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Molecular Shapes and Bond Angles (VSEPR)

Learn and apply Molecular Shapes and Bond Angles (VSEPR) in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Molecular Shapes and Bond Angles: Orientation

VSEPR questions are usually mark-scheme checklists: count regions, state shape, quote angle, then explain any deviation using lone pairs. This lesson trains that exact workflow.

Treat this as an extension of Atomic Structure (A Level), then use the Chemical Bonding hub to compare models across the topic.

Definitions (Must Know)

A. VSEPR theory

VSEPR theory predicts molecular shape by assuming electron pairs around a central atom repel and arrange to minimise repulsion.

B. Electron pair region (electron domain)

An electron pair region is a region of electron density around the central atom (a single bond, double bond, triple bond, or lone pair).

C. Bond pair vs lone pair

  • Bond pair (BP): a shared pair of electrons in a covalent bond.
  • Lone pair (LP): a non-bonding pair of electrons on the central atom.

D. Dipole moment

A dipole moment is a measure of charge separation in a bond or molecule. A molecule has a net dipole moment if its bond dipoles do not cancel due to the shape.

Detailed Explanations

A. Quick shape table (must know)

Electron pair regionsShape (no lone pairs)ExampleAngle (ideal)
2linearCO₂180°
3trigonal planarBF₃120°
4tetrahedralCH₄109.5°
6octahedralSF₆90°

Common lone pair shapes:

  • 4 regions, 1 lone pair: trigonal pyramidal (NH₃) ~107°
  • 4 regions, 2 lone pairs: bent (H₂O) ~104.5°
Six syllabus molecular shapes: linear carbon dioxide at 180 degrees, trigonal planar boron trifluoride at 120 degrees, tetrahedral methane at 109.5 degrees, trigonal pyramidal ammonia at about 107 degrees, bent water at about 104.5 degrees, and octahedral sulfur hexafluoride at 90 degrees
The six specified examples anchor analogous predictions: count regions around the central atom, account for lone pairs, then state shape and angle.

B. VSEPR workflow (how to answer fast)

  1. Draw the Lewis structure.
  2. Count electron pair regions around the central atom (double bonds count as 1).
  3. Use the table to state electron-pair geometry and molecular shape.
  4. Quote the angle (ideal or adjusted) and give the cause (lone pair repulsion).

Mini examples (fast checks):

  • CO₂: 2 regions around C → linear (180°).
  • BF₃: 3 regions around B → trigonal planar (120°).
  • CH₄: 4 regions around C → tetrahedral (109.5°).
  • NH₃: 4 regions around N → tetrahedral electron geometry; trigonal pyramidal molecular shape (~107°).
  • H₂O: 4 regions around O → tetrahedral electron geometry; bent molecular shape (~104.5°).
  • SF₆: 6 regions around S → octahedral (90° between adjacent bonds).

C. Multiple bonds and polarity (common mark traps)

  • A double bond is one region because the electron density occupies one “direction” from the central atom.
  • Symmetry matters for polarity: even if bonds are polar, dipoles can cancel (e.g. linear CO₂, trigonal planar BF₃).

To predict an analogous species, match its number of bonding regions and central-atom lone pairs to one of these models. For example, NH₄ + has four bonding regions and no lone pair on N, so it is tetrahedral like CH₄.

Worked Examples

Modelled example 1

Predict Ammonia Shape and Angle

Core

Problem

Predict the shape and bond angle of NH₃.

Study the worked solution

  1. Count electron-pair regions

    Method

    Count three N–H bond pairs and one lone pair around nitrogen.

    Reason

    VSEPR uses all bonding and lone-pair regions around the central atom.

    Working

    Four regions: 3 BP + 1 LP.

  2. Separate geometry from molecular shape

    Method

    Assign tetrahedral electron-pair geometry but name only atom positions for the molecular shape.

    Reason

    One tetrahedral position is occupied by a lone pair.

    Working

    Molecular shape: trigonal pyramidal.

  3. Adjust the angle

    Method

    Reduce the ideal tetrahedral angle from 109.5° to about 107°.

    Reason

    Lone pair–bond pair repulsion is stronger than bond pair–bond pair repulsion.

    Working

    H–N–H angle ≈ 107°.

Guided practice 2

Predict Carbon Dioxide Shape

About 5 min

Problem

Predict the shape and bond angle of CO₂.

Try this before viewing the solution

Hints

Hint 1: count regions, not bond lines
Each C = O double bond counts as one electron-density region.
Hint 2: separate two regions maximally
Two regions point in opposite directions.
View solution step by step

  1. Count the regions

    Method

    Count the two C=O bonds as two regions around carbon.

    Reason

    A multiple bond is one region for VSEPR shape prediction.

    Working

    Two bonding regions; no central lone pairs.

  2. State shape and angle

    Method

    Place the two regions opposite each other.

    Reason

    This maximises their separation and minimises repulsion.

    Working

    Linear; O–C–O angle = 180°.

Common misconception 3

Correct the Water–Ammonia Angle Comparison

Find and correct the mistake

Learner claim

Asked why the bond angle in H₂O is smaller than in NH₃, a learner says, “Water has more lone pairs, so the bonds are pushed farther apart and its angle is larger.” Identify the error and give the correct explanation.

Judge the lone-pair effect

More lone pairs around four regions make the bond angle

View solution step by step

  1. Establish the common geometry

    Method

    State that both central atoms have four electron-pair regions.

    Reason

    Both therefore begin from tetrahedral electron-pair geometry.

    Working

    NH₃:3 BP + 1 LP; H₂O:2 BP + 2 LP.

  2. Compare repulsions

    Method

    Use the extra lone pair in water.

    Reason

    Lone pair–lone pair and lone pair–bond pair repulsions exceed bond pair–bond pair repulsion.

    Working

    The O–H bond pairs are compressed more strongly.

  3. Correct the angles

    Method

    State the smaller water angle.

    Reason

    More lone-pair repulsion pushes the bond pairs closer together, not farther apart.

    Working

    H₂O ≈ 104.5° < NH₃ ≈ 107°.

Examiner practice 4

Compare Methane and Ammonia Polarity

4 marks

Problem

Explain why CH₄ is non-polar but NH₃ is polar. [4 marks]

Try this before viewing the solution

View solution step by step

  1. State methane's shape

    1 mark

    Method

    Identify tetrahedral, symmetrical CH₄.

    Reason

    Carbon has four bonding regions and no lone pairs.

    Working

    CH₄: tetrahedral.

  2. Resolve methane's dipoles

    1 mark

    Method

    State that the C–H bond dipoles cancel.

    Reason

    The equal dipoles are arranged symmetrically.

    Working

    Net dipole = 0; CH₄ is non-polar.

  3. State ammonia's shape

    1 mark

    Method

    Identify trigonal-pyramidal NH₃.

    Reason

    The nitrogen lone pair removes tetrahedral symmetry from the atom positions.

    Working

    NH₃: trigonal pyramidal.

  4. Resolve ammonia's dipoles

    1 mark

    Method

    State that the N–H dipoles do not cancel.

    Reason

    The pyramidal geometry gives a non-zero vector sum.

    Working

    NH₃ has a net dipole and is polar.

Challenge 5

Transfer from Ammonia to Ammonium

Minimal support

Problem

Ammonia uses its nitrogen lone pair to bond to H⁺ and form NH₄ +. Predict the shape and H–N–H bond angle of NH₄ +, and explain why they differ from those of NH₃.

Try this before viewing the solution

Hints

Hint 1: recount after donation
In ammonium, nitrogen has four N–H bonding regions and no lone pair.
Hint 2: remove the lone-pair compression
Four equivalent bonding regions adopt the ideal tetrahedral arrangement.
View solution step by step

  1. Count ammonium's regions

    Method

    Count four N–H bonding regions and no central lone pair.

    Reason

    The ammonia lone pair has become the shared pair in the fourth N–H bond.

    Working

    NH₄ + :4 BP + 0 LP.

  2. Predict shape and angle

    Method

    Assign the ideal four-region arrangement.

    Reason

    Four equivalent bonding regions maximise separation tetrahedrally.

    Working

    Tetrahedral; H–N–H angle 109.5°.

  3. Compare with ammonia

    Method

    Contrast tetrahedral ammonium with trigonal-pyramidal ammonia at about 107°.

    Reason

    Ammonium has no lone pair to compress its bond angles.

    Working

    109.5° > 107°.

Mind Stretchers

Mind stretcher 1Extension

BF₃ has polar B–F bonds. Explain why BF₃ has no net dipole moment.

Show Answer

Mark scheme:

  • BF₃ has 3 electron pair regions around B → trigonal planar shape.
  • The molecule is symmetrical, so the three B–F bond dipoles are equal and separated by 120°.
  • The dipoles cancel vectorially, so there is no net dipole moment.