VSEPR Theory and Molecular Geometry

Use VSEPR theory to predict molecular shapes, bond angles, and understand the relationship between electron geometry and molecular geometry.

VSEPR Theory and Molecular Geometry

What is VSEPR Theory?

VSEPR = Valence Shell Electron Pair Repulsion Theory

Key Principle: Electron pairs (bonding and lone pairs) repel each other and arrange themselves to be as far apart as possible, minimizing repulsion.

Result: Predicts 3D molecular shapes

Developed by: Ronald Gillespie and Ronald Nyholm (1957)

Electron Domains

Electron domain: A region of electron density around a central atom

Types of electron domains:

  1. Single bond: 1 electron domain
  2. Double bond: 1 electron domain (counts as one region)
  3. Triple bond: 1 electron domain (counts as one region)
  4. Lone pair: 1 electron domain

Key point: Multiple bonds count as ONE electron domain

Example: CO₂

\ceO=C=O\ce{O=C=O}

  • 2 double bonds = 2 electron domains
  • No lone pairs
  • Total: 2 electron domains

Steric Number

Steric Number (SN): Total number of electron domains around central atom

SN=(bonding domains)+(lone pairs)\text{SN} = (\text{bonding domains}) + (\text{lone pairs})

Alternative:

SN=(number of atoms bonded)+(lone pairs on central atom)\text{SN} = (\text{number of atoms bonded}) + (\text{lone pairs on central atom})

Examples:

CH₄: 4 bonds + 0 lone pairs = SN 4

NH₃: 3 bonds + 1 lone pair = SN 4

H₂O: 2 bonds + 2 lone pairs = SN 4

All have SN = 4, but different molecular shapes!

Electron Geometry vs. Molecular Geometry

Electron Geometry

Definition: Arrangement of ALL electron domains (bonds + lone pairs)

Determined by: Steric number only

Molecular Geometry

Definition: Arrangement of ATOMS only (ignores lone pairs)

Determined by: Steric number AND number of lone pairs

Key distinction: Lone pairs affect geometry but aren't "seen" in molecular shape

Five Basic Electron Geometries

Based on steric number:

| SN | Electron Geometry | Bond Angles | Example | |----|-------------------|-------------|---------| | 2 | Linear | 180° | CO₂ | | 3 | Trigonal planar | 120° | BF₃ | | 4 | Tetrahedral | 109.5° | CH₄ | | 5 | Trigonal bipyramidal | 90°, 120° | PCl₅ | | 6 | Octahedral | 90° | SF₆ |

Molecular Geometries by Steric Number

SN = 2 (2 electron domains)

Electron geometry: Linear

| Bonding | Lone Pairs | Molecular Geometry | Bond Angle | Example | |---------|------------|-------------------|------------|---------| | 2 | 0 | Linear | 180° | CO₂, BeH₂ |

Linear: All atoms in straight line

\ceO=C=O\ce{O=C=O} (180°)

SN = 3 (3 electron domains)

Electron geometry: Trigonal planar

| Bonding | Lone Pairs | Molecular Geometry | Bond Angle | Example | |---------|------------|-------------------|------------|---------| | 3 | 0 | Trigonal planar | 120° | BF₃, SO₃ | | 2 | 1 | Bent | <120° (~118°) | SO₂, O₃ |

Trigonal planar: Flat, three atoms around center

\ceF−B−F\ce{F - B - F} with third F above (all in plane) \ce∣\ce{ | } \ceF\ce{ F }

Bent (from trigonal planar): V-shaped

\ceO=S=O\ce{O=S=O} (with lone pair on S, angle ~119°)

SN = 4 (4 electron domains)

Electron geometry: Tetrahedral

| Bonding | Lone Pairs | Molecular Geometry | Bond Angle | Example | |---------|------------|-------------------|------------|---------| | 4 | 0 | Tetrahedral | 109.5° | CH₄, CCl₄ | | 3 | 1 | Trigonal pyramidal | <109.5° (~107°) | NH₃, PCl₃ | | 2 | 2 | Bent | <109.5° (~104.5°) | H₂O, H₂S |

Tetrahedral: 3D pyramid shape

CH₄: Carbon at center, 4 H atoms at corners of tetrahedron

Trigonal pyramidal: Pyramid with triangular base

NH₃: N at top, 3 H atoms form triangular base (lone pair on top)

Bent (from tetrahedral): V-shaped

H₂O: O at vertex, 2 H atoms, 2 lone pairs (angle 104.5°)

SN = 5 (5 electron domains)

Electron geometry: Trigonal bipyramidal

Two types of positions:

  • Axial: Top and bottom (3 at 90°)
  • Equatorial: Middle three (at 120°)

Lone pairs prefer equatorial positions (more space, less repulsion)

| Bonding | Lone Pairs | Molecular Geometry | Bond Angle | Example | |---------|------------|-------------------|------------|---------| | 5 | 0 | Trigonal bipyramidal | 90°, 120° | PCl₅ | | 4 | 1 | Seesaw | <90°, <120° | SF₄ | | 3 | 2 | T-shaped | <90° | ClF₃ | | 2 | 3 | Linear | 180° | XeF₂ |

Trigonal bipyramidal: Two pyramids joined at base

Seesaw: Like teeter-totter

T-shaped: Like letter T

Linear (from SN 5): Straight line with 3 lone pairs

SN = 6 (6 electron domains)

Electron geometry: Octahedral

All positions are equivalent (all 90° apart)

| Bonding | Lone Pairs | Molecular Geometry | Bond Angle | Example | |---------|------------|-------------------|------------|---------| | 6 | 0 | Octahedral | 90° | SF₆ | | 5 | 1 | Square pyramidal | <90° | BrF₅ | | 4 | 2 | Square planar | 90° | XeF₄ |

Octahedral: 8-sided figure, 6 atoms at corners of octahedron

Square pyramidal: Square base with atom above

Square planar: Flat square (lone pairs above and below)

Bond Angle Deviations

General Rules

Repulsion strength: Lone pair-lone pair > Lone pair-bond > Bond-bond

Effect: Lone pairs take up more space than bonding pairs

Result: Bond angles decrease when lone pairs present

Examples of Angle Changes

CH₄ (no lone pairs): 109.5° (ideal tetrahedral)

NH₃ (1 lone pair): ~107° (slightly compressed)

H₂O (2 lone pairs): ~104.5° (more compressed)

Trend: More lone pairs → smaller bond angles

Why Lone Pairs Compress Angles

  1. Lone pairs are closer to nucleus (not shared)
  2. Lone pairs have more electron density
  3. Greater repulsion pushes bonding pairs closer together
  4. Bond angles become smaller

Predicting Molecular Geometry - Step by Step

Step 1: Draw Lewis structure

Step 2: Count electron domains on central atom

  • Bonds (single, double, triple each = 1 domain)
  • Lone pairs (each = 1 domain)

Step 3: Determine steric number (SN)

Step 4: Identify electron geometry from SN

Step 5: Count lone pairs

Step 6: Determine molecular geometry

  • Use table based on SN and lone pairs

Step 7: Predict bond angles

  • Start with ideal angle for electron geometry
  • Decrease if lone pairs present

Example Walkthrough: NH₃

Step 1: Lewis structure

\ceH−N−H\ce{H - N - H} with lone pair on N \ce∣\ce{ | } \ceH\ce{ H }

Step 2: Count electron domains

  • 3 N-H bonds = 3 domains
  • 1 lone pair = 1 domain
  • Total: 4 domains

Step 3: Steric number SN = 4

Step 4: Electron geometry SN 4 → Tetrahedral electron geometry

Step 5: Count lone pairs 1 lone pair

Step 6: Molecular geometry SN 4, 1 lone pair → Trigonal pyramidal

Step 7: Bond angles

  • Ideal tetrahedral: 109.5°
  • With 1 lone pair: ~107° (slightly less)

Polarity and Molecular Geometry

Molecular polarity depends on:

  1. Bond polarity (electronegativity difference)
  2. Molecular geometry (symmetry)

Polar vs. Nonpolar Molecules

Nonpolar molecules:

  • Symmetrical geometry
  • Bond dipoles cancel out
  • Examples: CO₂ (linear), CCl₄ (tetrahedral), BF₃ (trigonal planar)

Polar molecules:

  • Asymmetrical geometry
  • Bond dipoles do NOT cancel
  • Examples: H₂O (bent), NH₃ (trigonal pyramidal), HCl (linear but different atoms)

Key Patterns

Always polar if:

  • Lone pairs on central atom with polar bonds
  • Asymmetric arrangement

Always nonpolar if:

  • All bonds are identical AND symmetrical
  • No lone pairs AND all surrounding atoms identical

Examples:

CO₂: O=C=O (linear, symmetrical) → Nonpolar

H₂O: H-O-H (bent due to lone pairs) → Polar

CCl₄: Tetrahedral, all Cl identical → Nonpolar

CHCl₃: Tetrahedral, but H ≠ Cl → Polar

NH₃: Trigonal pyramidal with lone pair → Polar

BF₃: Trigonal planar, all F identical → Nonpolar

Multiple Central Atoms

For molecules with multiple "central" atoms, analyze each separately:

Example: Ethanol (C₂H₅OH)

  • Left C: SN 4, tetrahedral
  • Right C: SN 4, tetrahedral
  • O: SN 4, bent (2 lone pairs)

Common Mistakes to Avoid

  1. Counting multiple bonds as multiple domains: Double/triple bonds = 1 domain
  2. Confusing electron geometry with molecular geometry: Lone pairs count for electron geometry only
  3. Forgetting lone pairs: Always include in steric number
  4. Wrong polarity determination: Check BOTH bond polarity AND geometry
  5. Ignoring lone pair repulsion: Lone pairs decrease bond angles

Summary Table

| SN | Electron Geometry | 0 LP | 1 LP | 2 LP | 3 LP | |----|-------------------|------|------|------|------| | 2 | Linear | Linear (180°) | - | - | - | | 3 | Trigonal planar | Trig planar (120°) | Bent (<120°) | - | - | | 4 | Tetrahedral | Tetrahedral (109.5°) | Trig pyramidal (<109.5°) | Bent (<109.5°) | - | | 5 | Trig bipyramidal | Trig bipyramidal (90°,120°) | Seesaw | T-shaped | Linear (180°) | | 6 | Octahedral | Octahedral (90°) | Sq pyramidal | Sq planar (90°) | - |

LP = Lone Pairs

Applications

  1. Predicting molecular properties: Shape affects polarity, boiling point, reactivity
  2. Drug design: Molecular shape determines how drugs fit receptors
  3. Materials science: Geometry affects crystal structure and material properties
  4. Understanding reactivity: Shape determines which atoms can interact

📚 Practice Problems

1Problem 1medium

❓ Question:

For the molecule XeF₄ (xenon tetrafluoride): (a) Draw the Lewis structure. (b) Determine the electron geometry and molecular geometry. (c) Predict the F-Xe-F bond angles. (d) Is the molecule polar or nonpolar?

💡 Show Solution

Solution:

(a) Lewis structure:

  • Xe: 8 valence electrons (noble gas)
  • 4 F: 4 × 7 = 28 valence electrons
  • Total: 36 electrons

Xe in center with 4 F atoms bonded, plus 2 lone pairs on Xe

(b) Geometries:

  • Electron pairs around Xe: 4 bonding + 2 lone = 6 electron pairs
  • Electron geometry: Octahedral
  • The 2 lone pairs occupy opposite positions (180° apart to minimize repulsion)
  • Molecular geometry: Square planar

(c) Bond angles:

  • F-Xe-F (adjacent): 90°
  • F-Xe-F (opposite): 180°

(d) Polarity: Nonpolar - Despite having polar Xe-F bonds, the square planar geometry is perfectly symmetric. The bond dipoles cancel out (4 F atoms arranged symmetrically around Xe in a plane), resulting in no net dipole moment.

2Problem 2easy

❓ Question:

Determine the molecular geometry and bond angle for carbon dioxide (CO₂).

💡 Show Solution

Solution:

Given: CO₂ Find: Molecular geometry and bond angle

Step 1: Draw Lewis structure

Total valence electrons: 4 (C) + 2(6) (O) = 16

\ceO=C=O\ce{O=C=O}

Each O has 2 double bonds and 2 lone pairs C has 2 double bonds (no lone pairs)

Step 2: Identify central atom

Central atom: C (carbon)

Step 3: Count electron domains on C

  • 1 double bond to left O = 1 electron domain
  • 1 double bond to right O = 1 electron domain
  • 0 lone pairs on C

Total electron domains: 2

Remember: Multiple bonds count as ONE domain!

Step 4: Determine steric number

SN=2\text{SN} = 2

Step 5: Identify electron geometry

SN = 2 → Linear electron geometry

Step 6: Count lone pairs on central atom

C has 0 lone pairs

Step 7: Determine molecular geometry

SN = 2, 0 lone pairs → Linear molecular geometry

Step 8: Predict bond angle

Linear geometry → 180°

Answer:

  • Molecular geometry: Linear
  • Bond angle: 180°
  • Structure: O=C=O (all atoms in straight line)

Verification:

  • 2 electron domains ✓
  • 0 lone pairs ✓
  • Electron geometry = Molecular geometry (no lone pairs) ✓
  • Maximum separation of 2 domains = 180° ✓

Note: CO₂ is also nonpolar because the two C=O dipoles point in opposite directions and cancel out due to the linear, symmetrical geometry.

3Problem 3medium

❓ Question:

For the molecule XeF₄ (xenon tetrafluoride): (a) Draw the Lewis structure. (b) Determine the electron geometry and molecular geometry. (c) Predict the F-Xe-F bond angles. (d) Is the molecule polar or nonpolar?

💡 Show Solution

Solution:

(a) Lewis structure:

  • Xe: 8 valence electrons (noble gas)
  • 4 F: 4 × 7 = 28 valence electrons
  • Total: 36 electrons

Xe in center with 4 F atoms bonded, plus 2 lone pairs on Xe

(b) Geometries:

  • Electron pairs around Xe: 4 bonding + 2 lone = 6 electron pairs
  • Electron geometry: Octahedral
  • The 2 lone pairs occupy opposite positions (180° apart to minimize repulsion)
  • Molecular geometry: Square planar

(c) Bond angles:

  • F-Xe-F (adjacent): 90°
  • F-Xe-F (opposite): 180°

(d) Polarity: Nonpolar - Despite having polar Xe-F bonds, the square planar geometry is perfectly symmetric. The bond dipoles cancel out (4 F atoms arranged symmetrically around Xe in a plane), resulting in no net dipole moment.

4Problem 4hard

❓ Question:

Compare and explain the bond angles in the following molecules: (a) CH₄ (109.5°), (b) NH₃ (107°), (c) H₂O (104.5°). All have 4 electron pairs around the central atom - why are the angles different?

💡 Show Solution

Solution:

All three molecules have tetrahedral electron geometry (4 electron pairs), but different molecular geometries:

(a) CH₄ - Tetrahedral, 109.5°

  • 4 bonding pairs, 0 lone pairs
  • All electron pairs are equivalent
  • Perfect tetrahedral angle: 109.5°

(b) NH₃ - Trigonal pyramidal, 107°

  • 3 bonding pairs, 1 lone pair
  • Lone pairs occupy more space than bonding pairs (electrons closer to nucleus)
  • Lone pair-bonding pair repulsion > bonding pair-bonding pair repulsion
  • Lone pair pushes bonding pairs together: 107° (smaller than 109.5°)

(c) H₂O - Bent, 104.5°

  • 2 bonding pairs, 2 lone pairs
  • Two lone pairs create even more repulsion
  • Both lone pairs push bonding pairs together
  • Even smaller angle: 104.5°

VSEPR Repulsion Order: Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair

Summary: As lone pairs replace bonding pairs, the bond angle decreases because lone pairs exert greater repulsive forces.

Angle trend: CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°)

5Problem 5hard

❓ Question:

Compare and explain the bond angles in the following molecules: (a) CH₄ (109.5°), (b) NH₃ (107°), (c) H₂O (104.5°). All have 4 electron pairs around the central atom - why are the angles different?

💡 Show Solution

Solution:

All three molecules have tetrahedral electron geometry (4 electron pairs), but different molecular geometries:

(a) CH₄ - Tetrahedral, 109.5°

  • 4 bonding pairs, 0 lone pairs
  • All electron pairs are equivalent
  • Perfect tetrahedral angle: 109.5°

(b) NH₃ - Trigonal pyramidal, 107°

  • 3 bonding pairs, 1 lone pair
  • Lone pairs occupy more space than bonding pairs (electrons closer to nucleus)
  • Lone pair-bonding pair repulsion > bonding pair-bonding pair repulsion
  • Lone pair pushes bonding pairs together: 107° (smaller than 109.5°)

(c) H₂O - Bent, 104.5°

  • 2 bonding pairs, 2 lone pairs
  • Two lone pairs create even more repulsion
  • Both lone pairs push bonding pairs together
  • Even smaller angle: 104.5°

VSEPR Repulsion Order: Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair

Summary: As lone pairs replace bonding pairs, the bond angle decreases because lone pairs exert greater repulsive forces.

Angle trend: CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°)

6Problem 6medium

❓ Question:

Water (H₂O) and methane (CH₄) both have a steric number of 4, yet they have different molecular geometries and bond angles. Explain why.

💡 Show Solution

Solution:

Given: H₂O and CH₄ both have SN = 4 Find: Explain different geometries and bond angles

Step 1: Draw Lewis structures

H₂O (Water):

\ceH−O−H\ce{H - O - H}

O has 2 bonding pairs and 2 lone pairs

CH₄ (Methane):

\ceH−C−H\ce{H - C - H} (with 2 more H atoms)

C has 4 bonding pairs and 0 lone pairs

Step 2: Calculate steric numbers

H₂O:

  • 2 O-H bonds = 2 domains
  • 2 lone pairs = 2 domains
  • SN = 4

CH₄:

  • 4 C-H bonds = 4 domains
  • 0 lone pairs = 0 domains
  • SN = 4

Both have SN = 4 ✓

Step 3: Determine electron geometry

Both have SN = 4 → Tetrahedral electron geometry

Step 4: Determine molecular geometry

H₂O:

  • SN = 4, 2 lone pairs
  • Molecular geometry: Bent
  • Only "see" the 2 H atoms (ignore lone pairs in molecular shape)

CH₄:

  • SN = 4, 0 lone pairs
  • Molecular geometry: Tetrahedral
  • "See" all 4 H atoms

Step 5: Explain bond angles

CH₄:

  • Ideal tetrahedral: 109.5°
  • No lone pairs → no compression
  • H-C-H angle = 109.5°

H₂O:

  • Ideal tetrahedral would be 109.5°
  • BUT: 2 lone pairs present
  • Lone pairs repel more than bonding pairs
  • Lone pairs push H atoms closer together
  • H-O-H angle = 104.5° (compressed)

Step 6: Visualize repulsions

Repulsion strength: LP-LP>LP-BP>BP-BP\text{LP-LP} > \text{LP-BP} > \text{BP-BP}

H₂O has:

  • 1 LP-LP repulsion (strongest)
  • 4 LP-BP repulsions
  • 1 BP-BP repulsion (weakest)

The strong LP-LP and LP-BP repulsions compress the bond angle.

CH₄ has:

  • 0 LP-LP repulsions
  • 0 LP-BP repulsions
  • 6 BP-BP repulsions (all equal)

Equal repulsions maintain ideal 109.5° angle.

Answer:

Why different despite same SN?

  1. Electron geometry is the same (tetrahedral for both)

  2. Molecular geometry differs due to lone pairs:

    • H₂O: Bent (2 LP compress bonding pairs)
    • CH₄: Tetrahedral (0 LP, all positions equivalent)

  3. Bond angles differ due to lone pair repulsion:

    • H₂O: 104.5° (compressed by 2 lone pairs)
    • CH₄: 109.5° (ideal tetrahedral, no lone pairs)

Key principle: Molecular geometry depends on BOTH steric number AND number of lone pairs. Lone pairs are "invisible" in the molecular shape but strongly affect bond angles through greater repulsion.

Summary:

| Property | H₂O | CH₄ | |----------|-----|-----| | Steric Number | 4 | 4 | | Electron Geometry | Tetrahedral | Tetrahedral | | Lone Pairs | 2 | 0 | | Molecular Geometry | Bent | Tetrahedral | | Bond Angle | 104.5° | 109.5° |

7Problem 7hard

❓ Question:

Xenon tetrafluoride (XeF₄) is a stable compound. Determine its molecular geometry, predict the bond angles, and explain whether the molecule is polar or nonpolar.

💡 Show Solution

Solution:

Given: XeF₄ (xenon tetrafluoride) Find: Molecular geometry, bond angles, polarity

Step 1: Draw Lewis structure

Count valence electrons:

  • Xe: 8 valence electrons (Group 18, noble gas)
  • F: 7 valence electrons (×4)
  • Total: 8 + 28 = 36 electrons

Skeletal structure: Xe is central

\ceF−Xe−F\ce{F - Xe - F} (with 2 more F atoms)

Connect with single bonds: 4 Xe-F bonds = 8 electrons used Remaining: 36 - 8 = 28 electrons

Complete octets of F atoms: Each F needs 6 more electrons (3 lone pairs) 4 F × 6 = 24 electrons Remaining: 28 - 24 = 4 electrons

Place remaining electrons on Xe: 4 electrons = 2 lone pairs on Xe

Lewis structure:

\ceF\ce{ F } \ce∣\ce{ | } \ceF−Xe−F\ce{F - Xe - F} \ce∣\ce{ | } \ceF\ce{ F }

Xe has 4 bonds and 2 lone pairs

Step 2: Count electron domains on Xe

  • 4 Xe-F single bonds = 4 domains
  • 2 lone pairs = 2 domains
  • Total: 6 electron domains

Step 3: Determine steric number

SN=6\text{SN} = 6

Step 4: Identify electron geometry

SN = 6 → Octahedral electron geometry

Step 5: Count lone pairs

Xe has 2 lone pairs

Step 6: Determine molecular geometry

SN = 6, 2 lone pairs → Square planar

Why square planar?

  • In octahedral geometry, all 6 positions are equivalent
  • Lone pairs occupy positions opposite each other (minimizes repulsion)
  • Remaining 4 F atoms form a square in a plane
  • Xe at center, lone pairs above and below

Step 7: Predict bond angles

F-Xe-F angles:

  • Adjacent F atoms: 90° (corners of square)
  • Opposite F atoms: 180° (across square)

Ideal octahedral angles: 90°

With 2 lone pairs:

  • LP-LP repulsion is strongest
  • LP-BP repulsion compresses slightly
  • But symmetry maintains approximately 90° and 180°

Step 8: Determine polarity

Check bond polarity:

  • Xe-F bonds are polar (ΔEN ≈ 2.0)
  • Each Xe-F bond has dipole toward F (F is more electronegative)

Check molecular symmetry:

\ceF\ce{ F } \ce↑\ce{ ↑ } \ceF←Xe→F\ce{F ← Xe → F} \ce↓\ce{ ↓ } \ceF\ce{ F }

Dipole analysis:

  • Top F dipole ↑ cancels with bottom F dipole ↓
  • Left F dipole ← cancels with right F dipole →
  • Perfect symmetry in square planar arrangement
  • Net dipole = 0

Answer:

Molecular geometry: Square planar

Bond angles:

  • F-Xe-F (adjacent): 90°
  • F-Xe-F (opposite): 180°

Polarity: Nonpolar

Explanation of polarity: Even though Xe-F bonds are polar, the square planar geometry is perfectly symmetrical. The four bond dipoles point toward the corners of a square and cancel out vectorially. The two lone pairs are positioned opposite each other (above and below the plane), maintaining symmetry.

Additional notes:

  1. Expanded octet: Xe (Period 5) can accommodate 12 electrons (6 electron domains) using d orbitals

  2. Lone pair placement: In octahedral geometry, lone pairs prefer to be 180° apart (trans position) to minimize LP-LP repulsion

  3. Why not other geometries?

    • If lone pairs were adjacent (90°), LP-LP repulsion would be much stronger
    • Square planar with trans lone pairs is most stable

  4. Other examples of square planar: ICl₄⁻, BrF₄⁻

Summary:

| Property | Value | |----------|-------| | Steric Number | 6 | | Electron Geometry | Octahedral | | Lone Pairs | 2 (opposite positions) | | Molecular Geometry | Square planar | | Bond Angles | 90°, 180° | | Polarity | Nonpolar (symmetrical) |

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