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
VSEPR Theory
Redneck Innovations
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.
 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.
The Single Molecular Shape Of The Linear
Electron-group Arrangement.
Examples:
CS2, HCN,
BeF2
The Two Molecular Shapes Of The Trigonal
Planar Electron-group Arrangement.
Class
Examples:
SO2, O3,
PbCl2,
SnBr2
Shape
Examples:
SO3, BF3, 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.
Effect of Double Bonds
H
1200
ideal
1200
C
O
H
greater
electron
density
1220
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
The Three Molecular Shapes Of The Tetrahedral
Electron-group Arrangement.
Examples:
CH4, SiCl4,
SO42-, ClO4-
NH3
H2O
PF3
OF2
ClO3
SCl2
H3O+
The Four Molecular Shapes Of The Trigonal
Bipyramidal Electron-group Arrangement.
PF5
SF4
AsF5
XeO2F2
SOF4
IF4+
IO2F2ClF3
XeF2
BrF3
I 3IF2-
The Three Molecular Shapes Of The Octahedral
Electron-group Arrangement.
SF6
IOF5
BrF5
TeF5XeOF4
XeF4
ICl4-
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
REDNECK INNOVATIONS: WHEN YOU
GOTTA GO…..
LET’S CONSIDER A FEW MORE
EXAMPLES:
6 TYPES OF MOLECULES WITH NO
UNSHARED PAIRS OF ELECTRONS
LINEAR
2
atoms attached to central atom
 0 unshared pairs (lone pairs)
 Bond
 Type:
angle = 180o
AB2
 Ex. : BeF2
3 EXCEPTIONS TO THE OCTET RULE
 Molecules
with an odd number of electrons
 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.
LINEAR
 Carbon
CO2
dioxide
TRIGONAL PLANAR
3
atoms attached to central atom
 0 lone pairs
 Bond
 Type:
angle = 120o
AB3
 Ex. : AlF3
TRIGONAL PLANAR
 Boron
Trifluoride
BF3
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TETRAHEDRAL
4
atoms attached to central atom
 0 lone pairs
 Bond
 Type:
angle = 109.5o
AB4
 Ex. : CH4
TETRAHEDRAL
 Carbon
tetrachloride
CCl4
TRIGONAL BIPYRAMIDAL
5
atoms attached to central atom
 0 lone pairs
 Bond
angle =
 equatorial -> 120o
 axial -> 90o
 Type:
AB5
 Ex. : PF5
TRIGONAL BIPYRAMIDAL
 Antimony
Pentafluoride
SbF5
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OCTAHEDRAL
6
atoms attached to central atom
 0 lone pairs
 Bond
 Type:
angle = 90o
AB6
 Ex. : SF6
OCTAHEDRAL
 Sulfur
hexafluoride
SF6
EXAMPLES OF MOLECULES WITH
BOTH PAIRED AND UNSHARED (LONE)
PAIRS OF ELECTRONS AROUND THE
CENTRAL ATOM.
TRIGONAL PLANAR
 Boron
Trifluoride
BF3
BENT
 Trigonal
Planar variation #1
 2 atoms attached to central atom
 1 lone pair
 Bond
 Type:
angle = <120
AB2E
 Ex. : SO2
TETRAHEDRAL
 Carbon
tetrachloride
CCl4
TRIGONAL PYRAMIDAL
 Tetrahedral
variation #1
 3 atoms attached to central atom
 1 lone pair
 Bond
 Type:
angle = 107o
AB3E
 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
AB2E2
 Ex. : H2O
BENT
 Chlorine
difluoride
ion
ClF2+
TRIGONAL BIPYRAMIDAL
 Antimony
Pentafluoride
SbF5
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
OCTAHEDRAL
 Sulfur
hexafluoride
SF6
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
BENT
 Nitrogen
NO2
dioxide
Redneck Innovations!!!!
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