ChemistryISC Class 11 · Chemical Bonding and Molecular Structure

VSEPR Theory, Molecular Shapes, Hybridisation and Polarity

Updated 17 October 2026 · Fact-checked

VSEPR theory says that electron pairs around a central atom arrange themselves to minimise repulsion, in the order lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. The steric number (bond pairs + lone pairs) gives the hybridisation and the arrangement; removing lone pairs from the arrangement gives the shape.

All three have four electron pairs (sp³); each lone pair repels bond pairs more strongly, so the H–X–H angle shrinks from 109.5° to 107° to 104.5°.

Illustration drawn for QuestionBuddy.

Understand VSEPR Theory, Shapes and Hybridisation

Method. (1) Count the valence electrons of the central atom (adjust for charge). (2) Number of bond pairs = number of atoms attached. (3) Lone pairs = (valence electrons − electrons used in bonding)/2. (4) Steric number SN = bond pairs + lone pairs fixes the geometry of electron pairs; the shape is described by the atoms only.

Geometries. SN 2: linear, sp (BeCl₂, CO₂). SN 3: trigonal planar, sp² (BF₃); with one lone pair the shape is bent (SO₂). SN 4: tetrahedral, sp³ (CH₄); one lone pair gives pyramidal (NH₃), two give bent (H₂O). SN 5: trigonal bipyramidal, sp³d (PCl₅); with one lone pair see-saw (SF₄), two lone pairs T-shaped (ClF₃), three lone pairs linear (XeF₂, ICl₂⁻). SN 6: octahedral, sp³d² (SF₆); one lone pair square pyramidal (BrF₅), two lone pairs square planar (XeF₄).

Bond angles. Lone pairs repel more than bond pairs and squeeze the bond angle: CH₄ 109.5°, NH₃ about 107°, H₂O about 104.5°. In hydrides of a group, the angle falls down the group (NH₃ > PH₃; H₂O > H₂S) because the less electronegative central atom holds the bond pairs farther away, so bond-pair repulsion is weaker and the lone pair occupies an orbital with more s-character.

Hybridisation. Mixing of atomic orbitals of similar energy to give equivalent hybrid orbitals: sp (180°), sp² (120°), sp³ (109.5°). The number of hybrid orbitals equals the number of orbitals mixed. Multiple bonds count as one region: σ bonds and lone pairs determine the hybridisation, with the π bonds formed by unhybridised p orbitals.

Dipole moment. μ = q × d (unit debye, D). Polar molecules have a non-zero vector sum of bond dipoles. Symmetrical molecules (CO₂, BF₃, CH₄, CCl₄, XeF₄) are non-polar even if the bonds are polar. For H₂O with bond dipole μ₀ and angle θ, μ = 2μ₀ cos(θ/2). Percentage ionic character = (observed μ/calculated μ for a 100% ionic bond) × 100.

VSEPR Theory, Shapes and Hybridisation: formulas and key facts

Lone pairs: $l o n e p ai r s = (v a l e n cee l ec t r o n s - e l ec t r o n s u se d inb o n d s) /2$; Adjust the electron count for any charge.
Steric number: $S N = b o n d p ai r s + l o n e p ai r s$; 2: sp | 3: sp² | 4: sp³ | 5: sp³d | 6: sp³d².
Dipole moment: $μ = q \times d ∣ μ (A B_{2} b e n t) = 2 μ_{0} cos (θ /2)$; 1 D = 3.34 × 10⁻³⁰ C m.
Percentage ionic character: $(μ o b ser v e d / μ i o ni c) \times 100$; μ ionic = e × bond length.

VSEPR Theory, Shapes and Hybridisation in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Shape of the water molecule

Suppose each O–H bond in water has a bond dipole of 1.5 D and the H–O–H angle is 104.5°.

Find the shape, hybridisation and the molecular dipole moment.

Try it first, then show the solutionSolution

O has 6 valence electrons, 2 bond pairs and 2 lone pairs, so SN = 4: sp³, tetrahedral arrangement, bent shape.
μ = 2 × 1.5 × cos(52.25°) = 3.0 × 0.613.
= 1.84 D, so the molecule is polar.

Answer: Bent; sp³; μ ≈ 1.84 D.

Memory hook: Two lone pairs push the bond angle below 109.5°.

Tricks to remember VSEPR Theory, Shapes and Hybridisation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Count pairs, then drop the lone pairs
Geometry follows the total pairs; shape follows the atoms only.
Use it when: Shape questions for SF₄, XeF₄ and ClF₃.
Lone pairs squeeze
Each lone pair reduces the bond angle by about 2 to 2.5° from the ideal.
Use it when: Comparing CH₄, NH₃ and H₂O.

VSEPR Theory, Shapes and Hybridisation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Shape of ammonia · 1 mark · MCQBoard level

The shape of the NH₃ molecule is:
(a) planar triangular
(b) tetrahedral
(c) pyramidal
(d) linear

Try it first, then show the solutionSolution

N has three bond pairs and one lone pair (SN = 4), so the shape is pyramidal.

Answer: (c) pyramidal

Example 2: Shapes and hybridisation · 3 marks · Short answerBoard level

Predict the shape and hybridisation of BF₃, NH₃ and H₂O, giving the bond pairs and lone pairs.

Try it first, then show the solutionSolution

BF₃: 3 bond pairs, 0 lone pairs, SN 3: trigonal planar, sp². (1 mark)
NH₃: 3 bond pairs, 1 lone pair, SN 4: pyramidal, sp³. (1 mark)
H₂O: 2 bond pairs, 2 lone pairs, SN 4: bent (V-shaped), sp³. (1 mark)

Answer: BF₃ trigonal planar (sp²); NH₃ pyramidal (sp³); H₂O bent (sp³).

Example 3: Shapes from the electron count · 4 marks · Short answerHarder board level

Find the number of lone pairs, the hybridisation and the shape of SF₄, XeF₄ and ICl₂⁻.

Try it first, then show the solutionSolution

SF₄: S has 6 electrons, 4 bonds, so (6 − 4)/2 = 1 lone pair; SN = 5, sp³d, see-saw. (1.5 marks)
XeF₄: Xe has 8 electrons, 4 bonds, so (8 − 4)/2 = 2 lone pairs; SN = 6, sp³d², square planar. (1.5 marks)
ICl₂⁻: I has 7 + 1 = 8 electrons, 2 bonds, so (8 − 2)/2 = 3 lone pairs; SN = 5, sp³d, linear. (1 mark)

Answer: SF₄: 1 lp, sp³d, see-saw. XeF₄: 2 lp, sp³d², square planar. ICl₂⁻: 3 lp, sp³d, linear.

Example 4: Dipole moment and ionic character · 4 marks · Short answerHarder board level

The H–Cl molecule has a dipole moment of 1.03 D and a bond length of 1.27 Å. Calculate the percentage ionic character. Take the electronic charge as 4.8 × 10⁻¹⁰ esu so that 1 D is 10⁻¹⁸ esu cm. Also say why CCl₄ is non-polar although each C–Cl bond is polar.

Try it first, then show the solutionSolution

μ for 100% ionic = e × d = 4.8 × 10⁻¹⁰ × 1.27 × 10⁻⁸ = 6.10 × 10⁻¹⁸ esu cm = 6.10 D. (1.5 marks)
Ionic character = 1.03/6.10 × 100 = 16.9%. (1.5 marks)
CCl₄ is tetrahedral, so the four bond dipoles cancel and the net dipole is zero. (1 mark)

Answer: About 16.9% ionic; CCl₄ is non-polar because the bond dipoles cancel by symmetry.

Example 5: Comparing bond angles · 5 marks · Long answerHOTS / competency

Explain why the bond angles decrease in the order CH₄ (109.5°) > NH₃ (about 107°) > H₂O (about 104.5°), and why the angle in PH₃ (about 93.5°) is much smaller than in NH₃.

Try it first, then show the solutionSolution

All three central atoms are sp³ hybridised, with a tetrahedral arrangement of electron pairs. (1 mark)
CH₄ has four bond pairs only. NH₃ has one lone pair, and H₂O two. (1 mark)
Lone pair-bond pair repulsion exceeds bond pair-bond pair repulsion, so each lone pair squeezes the bond angle further. (1 mark)
N is more electronegative and smaller than P, so the bond pairs lie closer to N, increasing bond-pair repulsion and keeping the angle larger. (1 mark)
In PH₃ the bond pairs are farther from P, so repulsion is weaker; P uses nearly pure p orbitals for bonding and keeps the lone pair in an s-rich orbital, giving an angle close to 90°. (1 mark)

Answer: The order follows the number of lone pairs; PH₃ has a smaller angle because P is less electronegative and larger than N.

Where students lose marks in VSEPR Theory, Shapes and Hybridisation

The slips examiners see most often, and what to write instead.

Mistake: Giving the geometry as the shape
Calling H₂O tetrahedral.
Do this: Give both the arrangement of electron pairs and the molecular shape: tetrahedral arrangement, bent shape.
Usually costs: 1 to 2 marks.
Mistake: Counting π bonds as hybrid regions
Counting each bond in a double bond separately.
Do this: Only σ bonds and lone pairs set the hybridisation.
Usually costs: 1 to 2 marks.
Mistake: Forgetting lone pair repulsion order
Writing bond pair-bond pair as the strongest.
Do this: lp-lp > lp-bp > bp-bp.
Usually costs: 1 mark.

Exam tips

Show the lone pair count calculation.
Mention the repulsion order when explaining bond angles.
Draw the shape with lone pairs for SF₄, ClF₃ and XeF₄.
For polarity, comment on the symmetry as well as the bond polarity.

Practice questions on VSEPR Theory, Shapes and Hybridisation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The hybridisation of carbon in ethene (C₂H₄) is:
(a) sp
(b) sp²
(c) sp³
(d) sp³d
Hint
Each carbon forms three σ bonds.
Answer
(b) sp²
Q2 · 2 marks · Short answerBoard level
Give the shapes of CO₂ and SO₂ and the reason for the difference.
Hint
Count lone pairs on the central atom.
Answer
CO₂ is linear (sp, no lone pairs on C). SO₂ is bent (sp², one lone pair on S).
Q3 · 3 marks · Short answerHarder board level
Count the σ and π bonds in ethene, ethyne and benzene.
Hint
A double bond is 1σ + 1π; a triple bond is 1σ + 2π.
Answer
Ethene: 5σ + 1π. Ethyne: 3σ + 2π. Benzene: 12σ + 3π (6 C–C σ + 6 C–H σ; three delocalised π bonds in the usual Kekulé count).
Q4 · 3 marks · Short answerHarder board level
Predict the hybridisation and shape of PCl₅, SF₆ and BrF₅.
Hint
Count bond pairs and lone pairs.
Answer
PCl₅: SN 5, sp³d, trigonal bipyramidal. SF₆: SN 6, sp³d², octahedral. BrF₅: 5 bonds + 1 lone pair, SN 6, sp³d², square pyramidal.
Q5 · 3 marks · Short answerHarder board level
Explain why NF₃ has a much smaller dipole moment than NH₃ even though both are pyramidal.
Hint
Compare the direction of the bond dipoles with the lone pair dipole.
Answer
In NH₃ the N–H bond dipoles point towards N, in the same direction as the lone-pair dipole, so they add. In NF₃ the N–F dipoles point towards F, opposing the lone-pair dipole, so the resultant is small.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a chemist studies XeF₂, XeF₄ and XeO₃ (Xe has 8 valence electrons).
(i) Give the number of lone pairs on Xe in each. (1 mark)
(ii) Give the hybridisation of Xe in each. (1 mark)
(iii) Give the shape of each. (1 mark)
(iv) Which of them are non-polar? (1 mark)
Hint
Count σ bonds; use the lone pairs formula.
Answer
(i) XeF₂: 3; XeF₄: 2; XeO₃: 1. (ii) sp³d, sp³d², sp³. (iii) Linear, square planar, pyramidal. (iv) XeF₂ and XeF₄ are symmetrical and non-polar; XeO₃ is polar.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Order of repulsion between electron pairs?Show answer
lp-lp > lp-bp > bp-bp.
2.Shape of SF₄?Show answer
See-saw.
3.Shape of XeF₄?Show answer
Square planar.
4.Hybridisation of S in SF₆?Show answer
sp³d².

VSEPR Theory, Shapes and Hybridisation: frequently asked questions

Why does NH₃ have a bond angle less than 109.5°?

The lone pair on N repels the bond pairs more strongly than they repel each other, pushing the N–H bonds closer together.

Can a molecule with polar bonds be non-polar?

Yes, if the bond dipoles cancel by symmetry, as in CO₂, BF₃, CH₄ and CCl₄.

Does hybridisation really occur?

Hybridisation is a model within valence bond theory that explains observed shapes and equal bond strengths; it is not a physical step the atom takes.