonal bipyramidal shape. This geometry has two distinct types of positions:

Axial vs. Equatorial Positions

• Equatorial (3 positions): Arranged in a flat triangle around the "equator" — 120° apart from each other
• Axial (2 positions): Located directly above and below the equatorial plane — 90° from equatorial positions and 180° from each other

Bond Angles

$\boxed{\text{Equatorial–Equatorial} = 120°}$ $\boxed{\text{Axial–Equatorial} = 90°}$ $\text{Axial–Axial} = 180°$

Why This Matters

🔑 Key Concept: In a trigonal bipyramid, the axial and equatorial positions are not equivalent. Lone pairs always prefer equatorial positions because they have fewer 90° neighbors (only 2 vs. 3).

Unlike tetrahedral or octahedral geometries, the positions in a trigonal bipyramid are NOT equivalent. This has important consequences:

• Lone pairs preferentially occupy equatorial positions (more room)
• Axial bonds are slightly longer than equatorial bonds
• The non-equivalence leads to several different molecular geometries when lone pairs are present

Example: $PCl_{5}$

Phosphorus pentachloride has 5 bonding pairs and 0 lone pairs:

• Steric number = 5
• Electron domain geometry = trigonal bipyramidal
• Molecular geometry = trigonal bipyramidal
• Bond angles: 90° (ax–eq) and 120° (eq–eq)

Test your understanding of the trigonal bipyramidal geometry.

Octahedral Geometry

When a central atom has 6 electron domains, they arrange in an octahedral shape — like two square pyramids joined at their bases.

Bond Angles

$\boxed{\text{Adjacent positions} = 90°}$ $\text{Opposite positions} = 180°$

Key Feature: All Positions Are Equivalent

🔑 Key Concept: Unlike the trigonal bipyramid, all 6 positions in an octahedron are equivalent — each has exactly 4 neighbors at 90° and 1 at 180°.

Unlike the trigonal bipyramid, all 6 positions in an octahedron are equivalent. Each position has exactly 4 neighbors at 90° and 1 neighbor at 180°.

Visualizing the Octahedron

Think of it as:

• 4 positions forming a square in the horizontal plane
• 1 position directly above
• 1 position directly below

Or equivalently: place atoms along the +x, −x, +y, −y, +z, and −z axes.

Examples

$SF_{6}$ is a classic example: sulfur forms 6 equivalent bonds to fluorine with all F–S–F angles = 90°.

Enter the bond angles for these geometries.

The Complete Set of Electron Domain Geometries

Pattern: Bond Angles Decrease as Steric Number Increases

More electron domains must share the space around the central atom, so each pair gets pushed closer together:

$\boxed{180° \to 120° \to 109.5° \to 90°}$

Dimensionality

• Steric number 2: 1D (line)
• Steric number 3: 2D (plane)
• Steric numbers 4, 5, 6: 3D (all extend into three dimensions)

Match each property to the correct geometry.

Check your understanding of expanded geometries.

👁️ Lone Pair Effects on Molecular Geometry

Part 4 of 7 — Bent, Trigonal Pyramidal, Seesaw, T-Shaped, Square Pyramidal, and Square Planar

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 4

• Understanding the core concepts covered in Part 4
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

Molecular Shapes from Steric Number 4

All of these have tetrahedral electron domain geometry but different molecular geometries:

Tetrahedral (0 lone pairs)

• Example: $CH_{4}$
• Bond angle: 109.5°
• 4 bonds, 0 lone pairs

Trigonal Pyramidal (1 lone pair)

• Example: $NH_{3}$
• Bond angle: ≈107° (compressed from 109.5°)
• 3 bonds, 1 lone pair
• Shape: like a tripod or a pyramid with a triangular base

Bent (2 lone pairs)

• Example: $H_{2}O$
• Bond angle: ≈104.5° (compressed further)
• 2 bonds, 2 lone pairs
• Shape: like a boomerang or V-shape

The Compression Pattern

🔑 Key Concept: Each additional lone pair on the central atom compresses bond angles by about 2–2.5° from the ideal value.

Each additional lone pair compresses the bond angle by about 2–2.5°.

Identify the molecular geometry for each scenario.

Molecular Shapes from Steric Number 5

Starting from trigonal bipyramidal electron domain geometry, lone pairs always go in equatorial positions first (fewer 90° repulsions).

💡 Tip: In a trigonal bipyramid, always place lone pairs in equatorial positions first — they have only 2 neighbors at 90° (vs. 3 for axial), minimizing repulsion.

Trigonal Bipyramidal (0 lone pairs)

• Example: $PCl_{5}$
• 5 bonds, 0 lone pairs
• Bond angles: 90° (ax–eq) and 120° (eq–eq)

Seesaw (1 lone pair, equatorial)

• Example: $SF_{4}$
• 4 bonds, 1 lone pair
• The lone pair occupies an equatorial position
• Shape looks like a seesaw or a distorted tetrahedron
• Bond angles: slightly less than 90° and 120°

T-Shaped (2 lone pairs, both equatorial)

• Example: $ClF_{3}$
• 3 bonds, 2 lone pairs
• Both lone pairs in equatorial positions
• Shape: like a capital letter T
• Bond angles: slightly less than 90°

Linear (3 lone pairs, all equatorial)

• Example: $XeF_{2}$
• 2 bonds, 3 lone pairs
• All 3 lone pairs fill the equatorial plane
• The 2 bonds are axial → linear molecular geometry
• Bond angle: 180°

Molecular Shapes from Steric Number 6

Starting from octahedral electron domain geometry:

Octahedral (0 lone pairs)

• Example: $SF_{6}$
• 6 bonds, 0 lone pairs
• All bond angles: 90°

Square Pyramidal (1 lone pair)

• Example: $BrF_{5}$
• 5 bonds, 1 lone pair
• The lone pair occupies one position, leaving 5 atoms in a square pyramid
• Bond angles: slightly less than 90°

Square Planar (2 lone pairs)

• Example: $XeF_{4}$
• 4 bonds, 2 lone pairs
• The 2 lone pairs are placed opposite each other (trans positions) to minimize LP–LP repulsion
• The 4 bonds form a flat square
• Bond angles: 90°

Why Trans for 2 Lone Pairs?

💡 Tip: In an octahedral arrangement with 2 lone pairs, they always adopt trans (180° apart) positions to minimize the very strong lone pair–lone pair repulsion.

If the 2 lone pairs were adjacent (cis), they would be only 90° apart — very strong repulsion. By placing them opposite (trans, 180° apart), LP–LP repulsion is minimized.

Match each molecule to its molecular geometry.

For each molecule, determine the number of lone pairs on the central atom.

Comprehensive lone pair effects quiz.

🧭 Predicting Molecular Geometry

Part 5 of 7 — From Lewis Structure to 3D Shape

Topics in This Part

🔑 Key Concept: Mastering this material will strengthen your foundation for both the AP Chemistry exam and more advanced chemistry topics.

What You'll Master in Part 5

• Understanding the core concepts covered in Part 5
• Applying these ideas to solve practice problems
• Building toward AP exam readiness for this topic

Worked Example: $SO_{2}$ (Sulfur Dioxide)

Problem: Predict the molecular geometry of $SO_{2}$.

Solution:

Step 1: Lewis Structure

• Total valence electrons: S(6) + 2 × O(6) = 18
• Sulfur is central; each O is bonded to S
• Best structure: S has one double bond to each O and one lone pair
• (Resonance structures exist, but the electron domain count is the same)

Step 2: Central Atom

• Sulfur (least electronegative, most bonds)

Step 3: Count Electron Domains

• 2 bonds (each double bond = 1 domain) + 1 lone pair = 3 electron domains

Step 4: Electron Domain Geometry

• Steric number 3 → Trigonal planar

Step 5: Molecular Geometry

• 3 electron domains minus 1 lone pair = 2 bonding positions visible
• Molecular geometry: Bent
• Bond angle: slightly less than 120° (lone pair compression)

Summary

$\boxed{\text{SO}_2: \quad \text{3 e⁻ domains} \to \text{trigonal planar (ED)} \to \text{bent (molecular)}}$

Worked Example: $XeF_{4}$ (Xenon Tetrafluoride)

Problem: Predict the molecular geometry of $XeF_{4}$.

Solution:

Step 1: Lewis Structure

• Total valence electrons: Xe(8) + 4 × F(7) = 36
• Xenon is central
• Xe forms 4 bonds to F, using 8 electrons
• Xe has 2 lone pairs (4 remaining electrons)
• Each F has 3 lone pairs

Step 2: Central Atom

• Xenon

Step 3: Count Electron Domains

• 4 bonds + 2 lone pairs = 6 electron domains

Step 4: Electron Domain Geometry

• Steric number 6 → Octahedral

Step 5: Molecular Geometry

• 2 lone pairs placed trans (opposite, 180° apart) to minimize repulsion
• 4 F atoms form a flat square
• Molecular geometry: Square planar
• Bond angles: 90°

Summary

$\boxed{\text{XeF}_4: \quad \text{6 e⁻ domains} \to \text{octahedral (ED)} \to \text{square planar (molecular)}}$

Apply the step-by-step method to predict molecular geometries.

For each molecule, determine the requested value. Use the Lewis structure to find bonds and lone pairs on the central atom.

Master Reference Chart

This chart is essential for the AP exam — memorize it!

🔑 Key Concept: This master chart covers every possible VSEPR molecular geometry. The molecular geometry depends on both the steric number and the number of lone pairs.

Select the correct molecular geometry for each molecule.

Test the full prediction method.

⚡ Polarity of Molecules

Part 6 of 7 — From Bond Dipoles to Molecular Dipoles

Practice Makes Perfect

This workshop features multi-step problems that mirror the AP Chemistry exam format. Each problem requires you to combine concepts from previous parts and show your work clearly.

🔑 Why this matters: The AP Chemistry exam rewards students who can apply concepts to unfamiliar problems — structured practice is the best preparation.

What You'll Master in Part 6

• Working through complete multi-step problems from start to finish
• Building problem-solving strategies you can apply on the AP exam
• Identifying which concepts to apply and in what order

Symmetry Is the Key

Nonpolar Molecules (Symmetric — Dipoles Cancel)

Even if individual bonds are polar, the molecule can be nonpolar if the geometry is symmetric and all outer atoms are the same:

Polar Molecules (Asymmetric — Dipoles Don't Cancel)

The Two Requirements for a Nonpolar Molecule

💡 Tip: A molecule is nonpolar only when both conditions are met: (1) the geometry is symmetric, and (2) all surrounding atoms are identical. If either fails, the molecule is polar.

1. The geometry must be symmetric
2. All surrounding atoms must be identical

If either condition fails, the molecule is polar.

Determine whether each molecule is polar or nonpolar.

Lone Pairs Guarantee Asymmetry

🔑 Key Concept: Any molecule with lone pairs on the central atom and polar bonds will be polar — the lone pairs create an asymmetric electron density distribution with no opposing dipole.

Any molecule with lone pairs on the central atom and polar bonds will be polar, because the lone pairs create an asymmetric distribution of electron density.

Why?

Lone pairs contribute to the electron density around the central atom but don't have a corresponding atom on the opposite side to balance them. This creates a region of high electron density with no opposing dipole.

Examples

Exception: Symmetric Lone Pair Arrangements

💡 Tip: Some molecules with lone pairs are still nonpolar — if the lone pairs are arranged symmetrically (e.g., $XeF_{2}$ with 3 equatorial lone pairs, $XeF_{4}$ with trans lone pairs), the overall dipole cancels.

Some molecules have lone pairs but are still nonpolar because the lone pairs are arranged symmetrically:

• $XeF_{2}$: 3 lone pairs (all equatorial) + 2 bonds (axial) → linear → nonpolar
• $XeF_{4}$: 2 lone pairs (trans) + 4 bonds → square planar → nonpolar

The key is whether the overall arrangement (bonds + lone pairs) produces a net dipole.

Classify each molecule as polar or nonpolar.

Why Polarity Matters

Molecular polarity directly affects physical properties:

Solubility

• "Like dissolves like"
• Polar molecules dissolve in polar solvents (e.g., water)
• Nonpolar molecules dissolve in nonpolar solvents (e.g., hexane)

Boiling Point

• Polar molecules have stronger intermolecular forces (dipole–dipole interactions)
• Higher polarity → higher boiling point (generally)
• Nonpolar molecules rely on weaker London dispersion forces

Intermolecular Forces Hierarchy

$\boxed{\text{Ion–ion} > \text{Hydrogen bonding} > \text{Dipole–dipole} > \text{London dispersion}}$

Polar molecules with N–H, O–H, or F–H bonds can form hydrogen bonds — the strongest type of intermolecular force (excluding ionic).

Example Comparison

The dramatic difference in boiling points is largely due to water's strong hydrogen bonding, which is possible because of its polar bent geometry.

Answer these questions about molecular polarity.

Test your understanding of molecular polarity.

🎯 Synthesis & AP Exam Review

Part 7 of 7 — Comprehensive Review

Bringing It All Together

This comprehensive review connects every concept from Parts 1–6 with AP-style problems. The questions are designed to mirror what you'll see on the actual exam — multi-step, multi-concept, and requiring clear written explanations.

🔑 Why this matters: AP Chemistry exam questions rarely test one concept in isolation — success requires connecting ideas across topics.

What You'll Master in Part 7

• Solving AP-style questions that integrate multiple concepts from this unit
• Writing clear, concise explanations using proper chemistry terminology
• Identifying and avoiding common AP exam traps and mistakes

Identify the molecular geometry of each species from its Lewis structure.

Predict the approximate bond angle for each molecule. Use the ideal angle for the geometry (don't worry about small lone pair compressions unless specified).

For each molecule, predict whether it is polar or nonpolar based on its geometry.

How to Answer VSEPR Free-Response Questions

AP Chemistry FRQs often ask you to:

1. Draw or describe the Lewis structure
2. Predict the molecular geometry
3. Explain whether the molecule is polar or nonpolar
4. Relate geometry/polarity to a physical property

Template Answer

💡 Tip: Use this template structure for VSEPR free-response answers — it hits every point the AP graders look for.

"The Lewis structure of [molecule] shows that the central atom has [X] bonding domains and [Y] lone pairs, giving a steric number of [X+Y]. The electron domain geometry is [ED geometry], and since there are [Y] lone pairs, the molecular geometry is [molecular geometry]. The bond angle is approximately [angle]°.

Because the [molecular geometry] shape is [symmetric/asymmetric], the individual bond dipoles [do/do not] cancel. Therefore, the molecule is [polar/nonpolar]."

Common Mistakes to Avoid

⚠️ Warning: These are the most frequent errors on the AP exam — review each one carefully.

1. Confusing electron domain and molecular geometry — always specify which one you mean
2. Forgetting lone pairs — they affect both geometry and polarity
3. Saying a molecule is nonpolar just because it has polar bonds — symmetry matters
4. Counting double bonds as 2 electron domains — a double bond is ONE domain
5. Forgetting to adjust for ions — add/subtract electrons for charges

🔑 Key Concept: The most common AP mistake is confusing electron domain geometry with molecular geometry — always specify which one you mean and remember that lone pairs make them different.

These questions mimic the style and difficulty of AP Chemistry exam questions.

For each molecule, provide the requested property.

Final comprehensive check. Get these right and you're AP exam ready!