SCH 4U
Valence Shell Electron Pair Repulsion Theory (VSEPR Theory) (pages 242 – 245)
This theory says that pairs of electrons in the valence shell of an atom stay as far apart as possible to minimize the repulsion of
their negative charges. Using this information helps us to predict the shape of a molecule around a central atom (an atom or
atoms in the molecule or ion that will make the largest number of bonds). It also helps us to predict the angles formed by bonds
with the central atom. The angles are particularly sensitive to the presence of non – bonding pairs of electrons or lone pairs of
electrons on the central atom because the orbitals for these electrons tend to flatten out and take up more room than do
those containing bonding electron pairs.
In the following table, A = central atom X = atoms attached to the central atom (by a single bond or a double bond
or a triple bond)
E = lone pairs of electrons on the central atom
Use Table 1, page 245 to fill in the following table.
No. BONDING ELECTRON
PAIRS ON CENTRAL ATOM
No. LONE PAIRS OF
ELECTRONS ON CENTRAL
ATOM
2
0
3
0
Four Items:
VSEPR shape
VSEPR formula
Hybridization for
central atom, A
Example
MODEL
and BOND ANGLES
(lone pairs need only be shown on the
central atoms)
linear
AX2
sp hybridization
for A
BeCl2
planar triangular
or trigonal planar
AX3
sp2
hybridization for A
BCl3
4
0
tetrahedral
AX4
sp3
hybridization for A
CH4
3
1
trigonal
pyramidal
AX3E1
sp3
hybridization for A
NH3
2
2
bent or nonlinear
AX2E2
sp3
hybridization for A
1
3
These are all
variations of
the
tetrahedral
shape
because they
all have 4
pairs of
electrons
around the
central
atom.
H 2O
Linear
AXE3
sp3
hybridization for A
HCl
NOTES:

The last 2 shapes (trigonal pyramidal and bent) are simply variations on the tetrahedral. ALL three of these shapes
require a TOTAL of 4 pairs of electrons around the central atom.

We’ll look at a few more VSEPR shapes next week.
Practice: Do
page 246#1, 2,
3; page 250 #1,
2, 3; page 289
#25, 26
SCH 4U
Valence Shell Electron Pair Repulsion Theory (VSEPR Theory) Continued (pages 242 – 250)
Other VSEPR shapes to know:
No. BONDING ELECTRON
PAIRS ON CENTRAL ATOM
1
No. LONE PAIRS OF
ELECTRONS ON CENTRAL
ATOM
3
Four Items:
VSEPR shape
VSEPR formula
Hybridization for
central atom, A
Example
Linear
AXE3
sp3
Name: _____________
Date: _____________
MODEL
and BOND ANGLES
(lone pairs need only be shown on the
central atoms)
Cl
H
hybridization for A
HCl
Note that lone
pairs spread out
and take up
more room than
bonding pairs
so they
decrease bond
angles between
bonding pairs.
Pg. 247
5
0
Trigonal
bipyramidal
AX5
sp3d
PCl5
4
1
Seesaw
AX4E1
sp3d
<
SF4
3
2
Notice: Lone pairs
found in the
equatorial plane.
Bond angles
<90o
T-shaped
AX3E2
sp3d
This shape should
have been on
Friday’s table
because it is a
version of a
tetrahedron with 4
pairs of electrons
around the central
atom.
These are all
versions of
the trigonal
bipyramidal
shape
because
there are 5
pairs of
electrons
around the
central
atom.
ClF3
2
3
Linear
AX2E3
sp3d
I3 Pg. 247
6
0
Octahedral
AX6
sp3d2
SF6
5
1
Square
pyramidal
AX5E1
sp3d2
Bond
angles
<90o
BrF5
4
2
Square planar
AX4E2
sp3d2
Lone
pairs stay
as far
apart as
possible.
Bond
angles
are 90o
XeF4
Although molecules with AX3E3 and AX2E4 are theoretically possible, no stable molecules with those shapes are known.
These are all
versions of the
octahedral
shape because
there are 6
pairs of
electrons
around the
central atom.
Lewis Structures, VSEPR Theory and Molecular Polarity
Part A: Lewis Structures
 Lewis structures are used to show which atoms are covalently bonded together in a molecule or in a polyatomic ion. Atoms
can share one pair of electrons (single bond), two pairs (double bond) or three pairs (triple bond).
 Most atoms will have a completed octet of valence electrons when the Lewis Structure is correct. As previously stated, H, B
and Be are exceptions to this rule because each of them will complete their valence shells with fewer than 8 electrons.
Many other atoms will also be exceptions under certain conditions but it will be because orbital hybridization has allowed
them to have more than an octet in their valence shell. Central atoms that do this are said to have an expanded valence or
an expanded octet. Valence electrons of elements of the third and subsequent rows of the Periodic Table may absorb
enough energy during bonding so that some or all are excited into their own unfilled d orbital (one electron per orbital).
This means that many more bonds then expected may form. eg. P with 5 valence electrons may form up to 5 bonds and S
with 6 valence electrons may form up to 6 bonds. Because of this compounds containing phosphorus or sulfur may have
more than one possible Lewis Structure. Many other atoms will also be exceptions under certain conditions but it will be
because orbital hybridization has allowed them to have more than an octet in their valence shell. Cl is an atom that also
often has an expanded valence.
 A co-ordinate covalent bond is a bond in which one atom contributes both electrons to the shared pair of electrons. These
bonds are no different than any other bonds, they are just thought to be formed slightly differently. This means that they
are not indicated any differently in a Lewis structure but you may be asked to identify them when they occur.
 Some atoms, ions and molecules have orbitals containing single unpaired electrons. These are called paramagnetic
because the unpaired electron(s) is/are affected by a magnetic field. Paramagnetic substances have an ODDnumber of
valence electrons so no matter how hard you try, your Lewis structure will always show an unpaired electron. eg. NO and
NO2
Steps for Drawing Lewis Structures
Step 1
Step 2
Decide which atoms are bonded together and arrange the atoms in a “skeleton structure” on your page to reflect your decision. It
helps to first determine one or more central atoms – one or more atoms to which many other atoms are bonded. Usually, the
central atom requires the largest number of bonds to complete its octet. It is often written first in the formula. It is often (but not
always) less electronegative than other atoms in the formula.
e.g. NF3
N needs the greatest number of bonds so it is the central atom to which each H is attached.
Skeleton formula:
F
N F
F
Count up the total number of valence electrons in the entire molecule
Example NF3
1 atom of N brings 5 valence electrons
+ 3 atoms of F, each bringing 7 valence electrons
26 valence electrons in total
NOTE: The charge on an ion indicates the gain of electron(s) (negative ion) or the loss of electrons (positive ion). So, SO42 – for
example contains 6 valence electrons from S, 4 atoms x 6 valence electrons each = 24 valence electrons from oxygen and 2 extra
electrons because of the – 2 charge. Total is 32 valence electrons.
NH41+ will contain 5 valence electrons from N and a total of 4 from the 4 hydrogen atoms. That gives 9 but the charge of 1+ means
we have to subtract one of those away so NH41+ has 8 valence electrons in total.
Step 3
Place 2 electrons (1 pair) of the total number of valence electrons into the spaces between atoms in the skeletal structure to
indicate single bonds.
e.g.
F
N
F
F
Step 4
Complete the octets of atoms attached to the central atom by adding electrons in pairs around each one (at north, south, east, west
as needed). In the example above, the atoms attached to the central atom, N, are F atoms:
F
N
F
F
Step 5
Place any remaining electrons on the central atom. In our NF3 example, we are supposed to be showing 26 valence electrons but
we only have 24 in the diagram, so the other two go on the central atom:
F
N
F
F
Step 6
If the central atom still does not have an octet, form double bonds. If necessary, form triple bonds. (In our NF3 example, we do not
need to use this step at all.)
Note that if all of the atoms attached to the central atom are identical, then it is possible for the multiple bond to be drawn to any
one of the atoms. Molecules and polyatomic ions with this set up are said to form resonance structures. Although we show multiple
Step 7
and single bonds in the diagram for this kind of molecule its bonds are actually a kind of hybrid between them. We say that triple
bonds have a bond order of 3, double bond have a bond order of 2 and single bonds have a bond order of 1. The higher the bond
order the more potential energy the bond contains and the shorter the bond length between the atoms in the bond.Often, in a
resonance structure, the bond order for what we draw as single bonds and as double bonds is actually found to be about 1.33, a
value in between a single bond and a double bond. We say that the electrons involved in the pi bond within the double bond are
delocalized and shared between the atoms within the structure. See step 7 for an example of how to show a resonance structure
for the molecule ozone, O3. Many molecules and polyatomic ions have resonance structures. The unique resonance structure for
benzene is also shown below.
Draw the final Lewis structure by replacing each shared pair of electrons with a stick. Polyatomic ions must be enclosed in square
brackets labeled with the ion charge outside the bracket in the upper right hand corner. If the molecule is a resonance structure,
show all of its possible structures as indicated below.
NF3 example
polyatomic ion, ClO21-
Resonance Structure for ozone, O3
In ozone, we cannot say definitively where the double bond is and all of the bonds have a bond order of 1.33 and so are acting as
intermediate bonds between single and double bonds. On a 2-D piece of paper, the way to describe ozone as a Lewis structure
would be:
This indicates that the ozone molecule is described by an average of the two Lewis structures (i.e. the resonance forms)
Resonance Structure for Benzene, C6H6
In benzene, each carbon atom forms 4 bonds but each C atom undergoes sp2 hybridization. This idea was first expressed with the
Kekule resonance structures for benzene:
Here are some problems with this representation:
a) It implies that benzene should be at least as reactive as
cyclohexene and be able to undergo addition reactions. Neither of
these ideas is supported experimentally.
b) Double and single bonds are of different lengths (C-C is about
0.154 nm and C=C about 0.134 nm) so the ring structure should be
uneven. It is a perfect hexagon, though.
c) The ring structure of benzene is very stable, much more stable
than the Kekule structure suggests.
It is believed that the 6 p-orbitals of the
6 carbon atoms form delocalized
orbital clouds in which electrons can be
shared equally between the carbon
atoms. This structure more clearly
supports the properties of benzene. To
represent this idea, the resonance
structures of benzene are usually
represented as one structure:
The important points to remember about resonance forms are:
 The molecule is not rapidly oscillating between different discrete forms.
 There is only one form of the ozone molecule, and the bond lengths between the oxygen atoms are intermediate between
characteristic single and double bond lengths between a pair of oxygen atoms.
 We draw two Lewis structures (in this case) because a single structure is insufficient to describe the real structure.
Part B: VSEPR Guidelines for Predicting Molecular Shape and Bond Angles
1.
Draw the Lewis structure of the molecule.
2.
Count the number of bonding pairs and the number of lone pairs around the central atom. Be sure to correct for any
charge for ions. Treat any double or triple bonds as though they were a single bonding pair.
3.
Determine the VSEPR Theory formula for the molecule. From this, you will be able to predict the hybridization of the
central atom and the VSEPR Theory shape for the molecule. (The five families of shapes available are linear (AX2);
trigonal planar (AX3); tetrahedral (AX4 or some variation); trigonal bipyramidal (AX5 or some variation) and octahedral
(AX6 or some variation). The actual shape of the molecule will depend on the VSEPR formula.)
4.
In predicting bond angles, note that a lone pair repels another lone pair or a bonding pair more strongly than a bonding
pair repels another bonding pair. Keep in mind that, in general, there is no easy way to accurately predict bond angles
when the central atom possesses one or more lone pairs but you can get a ballpark idea of whether the bond angles
are the same or different from the parent VSEPR Theory shape.
Part C: Guidelines for Predicting Molecular Polarity from Bond Dipoles and Molecular Shape
The guidelines for determining molecular polarity from bond dipoles and molecular shape are given on page 255 of the text in
the “Summary” Section. Please read over this section of the text and reproduce the summary in your notes as needed.
NOTE:
In general, for the molecules covered in this course, molecules will be polar if
a) the central atom has 1 or more lone pairs of electrons and/or
b) the atoms attached to the central atom are not identical to each other.
Molecules (and portions of molecules) not meeting at least one of these criteria are non-polar.
Practice:
Most of the following practice problems were given on Friday last week. Unless you read the text yourself and/or your grade 11 notes on
Lewis structures, these problems may’ve been challenging. Using the Lewis Structure notes above, try the questions now. Table headings have
been provided on the next page for you to record your work. You should be able to fill in all of the boxes for the assigned questions.
Practice: Do page 246#1, 2, 3; page 249, #11; page 250 #1, 2, 3; page 289 #25, 26