Chapter 10
The Shapes of Molecules
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
Figure 10.1
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
Give each
atom 8e(2e- for H)
Sum of
valence e-
:
: F:
: F:
:
Atom
placement
For NF3
:
Molecular
formula
N
Remaining
valence eLewis
structure
:
: 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:
Make bonds and fill in remaining valence
electrons placing 8e- around each atom.
F
:
F
:
: Cl :
:
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
:Cl C
:
Step 1: Carbon has the lowest EN and is the central atom.
The other atoms are placed around it.
Cl
: F:
F:
:
PLAN: Follow the steps outlined in Figure 10.1
:
PROBLEM:
Writing Lewis Structures for Molecules with
One Central Atom
SAMPLE PROBLEM 10.2
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
:
SOLUTION:
Write the Lewis structure for methanol (molecular formula
CH4O), an important industrial alcohol that is being used as a
gasoline alternative in car engines.
H
C
O
:
PROBLEM:
Writing Lewis Structure for Molecules with
More than One Central Atom
H
H
SAMPLE PROBLEM 10.3
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
PROBLEM:
PLAN:
For molecules with multiple bonds, there is a Step 5 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
:
H
C
C
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
.
:
:
:.
N
.
N
N
:
.:
N
.
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
A
O
O
O
C
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.
SAMPLE PROBLEM 10.4
PROBLEM:
PLAN:
Write resonance structures for the nitrate ion, NO3-.
After Steps 1-4, go to 5 and then see if other structures can be
drawn in which the electrons can be delocalized over more than
two atoms.
SOLUTION:
O
Writing Resonance Structures
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
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 of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
B
For OA
# valence
e-
O
# nonbonding
# bonding
e-
# valence e- = 6
O
=6
e-
For OC
=4
= 4 X 1/2 = 2
Formal charge = 0
A
For OB
O
# nonbonding e- = 6
C
# valence
# bonding e- = 2 X 1/2 = 1
Formal charge = -1
e-
=6
# 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 C
O
B
O
C
formal charges
-2
0
N C
+1
O
-1
0
N C
0
O
0
0
N C
-1
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.
SAMPLE PROBLEM 10.5
PROBLEM:
PLAN:
Writing Lewis Structures for Exceptions to the
Octet Rule.
Write Lewis structures for (a) H3PO4 and (b) BFCl2. In (a)
decide on the most likely structure.
Draw the Lewis structures for the molecule and determine if there is
an element which can be an exception to the octet rule. Note that
(a) contains P which is a Period-3 element and can have an
expanded valence shell.
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
0
O H 0
0
0 H O
0
O
0
P
O H 0
0
0 O
H more stable
0
lower formal charges
(b) BFCl2 will have only 1
Lewis structure.
F
B
Cl
Cl
10.2 Using Lewis Structures and Bond Energies to
Calculate Heats of Reaction
The enthalpy change required to break a particular bond in
one mole of gaseous molecules is the bond energy.
Bond Energy
DH0 = 436.4 kJ
H2 (g)
H (g) + H (g)
Cl2 (g)
Cl (g) + Cl (g) DH0 = 242.7 kJ
HCl (g)
H (g) + Cl (g) DH0 = 431.9 kJ
O2 (g)
O (g) + O (g) DH0 = 498.7 kJ
O
O
N2 (g)
N (g) + N (g) DH0 = 941.4 kJ
N
N
Bond Energies
Single bond < Double bond < Triple bond
Average bond energy in polyatomic molecules
H2O (g)
OH (g)
H (g) + OH (g) DH0 = 502 kJ
H (g) + O (g)
DH0 = 427 kJ
502 + 427
= 464.5 kJ
Average OH bond energy =
2
Bond Energies (BE) and Enthalpy changes in reactions
Imagine reaction proceeding by breaking all bonds in the
reactants and then using the gaseous atoms to form all the
bonds in the products.
DH0 = total energy input – total energy released
= SBE(reactants) – SBE(products)
Use bond energies to calculate the enthalpy change for:
H2 (g) + F2 (g)
2HF (g)
DH0 = SBE(reactants) – SBE(products)
Type of
bonds broken
H
H
F
F
Type of
bonds formed
H
F
Number of
bonds broken
Bond energy
(kJ/mol)
Energy
change (kJ)
1
1
436.4
156.9
436.4
156.9
Number of
bonds formed
Bond energy
(kJ/mol)
Energy
change (kJ)
2
568.2
1136.4
DH0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ
Figure 10.2
Using bond energies to calculate DH
0
rxn
Enthalpy, H
DH0rxn = DH0reactant bonds broken + DH0product bonds formed
DH01 = + sum of BE
DH0rxn
DH02 = - sum of BE
SAMPLE PROBLEM 10.6
PROBLEM:
Calculating Enthalpy Changes from Bond
Energies
Use Table 9.2 (button at right) 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:
H
Cl
H C H
+
3
Cl
H
Cl
H C Cl
+
3 H
Cl
bonds broken
bonds formed
Cl
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
Because there is no way to measure the absolute value of
the enthalpy of a substance, must I measure the enthalpy
change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of
formation (DH0f ) as a reference point for all enthalpy
expressions.
Standard enthalpy of formation (DH0f) is the heat change
that results when one mole of a compound is formed from
its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its
most stable form is zero.
DH0f (O2) = 0
DH0f (C, graphite) = 0
DH0f (O3) = 142 kJ/mol
DH0f (C, diamond) = 1.90 kJ/mol
0 ) as the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
Hess’s Law: When reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)
Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
DH0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. What is the heat released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ
= 2973 kJ/mol C6H6
2 mol
The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which could be used for
melting ice?
Which could be used for a
cold pack?
Valence shell electron pair repulsion
(VSEPR) model:
Valence shell electron pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic
repulsion between the electron (bonding and nonbonding) pairs.
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
B
Examples:
CS2, HCN, BeF2
B
Cl
Be
Cl
lone pairs
on to
central
atom
20
atoms
bonded
central
atom
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
0
trigonal
planar
trigonal
planar
AB3
3
Examples:
SO3, BF3, NO3-, CO32-
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
tetrahedral
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
Examples:
CH4, SiCl4, SO42-, ClO4-
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
Examples: PF5
AsF5
SOF4
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
AB6
6
0
octahedral
octahedral
SF6
IOF5
Figure 10.5
Electron-group repulsions and the five basic molecular shapes.
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.
H
Effect of Double Bonds
1200
ideal
1200
O
1160
C
H
greater
electron
density
real
Effect of Nonbonding Pairs
lone pairs repel bonding pairs
more strongly than bonding
pairs repel each other
H
larger EN
C
H
1220
Sn
Cl
Cl
950
O
lone-pair vs. lone pair
lone-pair vs. bonding
bonding-pair vs. bonding
>
>
repulsion
pair repulsion
pair repulsion
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB3
3
0
AB2E
2
1
Arrangement of
electron pairs
Molecular
Geometry
trigonal
planar
trigonal
planar
trigonal
planar
bent
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB4
4
0
AB3E
3
1
NH3
PF3
ClO3
H 3 O+
Arrangement of
electron pairs
Molecular
Geometry
tetrahedral
tetrahedral
tetrahedral
trigonal
pyramidal
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB4
4
0
Arrangement of
electron pairs
Molecular
Geometry
tetrahedral
tetrahedral
AB3E
3
1
tetrahedral
trigonal
pyramidal
AB2E2
2
2
tetrahedral
bent
O
H 2O
OF2
SCl2
H
H
VSEPR
Class
# of atoms
bonded to
central atom
AB5
5
AB4E
SF4
XeO2F2
IF4+
IO2F2-
4
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
0
trigonal
bipyramidal
trigonal
bipyramidal
1
trigonal
bipyramidal
distorted
tetrahedron
VSEPR
# of atoms
bonded to
central atom
Class
AB5
5
# lone
pairs on
central atom
0
AB4E
4
1
AB3E2
3
2
Arrangement of
electron pairs
Molecular
Geometry
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
distorted
tetrahedron
T-shaped
F
ClF3
BrF3
F
Cl
F
VSEPR
Class
AB5
# of atoms
bonded to
central atom
5
# lone
pairs on
central atom
0
AB4E
4
1
AB3E2
3
2
AB2E3
H2O
OF2
SCl2
2
3
Arrangement of
electron pairs
Molecular
Geometry
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
distorted
tetrahedron
trigonal
bipyramidal
T-shaped
linear
I
I
I
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
square
pyramidal
F
F
F
Arrangement of
electron pairs
Molecular
Geometry
BrF5
TeF5XeOF
4
Br
F
F
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
AB4E2
4
2
octahedral
square
pyramidal
square
planar
Arrangement of
electron pairs
Molecular
Geometry
F
XeF4
ICl4
-
F
Xe
F
F
Predicting Molecular Geometry
1. Draw Lewis structure for molecule.
2. Count number of lone pairs on the central atom and
number of atoms bonded to the central atom.
3. Use VSEPR to predict the geometry of the molecule.
What is the molecular geometry of SO2 and SF4?
O
S
AB2E
bent
F
O
F
S
F
AB4E
F
distorted
tetrahedron
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.
H
Effect of Double Bonds
1200
ideal
1200
O
1160
C
H
greater
electron
density
real
Effect of Nonbonding Pairs
lone pairs repel bonding pairs
more strongly than bonding
pairs repel each other
H
larger EN
C
H
1220
Sn
Cl
Cl
950
O
Figure 10.4
A balloon analogy for the mutual repulsion of electron groups.
Figure 10.7
The two molecular shapes of the trigonal planar electrongroup arrangement.
Class
Examples:
SO2, O3, PbCl2, SnBr2
Shape
Examples:
SO3, BF3, NO3-, CO32-
Figure 10.8
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 10.9
Lewis structures and molecular shapes
Figure 10.10
The four molecular shapes of the trigonal bipyramidal
electron-group arrangement.
PF5
SF4
AsF5
XeO2F2
SOF4
IF4+
IO2F2-
ClF3
XeF2
BrF3
I3 -
IF2-
Figure 10.11
The three molecular shapes of the octahedral electrongroup arrangement.
SF6
IOF5
BrF5
TeF5
-
XeOF4
XeF4
ICl4-
Figure 10.12
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 10.7
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
<109.50
The type of shape is
AX3E
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.
SAMPLE PROBLEM 10.7
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 10.8
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
Figure 10.13
The tetrahedral
centers of ethane.
Figure 10.13
The tetrahedral
centers of ethanol.
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
H
H C
H
O
C
H
C H
H
tetrahedral
trigonal planar
O
H
C
H C
C
H
HH
>1200
H
<1200
Dipole Moments and Polar Molecules
electron poor
region
electron rich
region
H
F
d+
d-
m=Qxr
Q is the charge
r is the distance between charges
1 D = 3.36 x 10-30 C m
Electric field OFF
Electric field ON
Which of the following molecules have a dipole moment?
H2O, CO2, SO2, and CH4
O
S
dipole moment
polar molecule
dipole moment
polar molecule
H
O
C
O
no dipole moment
nonpolar molecule
H
C
H
H
no dipole moment
nonpolar molecule
SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules
PROBLEM:
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
SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules
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