Chemical Bonding II
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Lewis Dot Symbols for the Representative Elements &
Noble Gases
2
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
lone pairs
F
F
lone pairs
single covalent bond
3
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
8e- 8e- 8e-
O
O
C
double bonds
double bonds
Triple bond – two atoms share three pairs of electrons
N N
8e- 8e-
triple bond
or
N
N
triple bond
4
Lengths of Covalent Bonds
Bond Lengths
Triple bond < Double Bond < Single Bond
5
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.
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
7
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 - Form double bond and re-check # of e-
O
C
O
O
2 single bonds (2x2) = 4
1 double bond = 4
8 lone pairs (8x2) = 16
Total = 24
8
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
of valence
electrons in
the free atom
-
total number
of nonbonding
electrons
-
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
H
-1
+1
C
O
formal charge
on an atom in
a Lewis
structure
H
=
C – 4 eO – 6 e2H – 2x1 e12 e-
total number
of valence
electrons in
the free atom
-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
of nonbonding
electrons
formal charge
on C
= 4 - 2 - ½ x 6 = -1
formal charge
on O
= 6 - 2 - ½ x 6 = +1
-
1
2
(
total number
of bonding
electrons
)
10
H
H
0
C
formal charge
on an atom in
a Lewis
structure
0
O
=
C – 4 eO – 6 e2H – 2x1 e12 etotal number
of valence
electrons in
the free atom
-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
of nonbonding
electrons
formal charge
on C
= 4 - 0 -½ x 8 = 0
formal charge
on O
= 6 - 4 -½ x 4 = 0
-
1
2
(
total number
of bonding
electrons
)
11
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
12
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
-
13
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
14
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
15
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
16
0 lone pairs on central atom
Cl
Be
Cl
2 atoms bonded to central atom
17
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
18
19
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
20
21
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
0
trigonal
bipyramidal
trigonal
bipyramidal
AB5
5
22
23
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
trigonal
bipyramidal
octahedral
AB5
5
0
trigonal
bipyramidal
AB6
6
0
octahedral
24
25
26
lone-pair vs. lone pair
repulsion
>
lone-pair vs. bonding
pair repulsion
>
bonding-pair vs. bonding
27
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
28
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB4
4
0
tetrahedral
tetrahedral
tetrahedral
trigonal
pyramidal
AB3E
3
1
29
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB4
4
0
tetrahedral
tetrahedral
AB3E
3
1
tetrahedral
trigonal
pyramidal
AB2E2
2
2
tetrahedral
bent
30
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
31
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
32
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB5
5
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
linear
T-shaped
33
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
square
pyramidal
34
VSEPR
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
AB4E2
4
2
octahedral
square
pyramidal
square
planar
35
36
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
37
How does Lewis theory explain the bonds in H2 and F2?
Sharing of two electrons between the two atoms.
Bond Enthalpy
Bond Length
Overlap Of
H2
436.4 kJ/mol
74 pm
2 1s
F2
150.6 kJ/mol
142 pm
2 2p
Valence bond theory – bonds are formed by sharing
of e- from overlapping atomic orbitals.
38
Change in Potential Energy of Two Hydrogen Atoms
as a Function of Their Distance of Separation
39
Change in electron density as two hydrogen atoms
approach each other.
40
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 90o
Actual H-N-H
bond angle is
107.3o
41
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
42
Formation of sp3 Hybrid Orbitals
43
Formation of Covalent Bonds in CH4
44
sp3-Hybridized N Atom in NH3
Predict correct
bond angle
45
Formation of sp Hybrid Orbitals
46
Formation of sp2 Hybrid Orbitals
47
How do I predict the hybridization of the central atom?
1.  Draw the Lewis structure of the molecule.
2.  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
48
49
sp2 Hybridization of Carbon
50
Unhybridized 2pz orbital (gray), which is perpendicular
to the plane of the hybrid (green) orbitals.
51
Bonding in Ethylene, C2H4
Sigma bond (σ) – electron density between the 2 atoms
Pi bond (π) – electron density above and below plane of nuclei
of the bonding atoms
52
Another View of π Bonding in Ethylene, C2H4
53
sp Hybridization of Carbon
54
Bonding in Acetylene, C2H2
55
Another View of the Bonding in Ethylene, C2H4
56
Describe the bonding in CH2O.
H
H
C
O
C – 3 bonded atoms, 0 lone pairs
C – sp2
57
Sigma (σ) and Pi Bonds (π)
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 σ and π bonds are in the acetic acid (vinegar)
molecule CH3COOH?
H
C
H
O
H
C
O
H
σ bonds = 6 + 1 = 7
π bonds = 1
58
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
-
59
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
δ+
e- rich
F
δ-
60
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
61
Dipole Moments and Polar Molecules
electron poor
region
electron rich
region
H
F
δ+
δ-
µ=Qxr
Q is the charge
r is the distance between charges
1 D = 3.36 x 10-30 C m
62
Behavior of Polar Molecules
field off
field on
63
Which of the following molecules have a dipole moment?
H2O, CO2, SO2, and CF4
O
S
dipole moment
polar molecule
dipole moment
polar molecule
F
O
C
O
no dipole moment
nonpolar molecule
C
F
F
F
no dipole moment
nonpolar molecule
64
Does BF3 have a dipole
moment?
65
Does CH2Cl2 have a
dipole moment?
66
67
The Ionic Bond
Ionic bond: the electrostatic force that holds ions together in an
ionic compound.
Li + F
1 22s22p5
1s22s1s
Li
LiF
e- +
Li+ +
Li+ F 1s2 1s22s22p6
[He]
[Ne]
Li+ + e-
F
F -
F -
Li+ F -
68
Electrostatic (Lattice) Energy
Lattice energy (U) is the energy required to completely separate
one mole of a solid ionic compound into gaseous ions.
E=k
Q+Qr
Lattice energy increases
as Q increases and/or
as r decreases.
E is the potential energy
Q+ is the charge on the cation
Q- is the charge on the anion
r is the distance between the ions
Compound Lattice Energy
(kJ/mol)
Q: +2,-1
MgF2
2957
Q: +2,-2
MgO
3938
LiF
1036
LiCl
853
r F- < r Cl69
Born-Haber Cycle for Determining Lattice Energy
°
ΔHoverall
= ΔH1°+ ΔH2 °+ ΔH3°+ ΔH4 °+ ΔH5 °
70
71
Free Electron Model for Metals



Metals are very good at conducting both heat and electricity.
Metals can be described as behaving like a set of nuclei forming
a lattice with a “sea of electrons” shared between all nuclei
(moving freely between them)
This is referred to as the free electron model for metals.
This model explains many of the properties of metals:




Electrical Conductivity: The mobile electrons carry current.
Thermal Conductivity: The mobile electrons can also carry heat.
Malleability and Ductility: Deforming the metal still leaves each
cation surrounded by a “sea of electrons”, so little energy is
required to either stretch or flatten the metal.
Opacity and Reflectance (Shininess): The electrons will have a wide
range of energies, so can absorb and re-emit many different
wavelengths of light.
72
Band Theory for Metals (and Other Solids)


Thus far, whenever we have seen electrons, they have been in
orbitals (atomic orbitals for atoms, molecular orbitals for
molecules). What about the electrons in a metal?
These solids can be treated in a way similar to molecular orbital
theory; however, instead of MOs, we will produce states.
Consider that, in a metal, there are no distinct molecules. You
could almost say that an entire piece of metal is a molecule.
That’s how we’ll be treating them:





We combine atomic orbitals from every atom in the sample to make
states which look rather like very large molecular orbitals.
As in LCAO-MO theory, the number of states produced must equal
the number of atomic orbitals combined.
The Pauli exclusion principle still applies, so each state can only hold
two electrons.
For a metal to conduct electricity, its electrons must be able to gain
enough extra energy to be excited into higher energy states.
The highest energy state when no such excitation has occurred (i.e.
in the ground state metal) is called the Fermi level.
73
Band Theory for Metals (and Other Solids)



So, how do states get formed, and
what do they look like?
Consider lithium. The figure at the
right shows the MOs produced by
linear combination of the 2s
orbitals in Li2, Li3 and Li4.
Note that, for every atom added:




An additional MO is formed
The energies of the MOs get closer
together
When a sample contains a very
large number of Li atoms (e.g.
6.022×1023 atoms in 6.941 g), the
MOs (now called states) produced
will be so close in energy that they
form a band of energy levels.
A band is named for the AOs from
which it was made (e.g. 2s band)
74
Image adapted from “Chemical Structure and Bonding” by R. L. DeKock and H. B. Gray
Band Theory for Metals (and Other Solids)

In an alkali metal, the valence s band is only half full.
e.g. sodium (band structure shown at right)





If there are N atoms of sodium in a sample, there
will be N electrons in 3s orbitals.
There will be N states made from 3s orbitals, each
able to hold two electrons.
As such, N /2 states in the 3s band will be full
and N /2 states will be empty (in ground state Na).
Like all other alkali metals, sodium conducts
electricity well because the valence band is
only half full. It is therefore easy for electrons
in the valence band to be excited into empty
higher energy states.
Since these empty higher energy states are in the
same band, we can say that the valence band for
sodium is also the conduction band.
75
Band Theory for Metals (and Other Solids)

In an alkaline earth metal, the valence s band is full.
e.g. beryllium (band structure shown at right)




If there are N atoms of beryllium in a sample, there
will be 2N electrons in 2s orbitals.
There will be N states made from 2s orbitals, each
able to hold two electrons.
As such, all states in the 2s band will be full and
none will be empty (in ground state Be).
So, why are alkaline earth metals conductors?



While the 2s band in beryllium is full, it overlaps
with the 2p band. (Recall the energy level diagram
on p.9 of Homonuclear Diatomics notes showed
that valence s and p AOs of metals were close.)
As such, some electrons in the valence band
can easily be excited into the conduction band.
In beryllium, the conduction band (band containing
the lowest energy empty states) is the 2p band.
76
Band Theory for Metals (and Other Solids)

What do the bands look like for something that doesn’t
conduct electricity? i.e. for an insulator
e.g. diamond (band structure shown at right)





If there are N atoms of carbon in a sample,
there will be 4N valence electrons.
The valence orbitals of the carbon atoms
will combine to make two bands, each
containing 2N states.
The lower energy band will therefore be
the valence band, containing 4N electrons
(in ground state diamond).
The higher energy band will be the
conduction band, containing no electrons
(in ground state diamond).
The energy gap between the valence band
and the conduction band is big enough that
it would be difficult for an electron in the
valence band to absorb enough energy to
be excited into the conduction band.
77
Band Theory for Metals (and Other Solids)

So, how big does a band gap have to be for a material to be
an insulator?




Our measuring stick is the temperature-dependent kB·T
-23 J/K
  kB is the Boltzmann constant: 1.38065 × 10
  T is the temperature in Kelvin
kB·T is a measure of the average thermal energy of particles in a
sample
As a rule of thumb:
  If the size of the band gap is much larger than kB·T, you
have an insulator. e.g. diamond: ~200×kB·T
  If the size of the band gap is smaller than (or close to) kB·T,
you have a conductor. e.g. sodium: 0×kB·T, tin: 3×kB·T
  If the size of the band gap is about ten times larger than
kB·T, you have a semiconductor. e.g. silicon: ~50×kB·T
Band gaps can be measured by absorption spectroscopy.
The lowest energy light to be absorbed corresponds to the
band gap.
78
Band Theory for Metals (and Other Solids)

There are two broad categories of semiconductors:


Intrinsic Semiconductors
  Naturally have a moderate band gap. A small fraction of the
electrons in the valence band can be excited into the
conduction band. They can carry current.
  The “holes” these electrons leave in the valence band can
also carry current as other electrons in the valence band can
be excited into them.
Extrinsic Semiconductors
  Have had impurities added in order to increase the amount
of current they can conduct. (impurities called dopants;
process called doping)
  The dopants can *either* provide extra electrons *or*
provide extra holes:
  A semiconductor doped to have extra electrons is an
n-type semiconductor (‘n’ is for ‘negative’)
  A semiconductor doped to have extra holes is a
p-type semiconductor (‘p’ is for ‘positive’)
79
Band Theory for Metals (and Other Solids)

How does an n-type semiconductor work?
e.g. silicon ([Ne]3s 23p 2) is doped with phosphorus ([Ne]3s 23p 3)


In silicon, the valence band is completely full and the conduction
band is completely empty.
The phosphorus provides an additional band full of electrons that is
higher in energy than the valence band of silicon. Electrons in this
donor band are more easily excited into the conduction band
(compared to electrons in the valence band of silicon).
80
Band Theory for Metals (and Other Solids)

How does a p-type semiconductor work?
e.g. silicon ([Ne]3s 23p 2) is doped with aluminium ([Ne]3s 23p 1)


In silicon, the valence band is completely full and the conduction
band is completely empty.
The aluminium provides an additional empty band that is lower in
energy than the conduction band of silicon. Electrons in the valence
band of silicon are more easily excited into this acceptor band
(compared to the conduction band of silicon).
81
Band Theory for Metals (and Other Solids)


Through careful choice of both dopant and concentration, the
conductivity of a semiconductor can be fine-tuned. There are
many applications of semiconductors and doping in electronics.
e.g. Diodes



An n-type and a p-type semiconductor are connected.
The acceptor band in the p-type semiconductor gets filled with the
extra electrons from the n-type semiconductor. The extra holes from
the p-type semiconductor thus “move” to the n-type semiconductor.
With negative charge moving one way and positive charge the other,
charge separation builds up and stops both electrons and holes
from moving *unless* the diode is connected
to a circuit:
  If a diode is connected such that the
electrons flow into the n-type
semiconductor, that replenishes the
electrons there and current can flow.
  If a diode is connected such that the
electrons flow into the p-type
semiconductor, electrons will pile up
82
there and the current will stop.
Band Theory for Metals (and Other Solids)

In a photodiode, the p-type semiconductor is exposed to light. This
can excite electrons from the former acceptor band into the
conduction band. They are then attracted to the neighbouring
ntype semiconductor (which has built up a slight positive charge).
This causes current to flow, and is how many solar cells work.
83