Chapter 10
The Shapes of Molecules
10-1
The Shapes of Molecules
10.1 Depicting Molecules and Ions with Lewis Structures
10.2 Using Lewis Structures and Bond Energies to
Calculate Heats of Reaction
10.3 Valence-Shell Electron-Pair Repulsion (VSEPR)
Theory and Molecular Shape
10.4 Molecular Shape and Molecular Polarity
10-2
Lewis Dot Structures
A bookkeeping way to keep track of valence
electrons and bonding in a molecule. Lines are used
to represent a pair of bonding electrons. Dots are
used to represent lone pair electrons.
In many cases (not H) the octet rule is followed,
having eight electrons (including bonding and lone
pairs) around each atom. We’ll see later that the octet
rule is a natural consequence of having four valence
orbitals (s, 3p’s) and the limitation of two electron’s
per orbital.
10-3
The steps in converting a molecular formula into a Lewis structure.
Molecular
formula
Step 1
Atom
placement
Place atom
with lowest
EN in center
Step 2
Sum of
valence e-
Add A-group
numbers
Step 3
Remaining
valence e-
Draw single bonds.
Subtract 2e- for each bond.
Step 4
Lewis
structure
10-4
Give each
atom 8e(2e- for H)
Sum of
valence e-
:
: F:
: F:
:
Atom
placement
For NF3
:
Molecular
formula
N
Remaining
valence eLewis
structure
10-5
:
: F:
N
5e-
F 7e- X 3 = 21eTotal
26e-
SAMPLE PROBLEM 10.1
Write a Lewis structure for CCl2F2, one of the compounds
responsible for the depletion of stratospheric ozone.
SOLUTION:
10-6
:
:
: Cl :
:Cl C
: F:
F:
:
Make bonds and fill in remaining valence
electrons placing 8e- around each atom.
F
F
:
Steps 2-4:
C has 4 valence e-, Cl and F each have 7. The
sum is 4 + 4(7) = 32 valence e-.
Cl C
:
Step 1: Carbon has the lowest EN and is the central atom.
The other atoms are placed around it.
Cl
:
PROBLEM:
Writing Lewis Structures for Molecules with
One Central Atom
SAMPLE PROBLEM 10.2
PROBLEM:
SOLUTION:
Writing Lewis Structure for Molecules with
More than One Central Atom
Write the Lewis structure for methanol (molecular formula
CH4O), an important industrial alcohol that is being used as a
gasoline alternative in car engines.
Hydrogen can have only one bond so C and O must be next
to each other with H filling in the bonds.
There are 4(1) + 4 + 6 = 14 valence e-.
C has 4 bonds and O has 2. O has 2 pair of nonbonding e-.
:
H
C
O
:
H
H
10-7
H
SAMPLE PROBLEM 10.3
PROBLEM:
PLAN:
Writing Lewis Structures for Molecules with
Multiple Bonds.
Write Lewis structures for the following:
(a) Ethylene (C2H4), the most important reactant in the
manufacture of polymers
(b) Nitrogen (N2), the most abundant atmospheric gas
For molecules with multiple bonds, there is an additional step which
follows the other steps in Lewis structure construction. If a central
atom does not have 8e-, an octet, then e- can be moved in to form a
multiple bond.
(a) There are 2(4) + 4(1) = 12 valence e-. H can have only
one bond per atom.
SOLUTION:
H
H
..
C
C
H
H
H
H
H
C
C
H
(b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make
the octet around each N.
N
.
:
N
.
:
:
:.
.:
10-8
N
.
N
N
:
N
.
Resonance: Delocalized Electron-Pair Bonding
Ozone, 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.
Not a single bond - double bond, but a bond and a half for both bonds
..
O
.
O
..
:
.
O
..
:
Resonance structures have the same relative atom placement but a
difference in the locations of bonding and nonbonding electron pairs.
A double headed arrow is used to indicate resonance structures.
10-9
10-10
Formal Charge: Selecting the Best Resonance Structure
An atom “owns” all of its nonbonding electrons and half of its bonding electrons.
Formal charge of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
For OC
For OA
=6
# bonding
e-
:
# nonbonding
e-
=4
= 4 X 1/2 = 2
Formal charge = 0
..
OB
OA
..
..
O
. .C
:
# valence
e-
# valence e- = 6
# nonbonding e- = 6
# bonding e- = 2 X 1/2 = 1
For OB
# valence
Formal charge = -1
e-
=6
# nonbonding e- = 2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
10-11
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.
10-12
Resonance (continued)
EXAMPLE: NCO- has 3 possible resonance forms -1
-1
.
.
.
.
..
:N
:N C O :
.. C O:
A
:N
B
C
..
O:
..
-1
C
formal charges
-2
..
:N
..
0
+1
C
O:
-1
-1
..
:N
0
C
0
..
O:
-1
0
:N
0
C
-1
..
O:
..
-1
Forms B and C have negative formal charges on N and O; this makes them
more important than form A.
Form C has a negative charge on O which is the more electronegative
element, therefore C contributes the most to the resonance hybrid.
10-13
Lewis Structures - Exceptions to the Octet Rule
10-14
Using bond energies to calculate DH0rxn
Enthalpy, H
DH0rxn = DH0reactant bonds broken + DH0product bonds formed
DH01 = + sum of BE
DH0rxn
10-15
DH02 = - sum of BE
Using bond energies to calculate DH0rxn for combustion of methane
BOND BREAKAGE
4BE(C-H)= +1652kJ
2BE(O2)= + 996kJ
DH0(bond breaking) = +2648kJ
Enthalpy,H
2[-BE(C O)]= -1598kJ
4[-BE(O-H)]= -1868kJ
DH0(bond forming) = -3466kJ
DH0rxn= -818kJ
10-16
BOND FORMATION
SAMPLE PROBLEM 10.6
PROBLEM:
Calculating Enthalpy Changes from Bond
Energies
Use average bond energies to calculate DH0rxn for the following
reaction:
CH4(g) + 3Cl2(g)
PLAN:
CHCl3(g) + 3HCl(g)
Write the Lewis structures for all reactants and products and
calculate the number of bonds broken and formed.
SOLUTION:
bonds broken
10-17
bonds formed
SAMPLE PROBLEM 10.6
Calculating Enthalpy Changes from Bond
Energies
continued
bonds broken
4 C-H
bonds formed
= 4 mol(413 kJ/mol) = 1652 kJ
3 C-Cl = 3 mol(-339 kJ/mol) = -1017 kJ
3 Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ
1 C-H = 1 mol(-413 kJ/mol) = -413 kJ
DH0bonds broken = 2381 kJ
3 H-Cl = 3 mol(-427 kJ/mol) = -1281 kJ
DH0bonds formed = -2711 kJ
DH0reaction = DH0bonds broken + DH0bonds formed = 2381 kJ + (-2711 kJ) = - 330 kJ
10-18
Valence-Shell Electron-Pair Repulsion Theory
(VSEPR)
VSEPR is a very good theory for predicting the
shape of molecules.
It involves any group of valence electrons around
an atom. These groups can be lone pairs, single
bonds, or multiple bonds. In essence, these groups
of negatively charge particles will be arranged as
far apart as possible around the atom.
10-19
Electron-group repulsions and the five basic
molecular shapes.
10-20
Looking at the Five Shapes in Detail
Examples:
CS2, HCN, BeF2
10-21
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
Multiple bonds count just as one group but are larger than a single bond.
The result is some compression of the other bond angles.
Effect of Nonbonding Pairs
Likewise lone pairs repel bonding pairs more strongly than
bonding pairs repel each other also compressing the other
angles.
10-22
The two molecular shapes of the trigonal planar electrongroup arrangement.
Class
Examples:
SO2, O3, PbCl2, SnBr2
Shape
Examples:
SO3, BF3, NO3-, CO32-
10-23
The three molecular shapes of the tetrahedral electrongroup arrangement.
Examples:
CH4, SiCl4,
SO42-, ClO4-
NH3
H 2O
PF3
OF2
ClO3
SCl2
H 3 O+
10-24
The four molecular shapes of the trigonal bipyramidal
electron-group arrangement.
PF5
SF4
AsF5
XeO2F2
SOF4
IF4+
IO2F2-
ClF3
XeF2
BrF3
I3 -
IF2-
10-25
The three molecular shapes of the octahedral electrongroup arrangement.
SF6
IOF5
BrF5
TeF5
-
XeOF4
10-26
XeF4
ICl4-
The steps in determining a molecular shape.
Molecular
formula
Step 1
Lewis
structure
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)
10-27
Lewis structures and molecular shapes
10-28
10-29
10-30
Molecular Shapes with More Than
One Central Atom
The tetrahedral
centers of ethane.
10-31
Molecular Shapes with More Than
One Central Atom
The tetrahedral
centers of ethanol.
10-32
SAMPLE PROBLEM 10.9
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
tetrahedral
trigonal planar
10-33
Molecular Polarity
Knowing the shape of the molecule, plus
knowing the polarity (dipole) of the individual
bonds allows the determination of the overall
polarity of the molecule.
10-34
SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules
PROBLEM:
From electronegativity (EN) values 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.
10-35
SAMPLE PROBLEM 10.10
10-36
Predicting the Polarity of Molecules
End of Chapter 10
10-37