VSEPR THEORY
Valence Shell Electron Pair
Repulsion
VSEPR THEORY:
AT THE CONCLUSION OF OUR TIME
TOGETHER, YOU SHOULD BE ABLE TO:
1. Use VSEPR to predict molecular shape
2. Name the 6 basic shapes that have no
unshared pairs of electrons
3. Name a few variations off of these basic
shapes
VSEPR Theory
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MOLECULAR SHAPES
Lewis structures show which atoms are
connected where, and by how many bonds,
but they don't properly show 3-D shapes of
molecules.
To find the actual shape of a molecule, first
draw the Lewis structure, and then use
VSEPR Theory.
VALENCE SHELL ELECTRON-PAIR
REPULSION THEORY OR VSEPR
► Molecular
Shape is determined by the
repulsions of electron pairs
 Electron pairs around the central atom stay
as far apart as possible.
► Electron Pair Geometry - based on number of
regions of electron density
 Consider non-bonding (lone pairs) as well as
bonding electrons. Unshared repel the most.
 Electron pairs in single, double and triple
bonds are treated as single electron clouds.
► Molecular Geometry - based on the electron
pair geometry, this is the shape of the molecule
Electron-group Repulsions And The
Five Basic Molecular Shapes.
LET’S CONSIDER THESE BASIC
SHAPES AND SOME VARIATIONS
OF THEM…
LINEAR
2
atoms attached to central atom
 0 unshared pairs (lone pairs)
 Bond
 Type:
angle = 180o
AX2
 Ex. : BeF2
LINEAR
 Carbon
CO2
dioxide
The Single Molecular Shape Of The Linear
Electron-group Arrangement.
Examples:
CO2, BeF2
TRIGONAL PLANAR
 Boron
Trifluoride
BF3
TRIGONAL PLANAR
3
atoms attached to central atom
 0 lone pairs
 Bond
 Type:
angle = 120o
AX3
 Ex. : AlF3
The Two Molecular Shapes Of The Trigonal
Planar Electron-group Arrangement.
Class
Shape
Examples:
H2CO, BCl3, NO3-,
CO32-
Factors Affecting Actual Bond Angles
Bond angles are consistent with theoretical angles
when the atoms attached to the central atom are the
same and when all electrons are bonding electrons of
the same order. Some exceptions follow:
Effect of Double Bonds
H
1200
ideal
1200
C
O
H
greater
electron
density
larger
EN
H
1160
C
H
real
O
Factors Affecting Actual Bond Angles
Effect of Nonbonding Pairs
Lone pairs repel
bonding pairs
more strongly than
bonding pairs
repel each other
Sn
Cl
Cl
950
TRIGONAL PLANAR VARIATION
The Second Molecular Shape Of The
Trigonal Planar Electron-group
Arrangement.
Examples:
SO2, O3
BENT
 Trigonal
Planar variation #1
 2 atoms attached to central atom
 1 lone pair
 Bond
 Type:
angle = <120
AX2E
 Ex. : SO2
REDNECK INNOVATIONS: WHEN YOU
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TETRAHEDRAL
4
atoms attached to central atom
 0 lone pairs
 Bond
 Type:
angle = 109.5o
AX4
 Ex. : CH4
TETRAHEDRAL
 Carbon
tetrachloride
CCl4
2 TETRAHEDRAL VARIATIONS
The Three Molecular Shapes Of The Tetrahedral
Electron-group Arrangement.
Examples:
CH4, SO42-
NH3
H2O
PF3
OF2
TRIGONAL PYRAMIDAL
 Tetrahedral
variation #1
 3 atoms attached to central atom
 1 lone pair
 Bond
 Type:
angle = 107o
AX3E
 Ex. : NH3
TRIGONAL PYRAMIDAL
 Nitrogen
trifluoride
NF3
BENT
 Tetrahedral
variation #2
 2 atoms attached to central atom
 2 lone pairs
 Bond
 Type:
angle = 104.5o
AX2E2
 Ex. : H2O
BENT
 Chlorine
difluoride
ion
ClF2+
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REMEMBER THE 3 EXCEPTIONS TO THE
OCTET RULE?
 Molecules
with atoms near the boundary
between metals and nonmetals will tend to
have less than an octet on the central atom.
(i.e. B, Be, Al, Ga)
 Molecules with a central atom with electrons
in the 3rd period and beyond will sometimes
have more than an octet on the central atom,
up to 12, called an extended or expanded octet.
 Molecules with an odd number of electrons
5 BOND SITES ON THE CENTRAL
ATOM
TRIGONAL BIPYRAMIDAL
5
atoms attached to central atom
 0 lone pairs
 Bond
angle =
 equatorial -> 120o
 axial -> 90o
 Type:
AX5
 Ex. : PF5
TRIGONAL BIPYRAMIDAL
 Antimony
Pentafluoride
SbF5
3 BIPYRAMIDAL VARIATIONS
The Four Molecular Shapes Of The Trigonal
Bipyramidal Electron-group Arrangement.
PF5
SF4
AsF5
XeO2F2
ClF3
XeF2
BrF3
I 3-
SEE SAW
 Trigonal
Bipyrimid
Variation #1
 Sulfur
tetrafluoride
SF4
T-SHAPED
 Trigonal
Bipyramid
Variation #2
 Chlorine tribromide
LINEAR
 Trigonal
Bipyramid
Variation #3
 Xenon difluoride
XeF2
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6 BOND SITES ON THE CENTRAL
ATOM
OCTAHEDRAL
6
atoms attached to central atom
 0 lone pairs
 Bond
 Type:
angle = 90o
AX6
 Ex. : SF6
OCTAHEDRAL
 Sulfur
hexafluoride
SF6
2 OCTAHEDRAL VARIATIONS
The Three Molecular Shapes Of The Octahedral
Electron-group Arrangement.
SF6
IOF5
IF5
XeOF4
XeF4
(BrF4)-
SQUARE PYRAMIDAL
 Octahedral
Variation #1
 Chlorine pentafluoride
ClF5
SQUARE PLANAR
 Octahedral
Variation #2
 Xenon tetrafluoride
XeF4
OCTAHEDRAL
 Do
not need to
know:
 T-shape
 Linear
LET’S REVIEW VSEPR THEORY
 Predicts
the molecular shape of a bonded
molecule
 Electrons around the central atom arrange
themselves as far apart from each other as
possible
 Unshared pairs of electrons (lone pairs) on the
central atom repel the most
 So only look at what is connected to the central
atom
REMEMBER THE 3 EXCEPTIONS TO THE
OCTET RULE?
 Molecules
with atoms near the boundary
between metals and nonmetals will tend to
have less than an octet on the central atom.
(i.e. B, Be, Al, Ga)
 Molecules with a central atom with electrons
in the 3rd period and beyond will sometimes
have more than an octet on the central atom,
up to 12, called an extended or expanded octet.
 Molecules with an odd number of electrons
BENT
 Nitrogen
NO2
dioxide
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VSEPR THEORY:
LET’S SEE IF YOU CAN:
1. Use VSEPR to predict molecular shape
2. Name the 6 basic shapes that have no
unshared pairs of electrons
3. Name a few variations off of these basic
shapes
Sample Problems
The Steps In Determining A
Molecular Shape.
Molecular
formula
Lewis
structure
Step 1
Electron-group
arrangement
Bond
angles
Count all e- groups
around central atom (A)
Step 2
Step 3
Step 4
Molecular
shape
(AXmEn)
Note lone pairs and
double bonds
Count bonding and
nonbonding e- groups
separately.
REVIEW OF LEWIS STRUCTURES


Step 1:
Count the number of valence electrons.
For a neutral molecule this is equal to
the number of valence electrons of the
constituent atoms.
Example (CH3NO2):
Each hydrogen contributes 1 valence
electron. Each carbon contributes 4,
nitrogen 5, and each oxygen 6 for a
total of 24.
REVIEW OF LEWIS STRUCTURES


Step 2:
Connect the atoms by a covalent bond
represented by a dash.
Example:
Methyl nitrite has the partial
structure:
H
H
C
H
O
N
O
REVIEW OF LEWIS STRUCTURES


Step 3:
Subtract the number of electrons in
bonds from the total number of valence
electrons.
Example:
24 valence electrons – 12 electrons in
bonds. Therefore, 12 more electrons to
assign.
REVIEW OF LEWIS STRUCTURES


Step 4: Add electrons in pairs so that
as many atoms as possible have 8
electrons. Start with the most
electronegative atom.
Example: The remaining 12 electrons
in methyl nitrite are added as 6 pairs.
H
H
C
H
..
O
..
N
..
..
:
O
..
REVIEW OF LEWIS STRUCTURES
Step 5: If an atom lacks an octet, use
electron pairs on an adjacent atom to
form a double or triple bond.

Example: There are 2 ways this can be
done.

H
H
C
H
H
O
..
N
..
..
:
O
..
H
C
H
..
O
..
N
..
..
O:
REVIEW OF LEWIS STRUCTURES
Step 6: Calculate formal charges.

Example: The left structure has
formal charges that are greater than 0.
Therefore it is a less stable Lewis
structure.

H
H
C
H
+
O
..
N
..
.. –
O
.. :
H
H
C
H
..
O
..
N
..
..
O:
SAMPLE
PROBLEM:
Predicting Molecular
Shapes with Two, Three,
or Four Electron Groups
PROBLEM:
Draw the molecular shape and predict
the bond angles (relative to the ideal
bond angles) of (a) PF3 and (b) COCl2.
SOLUTION: (a) For PF3 - there are 26 valence
electrons, 1 nonbonding pair
F
P
F
The shape is based upon the
tetrahedral arrangement.
F
The F-P-F bond angles should be
<109.50 due to the repulsion of the
nonbonding electron pair.
P
F
<109.50
F
F
The type of
shape is
AX3E
The final shape is
trigonal pyramidal.
(b) For COCl2, C has the lowest EN and will
be the center atom.
Cl
C
O
Cl
There are 24 valence e-, 3 atoms
attached to the center atom.
C does not have an octet; a pair of
nonbonding electrons will move in
from the O to make a double bond.
O
C
Cl
Cl
The shape for an atom with three atom
attachments and no nonbonding pairs
on the central atom is trigonal planar.
Type AX3
The Cl-C-Cl bond angle
will be less than 1200 due
to the electron density of
the C=O.
O
C
Cl
1110
124.50
Cl
SAMPLE
PROBLEM:
Predicting Molecular
Shapes with Five or Six
Electron Groups
PROBLEM:
Determine the molecular shape and
predict the bond angles (relative to the
ideal bond angles) of (a) SbF5 and (b)
BrF5.
SOLUTION:
(a) SbF5 - 40 valence e-; all
electrons around central atom
will be in bonding pairs; shape
is AX5 - trigonal bipyramidal.
F
F
F
F
F
Sb
F
F
Sb
F
F
F
(b) BrF5 - 42 valence e-; 5 bonding pairs and 1
nonbonding pair on central atom. Shape is
AX5E, square pyramidal.
F
F
F
Br
F
F
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