Why is the bond angle in H2O less than in NH3?

Water has two lone pairs on O, ammonia has only one on N. More lone pairs mean more lp-lp and lp-bp repulsion, pushing the bonded atoms closer together and reducing the bond angle from 107 degrees in NH3 to 104.5 degrees in H2O.

What is the hybridisation of IF7?

IF7 has 7 sigma bonds and no lone pairs, requiring 7 hybrid orbitals. Hybridisation is sp3d3 and the shape is pentagonal bipyramidal.

Valence Shell Electron Pair Repulsion (VSEPR) Theory

Q: What is the basic assumption of VSEPR theory?

Electron pairs in the valence shell of a central atom repel each other and arrange themselves in space to be as far apart as possible. Lone pairs repel more strongly than bond pairs.

Q: Why is XeF2 linear and not bent?

XeF2 has sp3d hybridisation with 2 bond pairs and 3 lone pairs. All three lone pairs occupy the equatorial positions of trigonal bipyramidal geometry, leaving the two F atoms at axial positions and giving a linear 180 degree arrangement.

Q: Why is XeF4 square planar?

XeF4 has 6 electron pairs on Xe (4 bond + 2 lone), giving octahedral electron geometry. The two lone pairs occupy trans positions to minimise lp-lp repulsion, leaving the four F atoms in a square plane.

Q: Why does SF4 have a see-saw shape?

SF4 has 5 electron pairs (4 bond + 1 lone), giving trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position to minimise 90 degree repulsions, giving a see-saw shape.

Q: Why is ClF3 T-shaped?

ClF3 has 5 electron pairs (3 bond + 2 lone). Both lone pairs occupy equatorial positions in the trigonal bipyramidal geometry, leaving the three F atoms in a T-shaped arrangement with two axial and one equatorial.

ChemistryChemical Bonding And Molecular StructureFor JEE aspirants

Valence Shell Electron Pair Repulsion (VSEPR) theory is a simple, powerful model for predicting the three-dimensional shapes of covalent molecules. Proposed by Sidgwick and Powell (1940) and refined by Nyholm and Gillespie (1957), it says that electron pairs in the valence shell of a central atom repel each other and arrange themselves in space to minimise this repulsion. Lone pairs repel more strongly than bond pairs, so the order of repulsion is lone pair-lone pair $>$ lone pair-bond pair $>$ bond pair-bond pair. From this single rule you can predict whether a molecule is linear, bent, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, or something in between - and explain bond angle variations like why $N H_{3}$ (107°) has a smaller angle than $C H_{4}$ (109.5°) and $H_{2} O$ (104.5°) smaller still.

Key Ideas - Quick Reference

Total electron pairs on central atom $=$ $σ$ -bond pairs $+$ lone pairs (multiple bonds count as one super-pair).
Repulsion order: lp-lp $>$ lp-bp $>$ bp-bp.
Number of hybrid orbitals rule: $Total valence electrons \div 8 =$ (quotient = $σ$ -bonds) remainder $=$ (non-bonded electrons); lone pairs $=$ non-bonded electrons $\div 2$ .
Regular geometry (no lone pairs): $A B_{2}$ linear, $A B_{3}$ trigonal planar, $A B_{4}$ tetrahedral, $A B_{5}$ trigonal bipyramidal, $A B_{6}$ octahedral.
In trigonal bipyramidal geometry, lone pairs always occupy equatorial positions (larger space, fewer 90° repulsions).
In octahedral geometry with two lone pairs, they occupy trans (opposite) positions.
Higher electronegativity of surrounding atoms $\Rightarrow$ smaller bond angle at the central atom.

1. The Postulates of VSEPR Theory

VSEPR is not a bonding theory - it does not explain why bonds form. It is a geometry-prediction recipe built on one physical idea: pairs of electrons in the valence shell of a central atom repel each other, and they arrange themselves to be as far apart as possible.

The main postulates are:

The shape of a molecule depends on the total number of valence-shell electron pairs (bonded and non-bonded) around the central atom.
Electron pairs repel each other because their clouds are negatively charged.
These pairs occupy positions in space that minimise repulsion and maximise the distance between them.
The valence shell is treated as a sphere on whose surface the electron pairs localise.
A multiple bond (double or triple) is treated as a single super-pair.
Where two or more resonance structures represent a molecule, VSEPR applies to any one of them.

Order of repulsion (memorise this):

lone pair-lone pair $>$ lone pair-bond pair $>$ bond pair-bond pair

Lone pairs occupy more angular space than bond pairs because lone pairs are held by only one nucleus, so their electron cloud spreads out more. This is the key to predicting distortions from ideal geometries.

2. Regular Geometry - No Lone Pairs on the Central Atom

When the central atom has only bond pairs (no lone pairs), the shape follows directly from the number of $σ$ -bonds. These are the "ideal" geometries.

Figure 1: The five regular VSEPR geometries. Bond angles are 180° (linear), 120° (trigonal planar), 109.5° (tetrahedral), 90° and 120° (trigonal bipyramidal, with axial and equatorial positions), and 90° (octahedral). Central atom brown, terminal atoms orange.

Electron pairs	Geometry	Bond angle	Hybridisation	Examples
2	Linear	180°	$s p$	$BeC l_{2}$ , $HgC l_{2}$ , $C O_{2}$
3	Trigonal planar	120°	$s p^{2}$	$B F_{3}$ , $AlC l_{3}$ , $BC l_{3}$
4	Tetrahedral	109.5°	$s p^{3}$	$C H_{4}$ , $N H_{4}^{+}$ , $Si F_{4}$
5	Trigonal bipyramidal	90°, 120°	$s p^{3} d$	$PC l_{5}$ , $P F_{5}$
6	Octahedral	90°	$s p^{3} d^{2}$	$S F_{6}$
7	Pentagonal bipyramidal	72°, 90°	$s p^{3} d^{3}$	$I F_{7}$

3. Rule for Determining Total Hybrid Orbitals

A quick numerical rule lets you find the number of hybrid orbitals (and therefore the geometry) for any main-group species:

Identify the central atom and peripheral atoms.
Count total valence electrons: central atom + peripheral atoms. Add electrons for negative charge, subtract for positive charge.
Divide the total by 8. The quotient gives the number of $σ$ -bonds; the remainder gives the non-bonded electrons.
Lone pairs $=$ non-bonded electrons $\div 2$ .
Total hybrid orbitals $=$ $σ$ -bonds $+$ lone pairs.

Solved Example 1

Predict the hybridisation and shape of

S F_{4}

Solution:

Total valence electrons $= 6 + (4 \times 7) = 34$ . Divide by 8: quotient $= 4$ , remainder $= 2$ . So there are $4$ $σ$ -bonds and $2 \div 2 = 1$ lone pair. Total hybrid orbitals $= 4 + 1 = 5$ , giving $s p^{3} d$ hybridisation and trigonal bipyramidal electron-pair geometry. With one lone pair in the equatorial plane, the molecular shape is see-saw.

Solved Example 2

Predict the hybridisation of the central atom in

POC l_{3}

OS F_{4}

, and

OI F_{5}

Solution:

$POC l_{3}$ : valence electrons $= 5 + 6 + 21 = 32$ ; $32/8 = 4$ , remainder $0$ . Hybridisation $= s p^{3}$ .

$OS F_{4}$ : valence electrons $= 6 + 6 + 28 = 40$ ; $40/8 = 5$ , remainder $0$ . Hybridisation $= s p^{3} d$ .

$OI F_{5}$ : valence electrons $= 6 + 7 + 35 = 48$ ; $48/8 = 6$ , remainder $0$ . Hybridisation $= s p^{3} d^{2}$ .

4. Irregular Geometry - Central Atom with Lone Pairs

When the central atom carries lone pairs, they distort the ideal geometry because lp-bp repulsion is stronger than bp-bp repulsion. The electron-pair geometry follows the total electron count, but the molecular shape only counts the bonded atoms.

Notation: $A B_{x} E_{y}$ where A is the central atom, B is a bonded atom, E is a lone pair.

Type	bp	lp	Electron geometry	Molecular shape	Example
$A B_{2} E$	2	1	Trigonal planar	Bent	$S O_{2}$ , $O_{3}$
$A B_{3} E$	3	1	Tetrahedral	Trigonal pyramidal	$N H_{3}$
$A B_{2} E_{2}$	2	2	Tetrahedral	Bent (V-shape)	$H_{2} O$
$A B_{4} E$	4	1	Trigonal bipyramidal	See-saw	$S F_{4}$
$A B_{3} E_{2}$	3	2	Trigonal bipyramidal	T-shaped	$Cl F_{3}$
$A B_{2} E_{3}$	2	3	Trigonal bipyramidal	Linear	$Xe F_{2}$ , $I_{3}^{-}$
$A B_{5} E$	5	1	Octahedral	Square pyramidal	$Br F_{5}$ , $XeO F_{4}$
$A B_{4} E_{2}$	4	2	Octahedral	Square planar	$Xe F_{4}$

Figure 2: Molecular geometries when the central atom carries lone pairs. Top row: one lone pair on the central atom. Bottom row: two or three lone pairs. Bond pairs shown as solid bonds to orange terminal atoms; lone pairs shown as pairs of orange dots at the end of dashed lines, indicating the geometric position they occupy on the parent electronic geometry.

Why lone pairs go equatorial in TBP

In a trigonal bipyramidal geometry, axial positions have three 90° neighbours, while equatorial positions have only two 90° neighbours. Since 90° repulsions are much stronger than 120° repulsions, placing a lone pair at equatorial minimises the strong lp-bp repulsion. This is why $S F_{4}$ is see-saw (not trigonal pyramidal), $Cl F_{3}$ is T-shaped (not trigonal planar), and $Xe F_{2}$ is linear (not bent).

Why lone pairs go trans in octahedral

All six positions in an octahedron are equivalent, so the first lone pair can go anywhere. But when a second lone pair is added, it must sit as far as possible from the first - directly opposite (trans, 180°). Adjacent (cis, 90°) placement would give a large lp-lp repulsion. This is why $Xe F_{4}$ is square planar with the two lone pairs above and below the plane.

5. Worked Examples - Predicting Shapes

Solved Example 3

Predict the shape of

NC l_{3}

Solution:

Valence electrons $= 5 + (3 \times 7) = 26$ . Divide by 8: quotient $= 3$ , remainder $= 2$ . So $3$ $σ$ -bonds + $1$ lone pair. Total $= 4 \Rightarrow s p^{3}$ . With one lone pair on a tetrahedral electron geometry, the molecular shape is trigonal pyramidal, like ammonia.

Solved Example 4

Predict the shape and hybridisation of

Xe F_{4}

Solution:

Valence electrons $= 8 + (4 \times 7) = 36$ . Divide by 8: quotient $= 4$ , remainder $= 4$ . So $4$ $σ$ -bonds + $2$ lone pairs. Total $= 6 \Rightarrow s p^{3} d^{2}$ , octahedral electron geometry. Both lone pairs must occupy trans positions (above and below the plane), so the four F atoms occupy a square. Shape: square planar.

Solved Example 5

Predict the shape of

Xe O_{2} F_{2}

Solution:

Valence electrons $= 8 + (2 \times 6) + (2 \times 7) = 34$ . Divide by 8: quotient $= 4$ , remainder $= 2$ . So $4$ $σ$ -bonds + $1$ lone pair. Total $= 5 \Rightarrow s p^{3} d$ , trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position. Shape: see-saw (distorted trigonal bipyramidal). The double bonds to oxygen are treated as single super-pairs.

Solved Example 6

P F_{2} B r_{3}

, where do the fluorine atoms go - axial or equatorial?

Solution:

The molecule is trigonal bipyramidal ( $s p^{3} d$ ). In TBP geometry with different substituents, the more electronegative atoms occupy the axial positions. Fluorine is more electronegative than bromine, so the two F atoms sit at the two axial positions and the three Br atoms occupy the equatorial plane. This minimises bp-bp repulsion because F pulls the bonded electrons away from the central atom more strongly, reducing electron density on the axial bonds where 90° repulsions are strongest.

6. Bond Angle Variations

VSEPR explains why real bond angles often deviate from the ideal geometry angles.

Effect of lone pairs on bond angle

Solved Example 7

Why is the H-C-H bond angle in methane 109.5°, while the H-N-H bond angle in ammonia is 107° and the H-O-H bond angle in water is 104.5°, even though C, N, and O are all

s p^{3}

hybridised?

Solution:

All three atoms have 4 electron pairs, so the electron-pair geometry is tetrahedral. But:

$C H_{4}$ : 4 bond pairs, 0 lone pairs. Only bp-bp repulsion. Angle $= 109.5°$ .
$N H_{3}$ : 3 bond pairs, 1 lone pair. The lp-bp repulsion is stronger than bp-bp, so the lone pair pushes the three bonds closer together. Angle drops to $107°$ .
$H_{2} O$ : 2 bond pairs, 2 lone pairs. Two lone pairs push harder still, with additional lp-lp repulsion. Angle drops further to $104.5°$ .

Effect of electronegativity of surrounding atoms

Rule: If the electronegativity of the peripheral atoms increases, the bond angle at the central atom decreases.

More electronegative peripheral atoms pull the bonded electron pair away from the central atom, reducing electron density near the central atom, weakening bp-bp repulsion, and letting bonds move closer.

Solved Example 8

The H-O-H bond angle in water is 104.5°, but the F-O-F angle in

O F_{2}

is only 102°. Explain.

Solution:

Both molecules are bent with two lone pairs on the central O. But F is far more electronegative than H, so in $O F_{2}$ the bonding electron pairs are pulled strongly towards the F atoms and away from the central O. This decreases bp-bp repulsion at oxygen and lets the bonds compress further. In $H_{2} O$ , the electron density stays closer to O (since H is not very electronegative), keeping bp-bp repulsion higher and the angle larger.

Solved Example 9

Compare the H-N-H bond angle in

N H_{3}

(107°) with the F-N-F bond angle in

N F_{3}

(102°).

Solution:

Same argument: F pulls electron density away from N, reducing bp-bp repulsion and shrinking the bond angle. In $N H_{3}$ , N holds the electron density more, keeping the angle larger.

7. Multiple Bonds and VSEPR

A double bond (or triple bond) is treated as a single super-pair when counting geometry. However, multiple bonds do contain more electrons than a single bond, so they occupy more angular space. This slightly compresses adjacent single-bond angles.

For example, in $H_{2} C = O$ (formaldehyde), the H-C-H angle is $116°$ (less than 120°) because the C=O double bond repels the C-H bonds more strongly.

Common Mistakes to Avoid

Watch out

Confusing electron-pair geometry with molecular shape. $N H_{3}$ has tetrahedral electron geometry but trigonal pyramidal molecular shape.
Forgetting the lone pair. Always calculate lone pairs from the total valence electrons, not from what "looks right" in a rough sketch.
Counting a double bond as two pairs for geometry. A double or triple bond counts as one super-pair for shape prediction (though it takes more space than a single bond).
Placing lone pairs axially in TBP. Lone pairs always go equatorial in trigonal bipyramidal geometry. Only $Xe F_{2}$ -type ions with three lone pairs look linear, but each lone pair is still equatorial.
Assuming ideal bond angles. Real molecules deviate from 109.5°, 120°, and 90° because of lone pairs and electronegativity of surrounding atoms.
Placing lone pairs cis in octahedral geometry with two lone pairs. They must go trans (opposite each other) to minimise lp-lp repulsion.
Using VSEPR for transition metal complexes. VSEPR works well for main-group molecules but fails for d-block complexes, where crystal field theory and hybridisation of d-orbitals are needed.

Frequently Asked Questions

Q1. What is the basic assumption of VSEPR theory?

The central assumption is that electron pairs in the valence shell of a central atom repel each other and arrange themselves in space to be as far apart as possible. Lone pairs repel more strongly than bond pairs because they are held by only one nucleus and spread out more.

Q2. Why is the bond angle in $H_{2} O$ (104.5°) less than in $N H_{3}$ (107°)?

Water has two lone pairs on the central oxygen, while ammonia has only one on nitrogen. Two lone pairs produce more lp-lp and lp-bp repulsion than one lone pair, pushing the O-H bonds closer together than the N-H bonds. Hence the H-O-H angle is smaller.

Q3. Why is $Xe F_{2}$ linear and not bent?

$Xe F_{2}$ has $s p^{3} d$ hybridisation with 2 bond pairs and 3 lone pairs. The three lone pairs occupy the equatorial positions of a trigonal bipyramidal geometry (where they have the most space), leaving the two F atoms at the axial positions. This gives a linear F-Xe-F arrangement of 180°.

Q4. Why is $Xe F_{4}$ square planar and not tetrahedral?

$Xe F_{4}$ has 6 electron pairs on Xe (4 bond + 2 lone), giving octahedral electron geometry ( $s p^{3} d^{2}$ ). The two lone pairs must sit trans to each other (above and below the plane) to minimise lp-lp repulsion. This leaves the four F atoms in a square in the equatorial plane.

Q5. Does VSEPR work for transition metal complexes?

Not reliably. VSEPR is designed for main-group compounds. Transition metals have partially filled d-orbitals whose effect on geometry is better described by crystal field theory or ligand field theory. VSEPR often fails to predict correct shapes for complexes like $[Ni (CN)_{4}]^{2 -}$ (square planar despite 4 ligands).

Q6. How does electronegativity of surrounding atoms affect bond angle?

More electronegative surrounding atoms pull bonding electron density away from the central atom. This reduces bp-bp repulsion at the central atom, allowing bond angles to decrease. For example, F-N-F in $N F_{3}$ (102°) is smaller than H-N-H in $N H_{3}$ (107°).

Q7. Why does $S F_{4}$ have a see-saw shape rather than trigonal bipyramidal?

$S F_{4}$ has 5 electron pairs on S (4 bond + 1 lone), giving trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position (where it has fewer 90° repulsions), so the four F atoms form a distorted see-saw around it, not a symmetric TBP.

Q8. Why is $Cl F_{3}$ T-shaped?

$Cl F_{3}$ has 5 electron pairs on Cl (3 bond + 2 lone). The electron geometry is trigonal bipyramidal ( $s p^{3} d$ ). Both lone pairs occupy equatorial positions (to minimise lp-lp and lp-bp repulsion), leaving three F atoms at two axial positions and one equatorial. The resulting Cl-F skeleton looks like a T.

Q9. What is the number of hybrid orbitals for $I F_{7}$ ?

Valence electrons $= 7 + (7 \times 7) = 56$ . Divide by 8: quotient $= 7$ , remainder $= 0$ . So there are 7 $σ$ -bonds and no lone pairs, giving 7 hybrid orbitals, $s p^{3} d^{3}$ hybridisation, and a pentagonal bipyramidal shape.

Previous year questions on Valence Shell Electron Pair Repulsion (VSEPR) Theory

29 questions from past papers, each with a step-by-step solution.

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1. The Postulates of VSEPR Theory

2. Regular Geometry - No Lone Pairs on the Central Atom

3. Rule for Determining Total Hybrid Orbitals

4. Irregular Geometry - Central Atom with Lone Pairs

Why lone pairs go equatorial in TBP

Why lone pairs go trans in octahedral

5. Worked Examples - Predicting Shapes

6. Bond Angle Variations

Effect of lone pairs on bond angle

Effect of electronegativity of surrounding atoms

7. Multiple Bonds and VSEPR

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the basic assumption of VSEPR theory?

Q2. Why is the bond angle in H2​O (104.5°) less than in NH3​ (107°)?

Q3. Why is XeF2​ linear and not bent?

Q4. Why is XeF4​ square planar and not tetrahedral?

Q5. Does VSEPR work for transition metal complexes?

Q6. How does electronegativity of surrounding atoms affect bond angle?

Q7. Why does SF4​ have a see-saw shape rather than trigonal bipyramidal?

Q8. Why is ClF3​ T-shaped?

Q9. What is the number of hybrid orbitals for IF7​?

Previous year questions on Valence Shell Electron Pair Repulsion (VSEPR) Theory

Ready to master Chemical Bonding And Molecular Structure?

Q2. Why is the bond angle in $H_{2} O$ (104.5°) less than in $N H_{3}$ (107°)?

Q3. Why is $Xe F_{2}$ linear and not bent?

Q4. Why is $Xe F_{4}$ square planar and not tetrahedral?

Q7. Why does $S F_{4}$ have a see-saw shape rather than trigonal bipyramidal?

Q8. Why is $Cl F_{3}$ T-shaped?

Q9. What is the number of hybrid orbitals for $I F_{7}$ ?