Chemical Bonding – VSEPR Theory & Molecular Shapes | Complete JEE Main Concept

Chemical Bonding – VSEPR Theory & Molecular Shapes

VSEPR Theory is one of the most important concepts in Chemical Bonding for Class 11 and JEE Main. It helps us predict the three-dimensional shape of molecules and ions by studying the arrangement of electron pairs around the central atom.

The basic idea is very simple:

Electron pairs repel each other and arrange themselves as far apart as possible.

Maximum separation gives minimum repulsion and therefore a more stable arrangement.

To solve VSEPR questions quickly, you need to understand just a few key ideas:

• Bond pairs
• Lone pairs
• Electron domains
• Steric number
• AX_mE_n notation
• Electron geometry
• Molecular shape
• Lone-pair repulsion

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1. What is VSEPR Theory? ⭐

VSEPR stands for:

Valence Shell Electron Pair Repulsion Theory

According to VSEPR theory, electron pairs present around the central atom repel one another.

Because electron pairs have the same negative charge, they try to stay as far apart as possible.

Therefore:

Electron-pair repulsion → Maximum separation → Minimum repulsion → Stable arrangement

For example, if a central atom has two electron domains, they arrange themselves in opposite directions:

Electron domain  ←  A  →  Electron domain

Angle = 180°

This produces a linear electron arrangement.

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2. Types of Electron Pairs

There are two important types of electron pairs around the central atom.

Bond Pair (BP)

A bond pair is an electron pair involved in forming a bond between the central atom and another atom.

For example, in CH₄, each C–H bond represents one bonding electron domain around carbon.

Lone Pair (LP)

A lone pair is an electron pair that is present on the central atom but is not involved in bonding.

Lone pairs occupy space around the central atom and therefore strongly influence molecular shape.

⭐ Repulsion Order

LP–LP > LP–BP > BP–BP

Why?

A lone pair is attracted only by the central nucleus and occupies a relatively larger region of space. Therefore, lone pairs produce stronger repulsion than bonding pairs.

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3. AX_mE_n Notation

VSEPR structures are often represented using AX_mE_n notation.

Symbol	Meaning
A	Central atom
X	Atoms bonded to the central atom
m	Number of bonded atoms/domains represented by X
E	Lone pair on the central atom
n	Number of lone pairs

Examples

CH₄ → AX₄

NH₃ → AX₃E

H₂O → AX₂E₂

This notation makes shape prediction much faster.

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4. Steric Number ⭐

The steric number (SN) tells us the total number of electron domains around the central atom.

For simple species:

SN = Number of σ bonds + Lone pairs

Each sigma bond contributes one electron domain. A multiple bond also counts as one electron domain for VSEPR purposes.

Steric Number	Electron Geometry	Hybridization
2	Linear	sp
3	Trigonal planar	sp²
4	Tetrahedral	sp³
5	Trigonal bipyramidal	sp³d
6	Octahedral	sp³d²

🔥 Shortcut:

SN → Electron Geometry → Molecular Shape

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5. Steric Number = 2

AX₂

There are:

• 2 bond pairs
• 0 lone pairs

Therefore:

SN = 2

Electron geometry = Linear

Molecular shape = Linear

Bond angle:

180°

Examples:

• BeCl₂
• CO₂

X — A — X

180°

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6. Steric Number = 3

AX₃ — Trigonal Planar

There are:

• 3 bond pairs
• 0 lone pairs

Therefore:

SN = 3

Shape = Trigonal planar

Bond angle:

120°

Example:

BF₃

AX₂E — Bent

Here:

• 2 bond pairs
• 1 lone pair

Total:

SN = 3

Electron geometry = Trigonal planar

Molecular shape = Bent / Angular

The bond angle becomes:

< 120°

Example:

SO₂

The important point is that electron geometry and molecular shape are not necessarily the same.

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7. Steric Number = 4

AX₄ — Tetrahedral

There are:

• 4 bond pairs
• 0 lone pairs

Electron geometry = Tetrahedral

Molecular shape = Tetrahedral

Bond angle:

109.5°

Example:

CH₄

AX₃E — Trigonal Pyramidal

There are:

• 3 bond pairs
• 1 lone pair

SN = 4

Electron geometry = Tetrahedral

Molecular shape = Trigonal pyramidal

Bond angle is approximately:

107°

Example:

NH₃

AX₂E₂ — Bent

There are:

• 2 bond pairs
• 2 lone pairs

Electron geometry = Tetrahedral

Molecular shape = Bent / V-shaped

Bond angle is approximately:

104.5°

Example:

H₂O

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8. CH₄, NH₃ and H₂O Comparison 🔥

These three molecules are extremely important for understanding the effect of lone pairs.

Molecule	AXmEn	BP	LP	Shape	Bond Angle
CH4	AX4	4	0	Tetrahedral	109.5°
NH3	AX3E	3	1	Trigonal pyramidal	≈107°
H2O	AX2E2	2	2	Bent	≈104.5°

Therefore, the bond-angle order is:

CH₄ > NH₃ > H₂O

or:

109.5° > 107° > 104.5°

Why does the angle decrease?

Because lone pairs create stronger repulsion.

LP–LP > LP–BP > BP–BP

As the number of lone pairs increases, they push bonding pairs closer together, reducing the bond angle.

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9. Steric Number = 5 🔥

For SN = 5, the electron geometry is:

Trigonal Bipyramidal (TBP)

There are two types of positions:

• Axial
• Equatorial

Important angles are:

• Equatorial–equatorial = 120°
• Axial–equatorial = 90°
• Axial–axial = 180°

⭐ Lone Pair Preference

Lone pairs prefer the:

Equatorial position

Why?

An equatorial lone pair experiences fewer 90° interactions than an axial lone pair.

• Equatorial LP → 2 × 90° interactions
• Axial LP → 3 × 90° interactions

Therefore:

LP prefers equatorial position.

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10. AX₅ — Trigonal Bipyramidal

There are:

• 5 bond pairs
• 0 lone pairs

Shape:

Trigonal bipyramidal

Example:

PCl₅

The molecule contains three equatorial positions and two axial positions.

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11. AX₄E — See-Saw

There are:

• 4 bond pairs
• 1 lone pair

SN = 5.

The lone pair prefers the equatorial position.

Molecular shape:

See-saw

Example:

SF₄

The lone pair occupies an equatorial position to minimize repulsion.

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12. AX₃E₂ — T-Shaped

There are:

• 3 bond pairs
• 2 lone pairs

Both lone pairs prefer equatorial positions.

The remaining three bonds form a:

T-shaped geometry

Example:

ClF₃

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13. AX₂E₃ — Linear

There are:

• 2 bond pairs
• 3 lone pairs

All three lone pairs occupy the three equatorial positions.

The two remaining axial positions are occupied by the bonded atoms.

F — Xe — F

    180°

Therefore:

AX₂E₃ → Linear

Example:

XeF₂

Bond angle:

180°

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14. Steric Number = 6

For SN = 6, the electron geometry is:

Octahedral

The six positions are arranged symmetrically around the central atom.

The important bond angle is:

90°

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15. AX₆ — Octahedral

There are:

• 6 bond pairs
• 0 lone pairs

Shape:

Octahedral

Example:

SF₆

Bond angle:

90°

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16. AX₅E — Square Pyramidal

There are:

• 5 bond pairs
• 1 lone pair

SN = 6.

Electron geometry:

Octahedral

Molecular shape:

Square pyramidal

Example:

BrF₅

The lone pair occupies one position of the octahedral arrangement, leaving five bonded atoms in a square-pyramidal arrangement.

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17. AX₄E₂ — Square Planar ⭐

There are:

• 4 bond pairs
• 2 lone pairs

SN = 6.

Electron geometry:

Octahedral

The two lone pairs occupy opposite positions.

The four bonded atoms then lie in one plane.

Therefore:

Molecular shape = Square planar

Example:

XeF₄

        F
        |
    F — Xe — F
        |
        F

Square planar arrangement

The important bond angles are:

90° and 180°

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18. Complete VSEPR Table 🔥

Type	BP	LP	Shape	Important Angle
AX2	2	0	Linear	180°
AX3	3	0	Trigonal planar	120°
AX2E	2	1	Bent	<120°
AX4	4	0	Tetrahedral	109.5°
AX3E	3	1	Trigonal pyramidal	≈107°
AX2E2	2	2	Bent	≈104.5°
AX5	5	0	Trigonal bipyramidal	90°, 120°
AX4E	4	1	See-saw	Approx. 90°, 120°
AX3E2	3	2	T-shaped	≈90°
AX2E3	2	3	Linear	180°
AX6	6	0	Octahedral	90°
AX5E	5	1	Square pyramidal	≈90°
AX4E2	4	2	Square planar	90°, 180°

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19. Electron Geometry vs Molecular Shape ⚠️

This is one of the most important conceptual distinctions in VSEPR.

Electron geometry considers:

Bond pairs + Lone pairs

But molecular shape considers the arrangement of the atoms only.

Example: NH₃

NH₃ has:

• 3 bond pairs
• 1 lone pair

Total electron domains = 4

Therefore:

Electron geometry → Tetrahedral

But the lone pair is not an atom. Considering only the positions of the bonded atoms:

Molecular shape → Trigonal pyramidal

Example: H₂O

H₂O has:

• 2 bond pairs
• 2 lone pairs

Therefore:

Electron geometry → Tetrahedral

But:

Molecular shape → Bent / V-shaped

Remember:

Electron geometry → BP + LP

Molecular shape → Position of atoms only

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20. Multiple Bonds — Important JEE Point 🔥

In VSEPR theory, a single bond, double bond or triple bond each represents one electron domain around the central atom.

Therefore:

Single bond = 1 electron domain

Double bond = 1 electron domain

Triple bond = 1 electron domain

Example: CO₂

Structure:

O = C = O

Carbon has two double bonds.

But each double bond represents one electron domain.

Therefore carbon has:

2 electron domains

So:

AX₂ → Linear

Bond angle:

180°

JEE Trap: Never count a double bond as two separate VSEPR electron domains.

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21. Lone Pair Position in Trigonal Bipyramidal Geometry

For steric number 5, lone pairs have a strong preference for equatorial positions.

Let's compare the 90° interactions.

Position	Number of 90° Interactions	Preference
Axial	3	Less preferred for LP
Equatorial	2	Preferred

Thus:

Lone pair → Equatorial position in TBP

This rule is extremely useful for questions involving SF₄, ClF₃ and XeF₂.

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22. JEE Shortcut for Molecular Shape 🔥

Whenever JEE Main asks you to determine the molecular shape, follow this fixed process.

Step 1: Draw the Lewis Structure

Determine how the atoms are connected and identify the central atom.

Step 2: Find the Central Atom

Identify the atom around which the electron pairs need to be arranged.

Step 3: Count Sigma Bonds and Lone Pairs

Calculate:

SN = σ bonds + LP

Step 4: Determine AX_mE_n

Write the VSEPR notation.

Step 5: Find Electron Geometry

Use the steric number.

Step 6: Find Molecular Shape

Now consider the positions of the bonded atoms, ignoring lone pairs from the name of the molecular shape.

Lewis Structure → BP + LP → SN → AX_mE_n → Shape

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23. JEE Main-Level Solved Question 🔥

Question: What is the molecular shape of XeF₄?

Step 1: Identify the central atom

The central atom is:

Xe

Step 2: Count bond pairs

There are four Xe–F bonds.

Therefore:

BP = 4

Step 3: Count lone pairs

Xe has two lone pairs in XeF₄.

Therefore:

LP = 2

Step 4: Write AX notation

AX₄E₂

Step 5: Find steric number

SN = 4 + 2 = 6

Step 6: Find electron geometry

SN = 6 corresponds to:

Octahedral electron geometry

Step 7: Place lone pairs

The two lone pairs occupy opposite positions in the octahedral arrangement.

The four F atoms therefore occupy the remaining four positions in one plane.

Final Answer

✅ XeF₄ → Square Planar

Important bond angles:

90° and 180°

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24. Common JEE Traps ⚠️

Trap 1: NH₃ is Tetrahedral in Molecular Shape

Incorrect.

NH₃ has tetrahedral electron geometry, but its molecular shape is:

Trigonal pyramidal

Trap 2: H₂O is Tetrahedral in Molecular Shape

Incorrect.

Its electron geometry is tetrahedral, but its molecular shape is:

Bent / V-shaped

Trap 3: Ignoring Lone Pairs

Lone pairs are included when finding electron-domain geometry.

SN = σ bonds + LP

Trap 4: Double Bond Counts as Two Electron Domains

Incorrect.

A double bond represents one electron domain in VSEPR.

Trap 5: Lone Pair Prefers Axial Position in TBP

Incorrect.

Lone pairs prefer equatorial positions because this minimizes 90° interactions.

Trap 6: XeF₄ is Tetrahedral

Incorrect.

XeF₄ is:

AX₄E₂ → Square planar

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25. Important Shapes to Memorize

Molecule	AXmEn	Shape
BeCl2	AX2	Linear
BF3	AX3	Trigonal planar
SO2	AX2E	Bent
CH4	AX4	Tetrahedral
NH3	AX3E	Trigonal pyramidal
H2O	AX2E2	Bent
PCl5	AX5	Trigonal bipyramidal
SF4	AX4E	See-saw
ClF3	AX3E2	T-shaped
XeF2	AX2E3	Linear
SF6	AX6	Octahedral
BrF5	AX5E	Square pyramidal
XeF4	AX4E2	Square planar

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26. JEE Main Quick-Solving Strategy

For a shape-based question, do not try to memorize every molecule separately.

Instead, remember the sequence:

Lewis Structure → Central Atom → BP + LP → SN → AX_mE_n → Shape

Then remember these five steric numbers:

• SN = 2 → Linear
• SN = 3 → Trigonal planar
• SN = 4 → Tetrahedral
• SN = 5 → Trigonal bipyramidal
• SN = 6 → Octahedral

After that, account for the lone pairs to obtain the actual molecular shape.

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27. One-Minute Revision 🚀

VSEPR: Valence Shell Electron Pair Repulsion Theory

Basic idea: Electron pairs repel and arrange as far apart as possible.

Repulsion order:

LP–LP > LP–BP > BP–BP

Steric Number:

SN = σ bonds + LP

SN 2 → Linear → sp

SN 3 → Trigonal planar → sp²

SN 4 → Tetrahedral → sp³

SN 5 → Trigonal bipyramidal → sp³d

SN 6 → Octahedral → sp³d²

CH₄ → AX₄ → Tetrahedral → 109.5°

NH₃ → AX₃E → Trigonal pyramidal → ≈107°

H₂O → AX₂E₂ → Bent → ≈104.5°

PCl₅ → AX₅ → TBP

SF₄ → AX₄E → See-saw

ClF₃ → AX₃E₂ → T-shaped

XeF₂ → AX₂E₃ → Linear

SF₆ → AX₆ → Octahedral

BrF₅ → AX₅E → Square pyramidal

XeF₄ → AX₄E₂ → Square planar

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28. Final Revision Box 🔥

⭐ VSEPR:

Electron pairs repel each other and arrange themselves with maximum separation.

REPULSION:

LP–LP > LP–BP > BP–BP

STERIC NUMBER:

SN = σ bonds + LP

GEOMETRY:

SN 2 → Linear

SN 3 → Trigonal planar

SN 4 → Tetrahedral

SN 5 → Trigonal bipyramidal

SN 6 → Octahedral

IMPORTANT SHAPES:

CH₄ → AX₄ → Tetrahedral → 109.5°

NH₃ → AX₃E → Trigonal pyramidal → ≈107°

H₂O → AX₂E₂ → Bent → ≈104.5°

PCl₅ → AX₅ → TBP

SF₄ → AX₄E → See-saw

ClF₃ → AX₃E₂ → T-shaped

XeF₂ → AX₂E₃ → Linear

SF₆ → AX₆ → Octahedral

BrF₅ → AX₅E → Square pyramidal

XeF₄ → AX₄E₂ → Square planar

TBP LONE PAIR RULE:

Lone pair prefers the equatorial position.

MULTIPLE BONDS:

Single, double or triple bond → one electron domain in VSEPR.

⭐ GOLDEN RULE: Electron Pairs Repel → Maximum Separation → Molecular Shape

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29. Practice Questions for JEE Main

1. Predict the molecular shape and bond angle of BeCl₂.
2. What are the electron geometry and molecular shape of NH₃?
3. Why is the bond angle of H₂O smaller than that of NH₃?
4. Determine the shape of SF₄ using VSEPR theory.
5. What is the molecular shape of ClF₃?
6. Explain why XeF₂ is linear despite having three lone pairs on Xe.
7. Find the molecular shape of BrF₅.
8. Why is XeF₄ square planar?
9. Determine the molecular geometry of CO₂ using the electron-domain concept.
10. Arrange CH₄, NH₃ and H₂O in decreasing order of bond angle.

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30. PDF Notes

You can add your detailed VSEPR Theory and Molecular Shapes PDF notes below for students who want to revise the chapter offline.

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31. Frequently Asked Questions (FAQs)

Q1. What does VSEPR stand for?

VSEPR stands for Valence Shell Electron Pair Repulsion Theory.

Q2. What is the basic principle of VSEPR theory?

Electron pairs around the central atom repel one another and arrange themselves as far apart as possible to minimize repulsion.

Q3. Which repulsion is strongest?

The order is:

LP–LP > LP–BP > BP–BP

Q4. What is steric number?

For simple species, steric number is the total number of sigma bonds plus lone pairs around the central atom.

Q5. What is the shape of NH₃?

NH₃ has AX₃E configuration and its molecular shape is trigonal pyramidal. Its electron geometry is tetrahedral.

Q6. What is the shape of H₂O?

H₂O is AX₂E₂ and has a bent or V-shaped molecular geometry.

Q7. Why do lone pairs prefer equatorial positions in TBP?

An equatorial lone pair experiences fewer 90° interactions than an axial lone pair. Therefore, the equatorial position minimizes repulsion.

Q8. How many electron domains does a double bond represent?

A double bond represents one electron domain in VSEPR theory.

Q9. What is the shape of XeF₄?

XeF₄ has AX₄E₂ configuration and a square planar molecular shape.

Q10. What is the difference between electron geometry and molecular shape?

Electron geometry considers both bond pairs and lone pairs, while molecular shape describes the arrangement of the bonded atoms.

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Final Thoughts

VSEPR Theory becomes much easier when you stop trying to memorize molecular shapes individually and instead follow a fixed method.

Start with the Lewis structure, identify the central atom, count the sigma bonds and lone pairs, calculate the steric number, write the AX_mE_n notation and finally determine the molecular shape.

The most important concepts to remember are the repulsion order:

LP–LP > LP–BP > BP–BP

and the steric-number sequence:

2 → Linear

3 → Trigonal planar

4 → Tetrahedral

5 → Trigonal bipyramidal

6 → Octahedral

Once these patterns are clear, molecules such as NH₃, H₂O, SF₄, ClF₃, XeF₂ and XeF₄ can be solved systematically instead of memorized randomly.

Understand the electron pairs → find their arrangement → predict the molecular shape → solve the JEE question.