The Shapes of Molecules

1
Chapter 10
The Shapes of Molecules
2
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
3
Figure 10.1
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
4
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
5
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.
6
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
7
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 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.
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.
8
Resonance Delocalized Electron-Pair Bonding
O3 can be drawn in 2 ways -
Neither structure is actually correct but can be
drawn to represent a structure which is a hybrid
of the two - a resonance structure.
Resonance structures have the same relative atom
placement but a difference in the locations of
bonding and nonbonding electron pairs.
9
SAMPLE PROBLEM 10.4
Writing Resonance Structures
PROBLEM
Write resonance structures for the nitrate ion,
NO3-.
PLAN
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
Nitrate has 1(5) 3(6) 1 24 valence e-
N does not have an octet a pair of e- will move
in to form a double bond.
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)
11
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.
12
Resonance (continued)
EXAMPLE NCO- has 3 possible resonance forms -
formal charges
-2
0
1
-1
0
0
0
0
-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.
13
SAMPLE PROBLEM 10.5
Writing Lewis Structures for Exceptions to the
Octet Rule.
PLAN
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
1
0
(b) BFCl2 will have only 1 Lewis structure.
0
0
0
0
0
0
0
0
0
0
more stable
0
0
lower formal charges
14
10.2 Using Lewis Structures and Bond Energies to
Calculate Heats of Reaction
15
The enthalpy change required to break a
particular bond in one mole of gaseous molecules
is the bond energy.
Bond Energy
16
Average bond energy in polyatomic molecules
17
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)
18
DH0 SBE(reactants) SBE(products)
DH0 436.4 156.9 2 x 568.2 -543.1 kJ
19
Figure 10.2
Using bond energies to calculate DH0rxn
DH0rxn DH0reactant bonds broken DH0product
bonds formed
Enthalpy, H
DH01 sum of BE
DH02 - sum of BE
DH0rxn
20
SAMPLE PROBLEM 10.6
Calculating Enthalpy Changes from Bond Energies
SOLUTION
bonds broken
bonds formed
21
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
22
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?
The standard enthalpy of formation of any element
in its most stable form is zero.
23
(No Transcript)
24
Hesss 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 doesnt matter
how you get there, only where you start and end.)
25
Calculate the standard enthalpy of formation of
CS2 (l) given that
1. Write the enthalpy of formation reaction for
CS2
2. Add the given rxns so that the result is the
desired rxn.
26
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.
27
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?
28
Valence shell electron pair repulsion (VSEPR)
model
29
Valence shell electron pair repulsion (VSEPR)
model
Predict the geometry of the molecule from the
electrostatic repulsion between the electron
(bonding and nonbonding) pairs.
AB2
2
0
Examples CS2, HCN, BeF2
30
(No Transcript)
31
VSEPR
AB2
2
0
linear
linear
AB3
3
0
Examples SO3, BF3, NO3-, CO32-
32
10.1
33
VSEPR
AB2
2
0
linear
linear
AB4
4
0
Examples CH4, SiCl4, SO42-, ClO4-
34
(No Transcript)
35
VSEPR
AB2
2
0
linear
linear
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
Examples PF5 AsF5
SOF4
36
(No Transcript)
37
VSEPR
AB2
2
0
linear
linear
AB4
4
0
tetrahedral
tetrahedral
AB6
6
0
SF6 IOF5
38
(No Transcript)
39
Figure 10.5
Electron-group repulsions and the five basic
molecular shapes.
40
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.
1200
larger EN
1200
ideal
greater electron density
lone pairs repel bonding pairs more strongly than
bonding pairs repel each other
950
41
(No Transcript)
42
VSEPR
trigonal planar
trigonal planar
AB3
3
0
AB2E
2
1
43
VSEPR
AB4
4
0
tetrahedral
tetrahedral
AB3E
3
1
NH3 PF3 ClO3 H3O
44
VSEPR
AB4
4
0
tetrahedral
tetrahedral
AB2E2
2
2
H2O OF2 SCl2
45
VSEPR
trigonal bipyramidal
trigonal bipyramidal
AB5
5
0
AB4E
4
1
SF4 XeO2F2 IF4 IO2F2-
46
VSEPR
trigonal bipyramidal
trigonal bipyramidal
AB5
5
0
AB3E2
3
2
ClF3 BrF3
47
VSEPR
trigonal bipyramidal
trigonal bipyramidal
AB5
5
0
AB2E3
2
3
H2O OF2 SCl2
48
VSEPR
AB5E
5
1
BrF5 TeF5- XeOF4
49
VSEPR
AB4E2
4
2
XeF4 ICl4-
50
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.

AB4E
AB2E
distorted tetrahedron
bent
51
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.
1200
larger EN
1200
ideal
greater electron density
lone pairs repel bonding pairs more strongly than
bonding pairs repel each other
950
52
Figure 10.4
A balloon analogy for the mutual repulsion of
electron groups.
53
Figure 10.7
The two molecular shapes of the trigonal planar
electron-group arrangement.
Examples SO2, O3, PbCl2, SnBr2
Examples SO3, BF3, NO3-, CO32-
54
Figure 10.8
The three molecular shapes of the tetrahedral
electron-group arrangement.
Examples CH4, SiCl4, SO42-, ClO4-
NH3 PF3 ClO3 H3O
H2O OF2 SCl2
55
Figure 10.9
Lewis structures and molecular shapes
56
Figure 10.10
The four molecular shapes of the trigonal
bipyramidal electron-group arrangement.
PF5 AsF5 SOF4
SF4 XeO2F2 IF4 IO2F2-
XeF2 I3- IF2-
ClF3 BrF3
57
Figure 10.11
The three molecular shapes of the octahedral
electron-group arrangement.
SF6 IOF5
BrF5 TeF5- XeOF4
XeF4 ICl4-
58
Figure 10.12
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)
59
SAMPLE PROBLEM 10.7
Predicting Molecular Shapes with Two, Three, or
Four Electron Groups
The shape is based upon the tetrahedral
arrangement.
The F-P-F bond angles should be lt109.50 due to
the repulsion of the nonbonding electron pair.
The final shape is trigonal pyramidal.
lt109.50
The type of shape is AX3E
60
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.
C does not have an octet a pair of nonbonding
electrons will move in from the O to make a
double bond.
Type AX3
The shape for an atom with three atom attachments
and no nonbonding pairs on the central atom is
trigonal planar.
The Cl-C-Cl bond angle will be less than 1200 due
to the electron density of the CO.
61
SAMPLE PROBLEM 10.8
Predicting Molecular Shapes with Five or Six
Electron Groups
(b) BrF5 - 42 valence e- 5 bonding pairs and 1
nonbonding pair on central atom. Shape is AX5E,
square pyramidal.
62
Figure 10.13
The tetrahedral centers of ethane.
63
Figure 10.13
The tetrahedral centers of ethanol.
64
SAMPLE PROBLEM 10.9
Predicting Molecular Shapes with More Than One
Central Atom
SOLUTION
65
Dipole Moments and Polar Molecules
electron rich region
electron poor region
m Q x r
Q is the charge
r is the distance between charges
1 D 3.36 x 10-30 C m
66
Electric field OFF
Electric field ON
67
(No Transcript)
68
dipole moment polar molecule
dipole moment polar molecule
no dipole moment nonpolar molecule
no dipole moment nonpolar molecule
69
(No Transcript)
70
SAMPLE PROBLEM 10.10
Predicting the Polarity of Molecules
(a) Ammonia, NH3
(b) Boron trifluoride, BF3
(c) Carbonyl sulfide, COS (atom sequence SCO)
The dipoles reinforce each other, so the overall
molecule is definitely polar.
ENN 3.0
ENH 2.1
molecular dipole
bond dipoles
71
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 (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.
1200
(c) COS is linear. C and S have the same EN
(2.0) but the CO bond is quite polar(DEN) so the
molecule is polar overall.
72
(No Transcript)
73
(No Transcript)