ChemistryCBSE Class 11 · Chemical Bonding and Molecular Structure

VSEPR Theory, Shapes and Hybridisation Class 11

Updated 17 October 2026 · Fact-checked

VSEPR theory predicts the shape from the number of electron-pair regions (bond pairs, lone pairs) around the central atom, which arrange themselves as far apart as possible: 2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral. Lone pairs repel more than bond pairs, reducing bond angles. The hybridisation (sp, sp², sp³, sp³d, sp³d²) matches the number of regions, and each double or triple bond counts as one region.

All three have four electron regions (sp³), but each added lone pair repels the bonds more and shrinks the bond angle.

Illustration drawn for QuestionBuddy.

Understand VSEPR Theory, Shapes and Hybridisation

Counting regions. For the central atom A in AB $_{x}$ : regions $= x$ (bonded atoms) $+$ lone pairs. Lone pairs $= \frac{( valence electrons of A \pm charge ) - ( electrons used in bonding )}{2}$ ; a multiple bond counts as one region. The arrangement of the regions gives the electron geometry; the positions of the atoms give the molecular shape.

2 regions: linear, sp
3 regions: trigonal planar, sp²
4 regions: tetrahedral, sp³
5 regions: trigonal bipyramidal, sp³d
6 regions: octahedral, sp³d²

Effect of lone pairs. Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. So $CH_{4}$ (109.5°), $NH_{3}$ (about 107°, 1 lone pair, pyramidal), $H_{2} O$ (about 104.5°, 2 lone pairs, bent). In trigonal bipyramidal geometry, lone pairs occupy equatorial positions (fewer 90° repulsions): $SF_{4}$ is see-saw, $ClF_{3}$ T-shaped, $XeF_{2}$ and $I_{3}^{-}$ linear. In octahedral geometry, two lone pairs sit opposite each other: $XeF_{4}$ is square planar, $BrF_{5}$ square pyramidal.

Central atom electronegativity. When the central atom is less electronegative (e.g. P versus N), bond pairs are farther from the centre, so lone-pair repulsion dominates and the angle is smaller ( $NH_{3} > PH_{3}$ ). When the terminal atoms are very electronegative (F), the bond pairs are pulled away, so the angle is smaller ( $NH_{3} > NF_{3}$ ).

Hybridisation and bonds. Mixing of atomic orbitals gives equivalent hybrids. Percentage s-character: sp 50%, sp² 33%, sp³ 25%; higher s-character gives shorter, stronger bonds and larger bond angles (180°, 120°, 109.5°). A single bond is one $σ$ bond; a double bond is one $σ$ + one $π$ ; a triple bond is one $σ$ + two $π$ . $π$ bonds use unhybridised p orbitals.

Dipole moment. $μ = q \times d$ (unit debye; 1 D $= 3.336 \times 1 0^{- 30}$ C m); a polyatomic molecule has the vector sum of its bond moments, which cancels in symmetric shapes (CO₂, BF₃, CCl₄, SF₆).

VSEPR Theory, Shapes and Hybridisation: formulas and key facts

Lone pairs: $l.p. = \frac{V \pm charge - electrons used in bonds}{2}$; V = valence electrons of the central atom.
Regions: $regions = bonded atoms + lone pairs$; Multiple bond = one region.
Dipole moment: $μ = q \times d$; 1 D = 3.336 × 10⁻³⁰ C m.
Bonds: $σ : 1 per bond; π : 1 per extra bond$; Double = σ + π.
Percent ionic character: $\frac{μ _{o b ser v e d}}{μ _{i o ni c}} \times 100$; Ionic: q = e.
Resultant bond moment: $μ = 2 μ_{b} cos \frac{θ}{2}$; Two equal bond moments at angle $θ$ .

How to solve VSEPR Theory, Shapes and Hybridisation questions

Count the regions around the central atom, assign geometry and hybridisation, then place lone pairs where they reduce repulsion most.

1Write the Lewis structure or count valence electrons of the central atom (adjust for charge).
2Count bonded atoms and lone pairs: regions = both.
3Match the regions to a geometry and hybridisation.
4Place lone pairs (equatorial in trigonal bipyramid; opposite in octahedral).
5Name the shape from the positions of the atoms only.
6Decide the polarity from the symmetry and the bond moments.

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.

Water and ice

Suppose a student asks why water is bent and not linear.

Use VSEPR to explain and state the hybridisation of oxygen.

Try it first, then show the solutionSolution

O has 6 valence electrons: 2 used in two O–H bonds, 2 lone pairs remain.
4 regions mean tetrahedral arrangement, sp³; the shape is bent (about 104.5°).

Answer: Bent; sp³.

Memory hook: Lone pairs bend the molecule.

Insulating gas in switchgear

Suppose a high-voltage switch is filled with SF₆, a very inert gas.

Predict its shape and polarity.

Try it first, then show the solutionSolution

S has 6 valence electrons, all used in 6 S–F bonds: 6 regions, no lone pair.
Octahedral, sp³d², symmetric so μ = 0.

Answer: Octahedral, sp³d², non-polar.

Memory hook: Symmetry cancels the bond moments.

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.

Regions to shape
2-3-4-5-6: linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral.
Use it when: Geometry.
Equatorial lone pairs
In trigonal bipyramid, lone pairs go equatorial.
Use it when: SF₄, ClF₃, XeF₂.
Double bond is one region
π bonds do not add regions.
Use it when: Counting.
s-character
sp 50%, sp² 33%, sp³ 25%.
Use it when: Bond lengths and angles.

VSEPR Theory, Shapes and Hybridisation: solved examples

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

Example 1: Linear versus bent · 2 marks · Very short answerBoard level

Predict the shape, hybridisation and bond angle of NO₂⁺ and NO₂⁻.

Try it first, then show the solutionSolution

NO₂⁺: N has 5 − 1 = 4 valence electrons, 2 double bonds, no lone pair: 2 regions, sp, linear, 180°. (1)
NO₂⁻: 5 + 1 = 6 electrons, 2 bonds + 1 lone pair: 3 regions, sp², bent (about 115° to 120°, less than 120°). (1)

Answer: NO₂⁺ linear, sp; NO₂⁻ bent, sp².

Example 2: Three oxo species · 3 marks · Short answerHarder board level

Find the number of bond pairs and lone pairs on the central atom, the shape and the hybridisation of (a) ClO₃⁻, (b) XeO₃ and (c) SF₆.

Try it first, then show the solutionSolution

(a) Cl (7 + 1) has 3 bonded O and 1 lone pair: 4 regions, sp³, trigonal pyramidal. (1)
(b) Xe (8 e) has 3 bonded O (double bonds) and 1 lone pair: 4 regions, sp³, trigonal pyramidal. (1)
(c) S has 6 bond pairs and no lone pair: octahedral, sp³d². (1)

Answer: (a) pyramidal, sp³; (b) pyramidal, sp³; (c) octahedral, sp³d².

Example 3: Allene · 3 marks · Short answerHarder board level

The molecule allene is CH₂=C=CH₂. Give the hybridisation of each carbon, count the σ and π bonds, and state whether the molecule is planar.

Try it first, then show the solutionSolution

The two end carbons have 3 regions each (sp²); the central carbon has 2 regions (sp). (1)
σ bonds: 4 C–H + 2 C–C = 6; π bonds: 2. (1)
The central sp carbon forms two perpendicular π bonds, so the two CH₂ planes are at 90°: the molecule is not planar. (1)

Answer: sp², sp, sp²; 6 σ and 2 π; not planar.

Example 4: Bond angles and dipole moments · 4 marks · Long answerHarder board level

(a) Give the shapes of NH₃, NF₃ and PH₃. (b) Arrange them in decreasing order of bond angle with reasons. (c) NH₃ has a dipole moment of 1.47 D and NF₃ only 0.24 D; explain.

Try it first, then show the solutionSolution

(a) All three are trigonal pyramidal (3 bond pairs and 1 lone pair, sp³). (1)
(b) NH₃ (107°) > NF₃ (102°) > PH₃ (93.5°). In NF₃ the electronegative F pulls the bond pairs away from N, reducing repulsion between them; in PH₃ the bond pairs are farther from the larger P, and the lone pair occupies more room. (1)
(c) In NH₃ the N–H bond moments point towards N, adding to the lone-pair moment. (1)
In NF₃ the N–F moments point towards F and oppose the lone-pair moment, so the net is small. (1)

Answer: (b) NH₃ > NF₃ > PH₃; (c) bond moments reinforce the lone pair in NH₃ and oppose it in NF₃.

Example 5: Hybridisation, bond length and polarity · 4 marks · Long answerHOTS / competency

(a) Give the s-character, the bond angle and the C–H bond length order (109, 108, 106 pm) of ethane, ethene and ethyne. (b) HF has μ = 1.91 D and a bond length of 92 pm: find the percentage ionic character (1 D = 3.336 × 10⁻³⁰ C m, e = 1.602 × 10⁻¹⁹ C). (c) Water has an O–H bond moment of 1.5 D and bond angle 104.5°: find the resultant moment.

Try it first, then show the solutionSolution

(a) sp³ 25% (109.5°, 109 pm), sp² 33% (120°, 108 pm), sp 50% (180°, 106 pm): more s-character, shorter C–H bond. (1)
(b) q = μ/d = 1.91 × 3.336 × 10⁻³⁰/92 × 10⁻¹² = 6.93 × 10⁻²⁰ C. (1)
Ionic character = 6.93 × 10⁻²⁰/1.602 × 10⁻¹⁹ = 43%. (1)
(c) μ = 2 × 1.5 × cos(52.25°) = 3.0 × 0.612 = 1.84 D. (1)

Answer: (a) 25%/33%/50%; (b) 43%; (c) 1.84 D.

Where students lose marks in VSEPR Theory, Shapes and Hybridisation

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

Mistake: Counting a double bond as two regions.
Treating π bonds as extra regions.
Do this: A multiple bond is one region.
Usually costs: The shape.
Mistake: Naming the shape by the electron geometry.
Including lone pairs in the name.
Do this: Name the shape from the atoms only (e.g. NH₃ is pyramidal).
Usually costs: 1 mark.
Mistake: Placing lone pairs axially in trigonal bipyramid.
Missing the 90° rule.
Do this: Lone pairs go equatorial.
Usually costs: The shape mark.
Mistake: Saying polar bonds always give a polar molecule.
Ignoring symmetry.
Do this: Check whether the bond moments cancel.
Usually costs: Reasoning mark.

Exam tips

Show the electron count for the central atom in the first line.
State the electron geometry and then the molecular shape.
Give the hybridisation with the number of regions.
Quote bond angles for lone-pair comparisons.
Justify dipole moments with a vector argument.

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 shape of the hydronium ion, H₃O⁺, is
(a) trigonal planar (b) trigonal pyramidal (c) tetrahedral (d) bent
Hint
3 bond pairs + 1 lone pair.
Answer
(b) trigonal pyramidal.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In SF₄, the lone pair of electrons occupies an equatorial position of the trigonal bipyramid.
Reason (R): An equatorial lone pair has fewer 90° repulsions with the bond pairs than an axial one.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Count 90° interactions.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Give the hybridisation of carbon in CO₂, CH₃⁺ and CH₃⁻.
Hint
Count regions.
Answer
CO₂: 2 regions, sp; CH₃⁺: 3 regions, sp²; CH₃⁻: 3 bond pairs + 1 lone pair = 4 regions, sp³.
Q4 · 3 marks · Short answerHarder board level
Predict the shape and hybridisation of (a) ClF₂⁺, (b) SOCl₂ and (c) POCl₃.
Hint
Count the lone pairs on the central atom.
Answer
(a) Cl: 7 − 1 = 6 electrons, 2 bonds, 2 lone pairs: bent, sp³. (b) S: 3 bonded atoms and 1 lone pair: trigonal pyramidal, sp³. (c) P: 4 bonded atoms, no lone pair: tetrahedral, sp³.
Q5 · 4 marks · Long answerHOTS / competency
The molecule CO₂ is linear with C–O bond moments of about 2.5 D. (a) Why is its dipole moment zero? (b) Why does SO₂ have a nonzero moment? (c) Compare the shapes and dipole behaviour of BF₃ and NF₃, giving hybridisation.
Hint
Add the bond moments as vectors.
Answer
(a) The two moments are equal and opposite on a straight line. (b) SO₂ is bent (S has a lone pair), so the moments do not cancel. (c) BF₃: sp², trigonal planar, μ = 0. NF₃: sp³, pyramidal, small nonzero μ because the lone-pair moment partly opposes the bond moments.

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.Regions for sp³?Show answer
Four.
2.Shape of SF₄?Show answer
See-saw.
3.Shape of XeF₄?Show answer
Square planar.
4.s-character in sp²?Show answer
33%.
5.1 D in C m?Show answer
3.336 × 10⁻³⁰.
6.Bonds in a triple bond?Show answer
One σ and two π.

VSEPR Theory, Shapes and Hybridisation: frequently asked questions

Why do lone pairs repel more than bond pairs?

A lone pair is held by only one nucleus and spreads over more space, so it repels neighbouring pairs more strongly.

Is the shape the same as the electron geometry?

Only when there are no lone pairs.

Why is water bent?

The oxygen has two lone pairs and two bond pairs, a tetrahedral arrangement with two positions empty.

Can second-period elements have sp³d hybridisation?

No; they have no d orbitals in the valence shell.

How do hybrid orbitals form?

Atomic orbitals of similar energy mix to give equivalent hybrid orbitals.