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Chemical Bonding I:
Basic Concepts
Chapter 9
Intermolecular Forces
Intermolecular forces are attractive forces between molecules.
Intramolecular forces hold atoms together in a molecule.
Intermolecular vs Intramolecular
•
41 kJ to vaporize 1 mole of water (inter)
•
930 kJ to break all O-H bonds in 1 mole of water (intra)
“Measure” of intermolecular force
Generally,
intermolecular
forces are much
weaker than
intramolecular
forces.
boiling point
melting point
DHvap
DHfus
DHsub
11.2
Valence electrons are the outer shell electrons of an
atom. The valence electrons are the electrons that
particpate in chemical bonding.
Group
e- configuration
# of valence e-
1A
ns1
1
2A
ns2
2
3A
ns2np1
3
4A
ns2np2
4
5A
ns2np3
5
6A
ns2np4
6
7A
ns2np5
7
9.1
The Ionic Bond
Li + F
1 22s22p5
1s22s1s
e- +
Li+ +
Li+ F [He]
1s
1s2[2Ne]
2s22p6
Li
Li+ + e-
F
F -
F -
Li+ F -
9.2
9.1
A covalent bond is a chemical bond in which two or more
electrons are shared by two atoms.
Why should two atoms share electrons?
F
+
7e-
F
F F
7e-
8e- 8e-
Lewis structure of F2
single covalent bond
lone pairs
F
F
lone pairs
single covalent bond
lone pairs
F F
lone pairs
9.4
Lewis structure of water
H
+
O +
H
single covalent bonds
H O H
or
H
O
H
2e-8e-2eDouble bond – two atoms share two pairs of electrons
O C O
or
O
O
C
double bonds
- 8e8e- 8ebonds
double
Triple bond – two atoms share three pairs of electrons
N N
triple
bond
8e-8e
or
N
N
triple bond
9.4
Lengths of Covalent Bonds
Bond
Type
Bond
Length
(pm)
C-C
154
CC
133
CC
120
C-N
143
CN
138
CN
116
Bond Lengths
Triple bond < Double Bond < Single Bond
9.4
9.4
Writing Lewis Structures
1. Draw skeletal structure of compound showing
what atoms are bonded to each other. Put least
electronegative element in the center.
2. Count total number of valence e-. Add 1 for
each negative charge. Subtract 1 for each
positive charge.
3. Complete an octet for all atoms except
hydrogen
4. If structure contains too many electrons, form
double and triple bonds on central atom as
needed.
9.6
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on N and F atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
F
N
F
F
9.6
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on C and O atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
Step 5 - Too many electrons, form double bond and re-check # of e-
O
C
O
2 single bonds (2x2) = 4
1 double bond = 4
8 lone pairs (8x2) = 16
Total = 24
O
9.6
Two possible skeletal structures of formaldehyde (CH2O)
H
C
O
H
H
C
H
O
An atom’s formal charge is the difference between the
number of valence electrons in an isolated atom and the
number of electrons assigned to that atom in a Lewis
structure.
formal charge
on an atom in
a Lewis
structure
=
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
)
The sum of the formal charges of the atoms in a molecule
or ion must equal the charge on the molecule or ion.
9.7
H
-1
+1
C
O
formal charge
on an atom in
a Lewis
structure
H
=
C – 4 eO – 6 e2H – 2x1 e12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
)
formal charge
= 4 -2 -½ x 6 = -1
on C
formal charge
= 6 -2 -½ x 6 = +1
on O
9.7
H
H
0
C
formal charge
on an atom in
a Lewis
structure
0
O
=
C – 4 eO – 6 e2H – 2x1 e12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
)
formal charge
= 4 - 0 -½ x 8 = 0
on C
formal charge
= 6 -4 -½ x 4 = 0
on O
9.7
Formal Charge and Lewis Structures
1. For neutral molecules, a Lewis structure in which there
are no formal charges is preferable to one in which
formal charges are present.
2. Lewis structures with large formal charges are less
plausible than those with small formal charges.
3. Among Lewis structures having similar distributions of
formal charges, the most plausible structure is the one in
which negative formal charges are placed on the more
electronegative atoms.
Which is the most likely Lewis structure for CH2O?
H
-1
+1
C
O
H
H
H
0
C
0
O
9.7
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 are the molecular geometries of SO2 and SF4?
O
S
AB2E
bent
F
O
F
S
F
AB4E
F
distorted
tetrahedron
10.1
Valence shell electron pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic
repulsions 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
B
10.1
Cl
Be
Cl
lone pairs
on to
central
atom
20
atoms
bonded
central
atom
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
0
trigonal
planar
trigonal
planar
AB3
3
10.1
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
10.1
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
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
10.1
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
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
AB6
6
0
octahedral
octahedral
10.1
10.1
10.1
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
10.1
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB4
4
0
AB3E
3
1
Arrangement of
electron pairs
Molecular
Geometry
tetrahedral
tetrahedral
tetrahedral
trigonal
pyramidal
10.1
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
H
10.1
VSEPR
Class
AB5
AB4E
# of atoms
bonded to
central atom
5
4
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
0
trigonal
bipyramidal
trigonal
bipyramidal
1
trigonal
bipyramidal
distorted
tetrahedron
10.1
VSEPR
Class
AB5
# of atoms
bonded to
central atom
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
F
Cl
F
10.1
VSEPR
Class
AB5
# of atoms
bonded to
central atom
5
# lone
pairs on
central atom
0
AB4E
4
1
AB3E2
3
2
AB2E3
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
10.1
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
Br
F
F
10.1
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
F
Xe
F
F
10.1
10.1
A resonance structure is one of two or more Lewis structures
for a single molecule that cannot be represented accurately by
only one Lewis structure.
O
O
+
-
-
O
O
+
O
O
What are the resonance structures of the
carbonate (CO32-) ion?
-
O
C
O
O
-
O
C
O
O
-
-
-
O
C
O
O
-
9.8
Exceptions to the Octet Rule
The Incomplete Octet
BeH2
BF3
B – 3e3F – 3x7e24e-
Be – 2e2H – 2x1e4e-
F
B
H
F
Be
H
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24
F
9.9
Exceptions to the Octet Rule
Odd-Electron Molecules
NO
N – 5eO – 6e11e-
N
O
The Expanded Octet (central atom with principal quantum number n > 2)
SF6
S – 6e6F – 42e48e-
F
F
F
S
F
F
F
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
9.9
Classification of bonds by difference in electronegativity
Difference
Bond Type
0
Covalent
2
0 < and <2
Ionic
Polar Covalent
Increasing difference in electronegativity
Covalent
Polar Covalent
share e-
partial transfer of e-
Ionic
transfer e-
9.5
Classify the following bonds as ionic, polar covalent,
or covalent: The bond in CsCl; the bond in H2S; and
the NN bond in H2NNH2.
Cs – 0.7
Cl – 3.0
3.0 – 0.7 = 2.3
Ionic
H – 2.1
S – 2.5
2.5 – 2.1 = 0.4
Polar Covalent
N – 3.0
N – 3.0
3.0 – 3.0 = 0
Covalent
9.5
Polar covalent bond or polar bond is a covalent
bond with greater electron density around one of the
two atoms
electron poor
region
H
electron rich
region
F
e- poor
H
d+
e- rich
F
d-
9.5
Electronegativity is the ability of an atom to attract
toward itself the electrons in a chemical bond.
Electron Affinity - measurable, Cl is highest
X (g) + e-
X-(g)
Electronegativity - relative, F is highest
9.5
9.5
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
10.2
10.2
10.2
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
10.2
Does CH2Cl2 have
a dipole moment?
10.2
Intermolecular Forces
Dipole-Dipole Forces
Attractive forces between polar molecules
Orientation of Polar Molecules in a Solid
11.2
Intermolecular Forces
Ion-Dipole Forces
Attractive forces between an ion and a polar molecule
Ion-Dipole Interaction
11.2
11.2
Intermolecular Forces
Dispersion Forces
Attractive forces that arise as a result of temporary
dipoles induced in atoms or molecules
ion-induced dipole interaction
dipole-induced dipole interaction
11.2
Intermolecular Forces
Dispersion Forces Continued
Polarizability is the ease with which the electron distribution
in the atom or molecule can be distorted.
Polarizability increases with:
•
greater number of electrons
•
more diffuse electron cloud
Dispersion
forces usually
increase with
molar mass.
11.2
What type(s) of intermolecular forces exist between
each of the following molecules?
HBr
HBr is a polar molecule: dipole-dipole forces. There are
also dispersion forces between HBr molecules.
CH4
CH4 is nonpolar: dispersion forces.
S
SO2
SO2 is a polar molecule: dipole-dipole forces. There are
also dispersion forces between SO2 molecules.
11.2
Intermolecular Forces
Hydrogen Bond
The hydrogen bond is a special dipole-dipole interaction
between they hydrogen atom in a polar N-H, O-H, or F-H bond
and an electronegative O, N, or F atom.
A
H…B
or
A
H…A
A & B are N, O, or F
11.2
Hydrogen Bond
11.2
Why is the hydrogen bond considered a
“special” dipole-dipole interaction?
Decreasing molar mass
Decreasing boiling point
11.2
Types of Crystals
Molecular Crystals
• Lattice points occupied by molecules
• Held together by intermolecular forces
• Soft, low melting point
• Poor conductor of heat and electricity
11.6
Types of Crystals
Metallic Crystals
• Lattice points occupied by metal atoms
• Held together by metallic bonds
• Soft to hard, low to high melting point
• Good conductors of heat and electricity
Cross Section of a Metallic Crystal
nucleus &
inner shell emobile “sea”
of e-
11.6
Types of Crystals
Ionic Crystals
• Lattice points occupied by cations and anions
• Held together by electrostatic attraction
• Hard, brittle, high melting point
• Poor conductor of heat and electricity
CsCl
ZnS
CaF2
11.6
Electrostatic (Lattice) Energy
Lattice energy (E) is the energy required to completely separate
one mole of a solid ionic compound into gaseous ions.
Q+Q E=k
r
Q+ is the charge on the cation
Q- is the charge on the anion
r is the distance between the ions
Lattice energy (E) increases
as Q increases and/or
as r decreases.
cmpd
MgF2
MgO
LiF
LiCl
lattice energy
2957 Q= +2,-1
3938 Q= +2,-2
1036
853
r F < r Cl
9.3
Types of Crystals
Covalent Crystals
• Lattice points occupied by atoms
• Held together by covalent bonds
• Hard, high melting point
• Poor conductor of heat and electricity
carbon
atoms
diamond
graphite
11.6
Types of Crystals
11.6
Hybridization – mixing of two or more atomic
orbitals to form a new set of hybrid orbitals.
1. Mix at least 2 nonequivalent atomic orbitals (e.g. s
and p). Hybrid orbitals have very different shape
from original atomic orbitals.
2. Number of hybrid orbitals is equal to number of
pure atomic orbitals used in the hybridization
process.
3. Covalent bonds are formed by:
a. Overlap of hybrid orbitals with atomic orbitals
b. Overlap of hybrid orbitals with other hybrid
orbitals
10.4
10.4
10.4
Predict correct
bond angle
10.4
Formation of sp Hybrid Orbitals
10.4
Formation of sp2 Hybrid Orbitals
10.4
How do I predict the hybridization of the central atom?
Count the number of lone pairs AND the number
of atoms bonded to the central atom
# of Lone Pairs
+
# of Bonded Atoms
Hybridization
Examples
2
sp
BeCl2
3
sp2
BF3
4
sp3
CH4, NH3, H2O
5
sp3d
PCl5
6
sp3d2
SF6
10.4
Valence Bond Theory and NH3
N – 1s22s22p3
3 H – 1s1
If the bonds form from overlap of 3 2p orbitals on nitrogen
with the 1s orbital on each hydrogen atom, what would
the molecular geometry of NH3 be?
If use the
3 2p orbitals
predict 900
Actual H-N-H
bond angle is
107.30
10.4
10.5
10.5
Sigma (s) and Pi Bonds (p)
1 sigma bond
Single bond
Double bond
1 sigma bond and 1 pi bond
Triple bond
1 sigma bond and 2 pi bonds
How many s and p bonds are in the acetic acid
(vinegar) molecule CH3COOH?
H
C
O
H
C
O
H
s bonds = 6 + 1 = 7
p bonds = 1
H
10.5
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
9.10
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 kJ
Average OH bond energy =
2
9.10
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)
9.10
H2 (g) + Cl2 (g)
2HCl (g)
2H2 (g) + O2 (g)
2H2O (g)
9.10
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
9.10