Ch 10

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Chapter 10 The Shapes of Molecules
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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
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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. Well
see later that the octet rule is a natural
consequence of having four valence orbitals (s,
3ps) and the limitation of two electrons per
orbital.
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The steps in converting a molecular formula into
a Lewis structure.
Place atom with lowest EN in center
Molecular formula
Step 1
Atom placement
Add A-group numbers
Step 2
Sum of valence e-
Draw single bonds. Subtract 2e- for each bond.
Step 3
Give each atom 8e- (2e- for H)
Remaining valence e-
Step 4
Lewis structure
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Molecular formula
For NF3
Atom placement


N 5e-
F
F





Sum of valence e-
F 7e-
X 3 21e-
N
Total 26e-
F



Remaining valence e-
Lewis structure
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SAMPLE PROBLEM 10.1
Writing Lewis Structures for Molecules with One
Central Atom
SOLUTION
Cl
Step 1 Carbon has the lowest EN and is the
central atom. The other atoms are placed
around it.
C
Cl
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-.
Make bonds and fill in remaining valence
electrons placing 8e- around each atom.
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SAMPLE PROBLEM 10.2
Writing Lewis Structure for Molecules with More
than One Central Atom
SOLUTION
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

H
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SAMPLE PROBLEM 10.3
Writing Lewis Structures for Molecules with
Multiple Bonds.
PROBLEM
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
PLAN
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.
SOLUTION
(a) There are 2(4) 4(1) 12 valence e-. H
can have only one bond per atom.
. .
(b) N2 has 2(5) 10 valence e-. Therefore a
triple bond is required to make the octet around
each N.
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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.
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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)
. .
. .
OA

OB

OC
. .
. .
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Resonance (continued)
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.
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Resonance (continued)
EXAMPLE NCO- has 3 possible resonance forms -
-1
-1
-1
. .
. .
. .
. .
N C O
N C O
N C O






. .
. .
B
C
formal charges
-2
0
1
-1
0
0
0
0
-1
-1
-1
-1
. .
. .
. .
. .
N C O
N C O
N C O






. .
. .
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.
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Lewis Structures - Exceptions to the Octet Rule
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Using bond energies to calculate DH0rxn
DH0rxn DH0reactant bonds broken DH0product
bonds formed
Enthalpy, H
DH01 sum of BE
DH02 - sum of BE
DH0rxn
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Using bond energies to calculate DH0rxn for
combustion of methane
BOND BREAKAGE
4BE(C-H) 1652kJ
2BE(O2) 996kJ
DH0(bond breaking) 2648kJ
BOND FORMATION
4-BE(O-H) -1868kJ
Enthalpy,H
DH0(bond forming) -3466kJ
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SAMPLE PROBLEM 10.6
Calculating Enthalpy Changes from Bond Energies
SOLUTION
bonds broken
bonds formed
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SAMPLE PROBLEM 10.6
Calculating Enthalpy Changes from Bond Energies
continued
bonds broken
bonds formed
4 C-H 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
3 H-Cl 3 mol(-427 kJ/mol) -1281 kJ
DH0bonds broken 2381 kJ
DH0bonds formed -2711 kJ
DH0reaction DH0bonds broken DH0bonds formed
2381 kJ (-2711 kJ) - 330 kJ
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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.
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Electron-group repulsions and the five basic
molecular shapes.
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Looking at the Five Shapes in Detail
Examples CS2, HCN, BeF2
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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.
Multiple bonds count just as one group but are
larger than a single bond. The result is some
compression of the other bond angles.
Likewise lone pairs repel bonding pairs more
strongly than bonding pairs repel each other also
compressing the other angles.
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The two molecular shapes of the trigonal planar
electron-group arrangement.
Examples SO2, O3, PbCl2, SnBr2
Examples SO3, BF3, NO3-, CO32-
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The three molecular shapes of the tetrahedral
electron-group arrangement.
Examples CH4, SiCl4, SO42-, ClO4-
NH3 PF3 ClO3 H3O
H2O OF2 SCl2
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The four molecular shapes of the trigonal
bipyramidal electron-group arrangement.
PF5 AsF5 SOF4
SF4 XeO2F2 IF4 IO2F2-
XeF2 I3- IF2-
ClF3 BrF3
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The three molecular shapes of the octahedral
electron-group arrangement.
SF6 IOF5
BrF5 TeF5- XeOF4
XeF4 ICl4-
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The steps in determining a molecular shape.
Molecular formula
Step 1
Lewis structure
Count all e- groups around central atom (A)
Step 2
Electron-group arrangement
Note lone pairs and double bonds
Step 3
Count bonding and nonbonding e- groups separately.
Bond angles
Step 4
Molecular shape (AXmEn)
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Lewis structures and molecular shapes
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Molecular Shapes with More Than One Central Atom
The tetrahedral centers of ethane.
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Molecular Shapes with More Than One Central Atom
The tetrahedral centers of ethanol.
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SAMPLE PROBLEM 10.9
Predicting Molecular Shapes with More Than One
Central Atom
SOLUTION
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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.
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SAMPLE PROBLEM 10.10
Predicting the Polarity of Molecules
(a) Ammonia, NH3
(b) Boron trifluoride, BF3
(c) Carbonyl sulfide, COS (atom sequence SCO)
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SAMPLE PROBLEM 10.10
Predicting the Polarity of Molecules
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End of Chapter 10