Chapter 10
The Shapes of Molecules
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Lewis Structures
Mostly using the Octet Rule, but there
are exceptions
 The best rules are in your handouts as
well as in your lecture notes and
following. Use them!

10-2
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THE BEST RULES for writing Lewis Electron Dot
structures for covalent compounds:
1. Write the Lewis structure for each atom, then total the
number of valence electrons. DO NOT worry about
which electrons go with which atom! For example, CO2
would have a total of 16 valence electrons.
2. Put the atom that is "leftist &/or lowest" as the central
atom. Arrange the other atoms around it. Then draw
one single covalent bond of 2 electrons from the central
atom to each of the other atoms around it. Subtract
the number of electrons in bonds from the total found
in step 1. In the CO2 example, C is central and O’s are
terminal. The electron counting is: 16 - 4 = 12.
3. Put the remaining electrons around the terminal atoms
as lone pairs of electrons until they are “happy,” except
for H. (Why?) If there are still some electrons left they
go on the central atom.
10-3
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THE BEST RULES for writing Lewis Electron Dot
structures for covalent compounds:
4. See if each atom has 8 electrons around it. If not, you
take one lone pair on a terminal atom and make it a
bond pair, then another, etc., until all the atoms are
happy. (This is what makes multiple bonds, double or
triple is possible, quadruple is not!) In CO2 our O's are
happy but C is not. Take a LP from each terminal O and
make it a bond pair.
5. Check again that you have a total of 16 electrons and
that all atoms “think” that they have 8 electrons.
If it is a polyatomic ion, you must account for the charge by
adding in extra electrons in step 1 if it is an anion, or
subtracting if it is a cation, like NH4+.
The bracket must be placed around the structure and the
charge indicated.
10-4
SAMPLE PROBLEM 10.1
Write a Lewis structure for CCl2F2.
SOLUTION:
Cl C
:
:
: Cl :
:Cl C
:
: F:
:
Steps 4 & 5: Each atom has 8 e-s, and the total is still
32.
F
F
:
Steps 1, 2 & 3: There are 32 total valence e-s. Put C in
center & arrange other terminal atoms around it. Each
bond is 2 e-s, or 8 e-s. 32-8=24 e-s to go as LP’s. There
is room enough for all 24 e’s, so none go on the central
atom.
Cl
Note that bonds can be shown as lines, but lone pairs of
electrons are always dots!
F:
:
PROBLEM:
Writing Lewis Structures for Molecules with
One Central Atom
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More Complex Lewis Structures
Many compounds have more than one central atom,
especially organic compounds where all the carbon
atoms are “central” and all H’s are terminal.
With multiple bonds, consider how many single bonds
each atom would usually “like” to make:
C = 4, O = 2, N = 3, H and F always = 1, halogens
“like” 1
H2O2 has two O atoms as central. Draw it that way and
distribute the H’s equally around the two central atoms.
Keep using the rules!
.. ..
H-O-O-H
.. ..
10-6
SAMPLE PROBLEM 10.2 Writing Lewis Structure for Molecules with
More than One Central Atom
Hydrogen can have only one bond so C and O must be next
to each other with H’s around them.
There are 4(1) + 4 + 6 = 14 valence e-s.
C likes to have 4 bonds and O likes to have 2, therefore O
gets the 2 Lone Pairs of e-s.
H
:
SOLUTION:
Write the Lewis structure for methanol (molecular formula
CH3OH).
H
C
O
:
PROBLEM:
H
H
SAMPLE PROBLEM 10.3
PROBLEM:
SOLUTION:
H
(a) There are 2(4) + 4(1) = 12 valence e-. H can have only
one bond per atom. C wants to make 4 bonds.
:
H
Writing Lewis Structures for Molecules with
Multiple Bonds.
Write Lewis structure for Ethene (C2H4), the most important
reactant in the manufacture of polymers and also for N2.
C
C
H
H
H
H
H
C
C
H
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Diatomic nitrogen
N2 has 10 valence e-s
N-N 10-2=8 e-s remain
.. ..
:N-N: 8 – 8 = 0 e-s, but neither N happy
Move one LP from each N in to become a
shared pair or covalent bond:
:N=N: Triple bond makes N happy
10-9
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Lewis Structures
PRACTICE: OF2, H2CO, NO2, O3, I3-, HCOOH,
CH3CH2OH, SF4, NO, C6H6 in a ring
EXCEPTIONS:
H2 has only two e-s each - not an octet. Why is
that OK?
I3- and SF4 have EXPANDED valence shell central atom has 10 e-s
NO has a lone e- called a FREE RADICAL, makes
it a very reactive compound
10-10
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Lewis Structures
O3 and C6H6 can be written in two
different ways: which is the correct way?
BOTH ARE!
RESONANCE STRUCTURES: resonance is
an OLD concept: e-s don't bounce around
bond to bond, but are continually being
shared.
10-11
Resonance: Delocalized Electron-Pair Bonding
O3 can be drawn in 2 ways -
O
O
O
O
O
O
Neither structure is actually correct but can be drawn to represent a structure
which is a hybrid of the two - a resonance structure.
B
B
O
O
O
A
O
C
O
O
O
O
A
O
C
Resonance structures have the same relative atom placement but a
difference in the locations of bonding and nonbonding electron pairs.
is used to indicate that resonance occurs.
SAMPLE PROBLEM 10.4
Writing Resonance Structures
PROBLEM:
Write resonance structures for the nitrate ion, NO3-.
SOLUTION:
Nitrate has 1(5) + 3(6) + 1 = 24 valence e24 – 6 = 18
O
18 – 18 = 0
O
O
O
N
N
N
O
O
O
O
O
N does not have an
octet; a pair of ewill move in to form
a double bond.
O
O
O
O
N
N
N
O
O
O
O
O
Formal Charge: Selecting the Best Resonance Structure
An atom “owns” all of its nonbonding electrons and half of its bonding electrons.
Formal charge is the charge an atom would have if the bonding electrons were
shared equally.
Formal charge of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
B
# valence
e-
=6
O
e-
A
# nonbonding
# bonding
For OC
O
For OA
e-
=4
= 4 X 1/2 = 2
O
C
For OB
# valence e- = 6
Formal charge = 0
# valence e- = 6
# nonbonding e- = 6
# bonding e- = 2 X 1/2 = 1
Formal charge = -1
# nonbonding e- = 2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
Resonance (continued)
Three criteria for choosing the more important resonance structure:
Smaller formal charges (either positive or negative) are preferable
to larger charges.
Avoid like charges (+ + or - - ) on adjacent atoms.
A more negative formal charge should exist on an atom with a
larger EN value.
Resonance (continued)
EXAMPLE: NCO- has 3 possible resonance forms -
N
C
O
N C
A
N
O
B
C
O
C
formal charges
-2
N
0
+1
C
O
-1
0
N C
0
O
0
N
0
C
-1
O
Forms B and C have negative formal charges on N and O; this makes them
more preferred than form A.
Form C has a negative charge on O which is the more electronegative
element, therefore form C contributes the most to the resonance hybrid.
SAMPLE PROBLEM 10.5 Writing Lewis Structures for Octet Rule Exceptions
PROBLEM:
Write Lewis structures for (a) H3PO4 (pick the most likely
structure); (b) BFCl2.
SOLUTION:
(a) H3PO4 has two resonance forms and formal charges
indicate the more important form.
-1
0
O
0 H O
P
O
0
H
0
+1
O H 0
0
0
0 H O
0
O
0
P
O H 0
0 O
H
0
(b) BFCl2 will have only 1
Lewis structure. B is edeficient!
F
0
more stable
Lower formal charges (plus O likes to make
2 bonds). P has expanded valence shell.
B
Cl
Cl
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Localized electron bonding model:
Assumes that a molecule is composed of atoms
that are bound together by sharing pairs of e-s
using the atomic orbitals of the bound atoms
- shared pairs are called bonding pairs
- those found on one atom are called lone pairs
This basic model was seen with our drawing of
Lewis structures
It is also the basis of VSEPR Theory
10-18
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VSEPR THEORY: VALENCE SHELL ELECTRON PAIR
REPULSION THEORY
Methane’s Lewis dot structure is 2-D and looks flat with
90o bond angles.
Methane in reality, in 3-D, is a regular tetrahedron
with bond angles = 109°
Are CH3Cl and CH2Cl2 equivalent geometric
structures?
Molecular Shape = function of repulsion of electron pairs
in valence shell of central atom(s)
Electron pairs try to arrange themselves as far apart as
possible to minimize repulsions
- this includes LP's as well as BP's - to minimize
electrostatic repulsion.
10-19
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VSEPR THEORY: VALENCE SHELL ELECTRON
PAIR REPULSION THEORY
This leads to five major families of e- pair
arrangements: linear, trigonal planar,
tetrahedral, trigonal bipyramidal,
octahedral.
When drawing these we use AXbEc where:
A= central atom
X (or B) = surrounding atom(s)
E = nonbonding valence e- group (lone
pair)
10-20
Figure 10.2
Electron-group repulsions and the five basic molecular shapes.
linear
trigonal bipyramidal
tetrahedral
trigonal planar
octahedral
Figure 10.3
The single molecular shape of the linear electron-group
arrangement.
Examples:
CS2, HCN, BeF2
Figure 10.4
The two molecular shapes of the trigonal planar electron-group
arrangement.
Examples:
SO2, O3, PbCl2, SnBr2
Examples:
SO3, BF3, NO3-, CO32-
Figure 10.5
The three molecular shapes of the tetrahedral electrongroup arrangement.
Examples:
CH4, SiCl4,
SO42-, ClO4-
NH3
H 2O
PF3
OF2
ClO3
SCl2
H 3 O+
Figure 10.7
The four molecular shapes of the trigonal bipyramidal
electron-group arrangement.
PF5
SF4
AsF5
XeO2F2
SOF4
IF4+
IO2F2-
XeF2
ClF3
I3 -
BrF3
IF2-
Figure 10.8
The three molecular shapes of the octahedral electrongroup arrangement.
SF6
IOF5
BrF5
TeF5
-
XeOF4
XeF4
ICl4-
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Atomic bond angles:
Bond angles are important in determining a
compound's chemical behavior
If all "ligands" are equal, each structural shape has
specific bond angles
Bond angles are affected if there's a multiple bond,
a variety of ligands, and by lone pairs
LP's affect bond angles and the shape of the
molecule
Stronger electrostatic repulsions: LP-LP > LPBP > BP-BP
10-27
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“Electron Pair Families - Minor”
TRIGONAL PLANAR GROUP:
- each has 3 e- pairs around central atom, at
corners of a trigonal planar arrangement
- includes AX3 (trigonal planar), AX2E (bent)
LINEAR GROUP:
- each has 2 e- pairs around central atom in
linear arrangement
- includes AX2 (linear) and ABX (linear)
10-28
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“Electron Pair Families”
TETRAHEDRAL GROUP:
- each has 4 e- pairs around central atom,
at corners of a tetrahedron
- includes AX4 (tetrahedral), AX3E (trig
pyr), AX2E2 (bent)
TRIGONAL BIPYRAMIDAL GROUP:
- each has 5 e- pairs around central atom,
3 on equatorial and 2 on axial
- includes AX5 (trig bipyr), AX4E (see-saw),
AX3E2 (T-shaped), AX2E3 (linear)
10-29
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“Electron Pair Families”
OCTAHEDRAL GROUP:
- each has 6 e- pairs around central atom,
at equal angles and distances
- includes AX6 (octahedral), AX5E (square
pyramidal), AX4E2 (square planar), and
AX3E3 (T-shaped).
10-30
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Bonding and Shape
Multiple Bonds: a double or triple bonded
molecule has a predictable shape because they
are counted as one group of e-s, connecting B
to A.
Because there are more e-s, they will repel more
than single BP's.
Roughly LP > TBP > DBP > BP
If a molecule has a resonance structure, bonds
sharing the resonance are equal in repelling LP's
or BP's
10-31
Figure 10.9
from 4th ed.
A summary of common molecular shapes with two to six electron groups.
Figure 10.10
Molecular
formula
The steps in determining a molecular shape.
Step 1
Lewis
structure
See Figure
10.1
Step 2
Electron-group
arrangement
Count all e- groups around central
atom (A)
Step 3
Bond
angles
Note lone pairs and double
bonds
Count bonding and
Step 4
nonbonding egroups separately.
Molecular
shape
(AXmEn)
SAMPLE PROBLEM 10.6
PROBLEM:
Predicting Molecular Shapes with Two, Three,
or Four Electron Groups
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
The shape is based upon the tetrahedral arrangement.
F
P
F
F
P
F
F
F
The F-P-F bond angles should be <109.50 due
to the repulsion of the nonbonding electron
pair.
The final shape is trigonal pyramidal.
<109.50
The type of shape is
AX3E
When I ask for molecular geometry,
write the name not the type!
SAMPLE PROBLEM 10.6
Predicting Molecular Shapes with Two, Three,
or Four Electron Groups
continued
(b) For COCl2, C has the lowest EN and will be the center atom.
There are 24 valence e-, 3 atoms attached to the center atom.
Cl
C
O
Cl
The shape for an atom with three atom
attachments and no nonbonding pairs on the
central atom is trigonal planar.
O
O
C
Cl
C does not have an octet; a pair of nonbonding
electrons will move in from the O to make a
double bond.
Cl
The Cl-C-Cl bond angle will
be less than 1200 due to
the electron density of the
C=O.
124.50
C
Cl
1110
Cl
Type AX3
SAMPLE PROBLEM 10.7
PROBLEM:
SOLUTION:
Predicting Molecular Shapes with Five or Six
Electron Groups
Determine the molecular shape and predict the bond angles
(relative to the ideal bond angles) of (a) SbF5 and (b) BrF5.
(a) SbF5 - 40 valence e-; all electrons around central
atom will be in bonding pairs; shape is AX5 - trigonal
bipyramidal.
F
F
F
F
Sb
F
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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Multiple Central Atoms
Molecules with more than one central atoms, like
hydrocarbons, alcohols, thiosulfate ion, etc:
apply VSEPR to each central atom and
determine the shape around it
Try methanol CH3OH (Note: methanol is organic
and the formula is written to help perceive the
structure, so the formula tells you that C has 3
H’s with it and O has 1 H.)
10-37
SAMPLE PROBLEM 10.8
PROBLEM:
PLAN:
Predicting Molecular Shapes with More Than
One Central Atom
Determine the shape around each of the central atoms in
acetone, (CH3)2C=O.
Find the shape of one atom at a time after writing the Lewis
structure.
SOLUTION:
tetrahedral
H
H C
H
O
H
C C H
tetrahedral
H
trigonal planar
O
>1200
H
C
H
H C
C
H
HH
<1200
Figure 10.10
The tetrahedral centers of ethane and ethanol.
ethane
ethanol
CH3CH3
CH3CH2OH
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You practice finding Lewis structure, e- pair
arrangement and molecular geometry

10-40
Try CO2, SO2, SO3, SF4, SF6, PCl3, PCl5,
XeF4, I3-
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Dipole Moments & Electronegativity
Refer to handout and recall
electronegativity and polar covalent
bonding from last chapter
If a bond is polar covalent, one end is
slightly negative and the other is slightly
positive
Leads to a dipole moment, which can
respond to an external electrical field
10-41
Figure 10.11
The orientation of polar molecules in an electric field.
Electric field OFF
Electric field ON
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Molecular Dipole Moment
DIPOLE MOMENT: polar covalent bonds
cause a molecule to have a dipole
moment, unless the polarity vectors
cancel each other in a symmetrical
arrangement (see handout)
Nonpolar (CCl4, PCl5, SF6)
If a molecule is not symmetric, it will
probably have a dipole moment (BrF3,
OF2, NH3, SF4)
10-44
SAMPLE PROBLEM 10.9
PROBLEM:
Predicting the Polarity of Molecules
From electronegativity (EN) values (button) and their periodic
trends, predict whether each of the following molecules is polar
and show the direction of bond dipoles and the overall
molecular dipole when applicable:
(a) Ammonia, NH3
(b) Boron trifluoride, BF3
(c) Carbonyl sulfide, COS (atom sequence SCO)
PLAN: Draw the shape, find the EN values and combine the concepts to
determine the polarity.
SOLUTION:
(a) NH3
The dipoles reinforce each
other, so the overall
molecule is definitely polar.
ENN = 3.0
H
ENH = 2.1
N
H
H
H
N
H
H
bond dipoles
H
N
H
H
molecular
dipole
continued
(b) BF3 has 24 valence e- and all electrons around the B will be involved in
bonds. The shape is AX3, trigonal planar.
F
B
F
F
1200
F (EN 4.0) is more electronegative
than B (EN 2.0) and all of the dipoles
will be directed from B to F. Because
all are at the same angle and of the
same magnitude, the molecule is
nonpolar.
(c) COS is linear. C and S have the same EN (2.0) but the C=O bond is
quite polar(DEN) so the molecule is polar overall.
S
C
O