Covalent Bonding: Orbitals
Chapter 09
1
The four bonds around C are of equal length and Energy
2
Can you explain this based on your knowledge
of electron energy levels?
6C
1s2 2s2 2p2
Bonding from s is
Different from bonding
From p. In addition the
Angles should be within
900!
3
How to generate four equal orbitals?
4
Hint:
A key to wave mechanics is
superposition
Which is creating new waves from interference of old ones
5
Let’s do some mixing
6
7
8
9
Cross
section of
3
an sp
orbital.
10
An energy-level diagram showing the
3
formation of four sp orbitals.
11
Hybridization
Ground state of C
Promote electron at n=2
Hybridize at n=2
1s2 2s2 2p2
s
px
py
pz
s
px
py
pz
sp3
sp3
2s12p3
sp3 sp3 sp3 sp3
sp3 sp3
4sp3 orbitals of equal length, energy and in tetrahedral shape
12
Hybridization
The mixing of atomic orbitals to form
special orbitals for bonding.
The atoms are responding as needed to give
the minimum energy for the molecule.
13
Valence Bond Theory and NH3
N – 1s22s22p3
3 H – 1s1
2s
2px
2py 2pz
If use the
3 2p orbitals
predict 900
Actual H-N-H bond angle is 107.30
How do you explain this?
14
Consider the n=2 for N
Original
2s
Mix 1s and 3p
And generate four
2px 2py 2pz
sp3 sp3 sp3
sp3
Equivalent sp3
Hybridized orbitals
1 sp3
lone
pair
3 bonding orbitals
Which can accommodate
The 1s1 electron from
hydrogen
15
3
The nitrogen atom in ammonia is sp
hybridized.
16
17
2
An orbital energy-level diagram for sp
hybridization. Note that one p orbital
remains unchanged.
18
19
When an s and two p orbitals are mixed to form a set
2
of three sp orbitals, one p orbital remains unchanged
and is perpendicular to the plane of the hybrid orbitals.
20
21
Figure 9.13: (a) The orbitals used to form
the bonds in ethylene. (b) The Lewis
structure for ethylene.
22
23
Pi bond (p) – electron density above and below plane of nuclei
Sigma bond (s) – electron density between the 2 atoms 24
of the bonding atoms
25
The s bonds in ethylene. Note that for each bond the
shared electron pair occupies the region directly between
the atoms.
26
A sigma (s) bond centers along the
internuclear axis.
A pi (p) bond occupies the space above and
below the internuclear axis.
H
p
H
C sC
H
H
27
(a)The orbitals used
to form the bonds
in ethylene.
(b) The Lewis
structure for
ethylene.
28
The orbital energy-level diagram for the
formation of sp hybrid orbitals on carbon.
29
30
When one s orbital and one p orbital are
hybridized, a set of two sp orbitals oriented at
180 degrees results.
31
The orbitals of an sp hybridized
carbon atom.
32
33
2
The orbital arrangement for an sp hybridized
oxygen atom to form CO2.
8O
1s22s22p4
34
The hybrid orbitals in the CO2 molecule.
35
(a) The orbitals used to form the bonds in carbon dioxide. Note
that the carbon-oxygen double bonds each consist of one s bond
and one p bond. (b) The Lewis structure for carbon dioxide.
2sp orbitals from C to form two double bonds
36
(a) An sp hybridized nitrogen atom. (b) The s bond in the N2
molecule. (c) The two p bonds in N2 are formed when electron
pairs are shared between two sets of parallel p orbitals. (d) The
total bonding picture for N2.
37
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
H
O
H
C
O
H
s bonds = 6 + 1 = 7
p bonds = 1
38
10.5
How to generate more than 4
bonds?
Remember the PCl5, SF6, etc…
The s and 3p can generate 4 orbitals
Include d orbitals to generate more!
39
s
p
p
p
d
d
d
d
d
Hybridize 1s and 3p and 1d
Result in 5 dsp3 orbitals
1
2
dsp3 dsp3
3
4
dsp3 dsp3
5
dsp3
40
A set of dsp3 hybrid orbitals on a phosphorus atom. Note
that the set of five dsp3 orbitals has a trigonal
bipyramidal arrangement. (Each dsp3 orbital also has a
small lobe that is not shown in this diagram.)
41
(a) The PCl5 molecule. (b) The orbitals used to form the bonds in
PCl5. The phosphorus uses a set of five dsp3 orbitals to share
electron pairs with sp3 orbitals on the five chlorine atoms. The
other sp3 orbitals on each chlorine atom hold lone pairs.
42
How to form 6 orbitals?
s
p
p
p
d
d
d
d
d
Hybridize 1s and 3p and 2d
1
Result in 6 d2sp3 orbitals
2
3
4
5
d2sp3 ds2p3
d2sp3
6
d2sp3 d2sp3 d2sp3
43
The relationship of
the number of
effective
pairs,
their
spatial
arrangement, and
the hybrid orbital
set required.
44
The Localized Electron Model
 Draw
the Lewis structure(s)
 Determine
the arrangement of electron pairs
(VSEPR model).
 Specify
the necessary hybrid orbitals.
45
Molecular Orbitals (MO)
Analagous to atomic orbitals for atoms,
MOs are the quantum mechanical
solutions to the organization of valence
electrons in molecules. Remember
bonds are waves and wave may be
arranged in constructive or destructive
interference.
46
47
Types of MOs
bonding: lower in energy than the
atomic orbitals from which it is
composed.
antibonding: higher in energy (unstable)
than the atomic orbitals from which it is
composed.
48
The combination of hydrogen 1s atomic orbitals
to form molecular orbitals.
49
50
Bonding and antibonding molecular
orbitals (MOs).
51
(a) The molecular
orbital energy-level
diagram
for the H2 molecule.
(b) The shapes
of the molecular
orbitals are obtained
by squaring the
wave functions for
MO1 and MO2.
52
A molecular
orbital energylevel diagram
for the H2
molecule.
53
The molecular orbital energy-level
diagram for the H2 ion.
54
The molecular orbital energy-level
diagram for the He2 molecule.
55
The molecular orbital energy-level
diagram for the Li2 molecule.
56
57
The expected molecular orbital energy-level
diagram resulting from the combination of the 2p
orbitals on two boron atoms.
58
The expected
molecular orbital
energy-level diagram
for the B2 molecule.
However, B2 is found to
be Paramagnetic!
59
The correct molecular
orbital energy-level
diagram for the B2
molecule. When p-s
mixing is allowed, the
energies of the s2p and π2p
orbitals are reversed. The
two electrons from the B
2p orbitals now occupy
separate, degenerate π2p
molecular orbitals and thus
have parallel spins.
Therefore, this diagram
explains the observed
paramagnetism of B2.
60
(a) The three mutually perpendicular 2p orbitals on two
adjacent boron atoms. Two pairs of parallel p orbitals
can overlap as shown in (b) and (c), and the third pair
can overlap head-on as shown in (d).
61
(a) The two p orbitals on the boron atom that overlap
head-on produce two s molecular orbitals, one bonding
and one antibonding. (b) Two p orbitals that lie parallel
overlap to produce two p molecular orbitals, one bonding
and one antibonding.
62
The molecular orbital energy-level diagrams, bond orders, bond
energies, and bond lengths for the diatomic molecules B2 through
F2. Note that for O2 and F2 the s 2p orbital is lower in energy than
the π2p orbitals.
63
Paramagnetic
Diamagnetic
64
Note that O2 is paramagnetic
When liquid oxygen is
poured into the space
between the poles of a
strong magnet, it
remains there until it
boils away. This
attraction of liquid
oxygen for the
magnetic field
demonstrates the
paramagnetism of the
O2 molecule.
65
Paramagnetism
 unpaired
electrons
 attracted
to induced magnetic field
 much
stronger than diamagnetism
66
Bond Order (BO)
Difference between the number of bonding
electrons and number of antibonding
electrons divided by two.
# bonding electrons  # antibonding electons
BO =
2
67
1
bond order =
2
bond
order
½
(
Number of
electrons in
bonding
MOs
1
-
½
Number of
electrons in
antibonding
MOs
)
0
68
Molecular Orbital (MO) Configurations
1. The number of molecular orbitals (MOs) formed is always
equal to the number of atomic orbitals combined.
2. The more stable the bonding MO, the less stable the
corresponding antibonding MO.
3. The filling of MOs proceeds from low to high energies.
4. Each MO can accommodate up to two electrons.
5. Use Hund’s rule when adding electrons to MOs of the
same energy.
6. The number of electrons in the MOs is equal to the sum of
all the electrons on the bonding atoms.
69
Outcomes of MO Model
1.
As bond order increases, bond energy
increases and bond length decreases.
2.
Bond order is not absolutely associated with a
particular bond energy.
3.
N2 has a triple bond, and a correspondingly
high bond energy.
4.
O2 is paramagnetic. This is predicted by the
MO model, not by the LE model, which
predicts diamagnetism.
70
Heteronuclear Molecules and similar?
The molecular
orbital energy-level
diagram for the NO
molecule. We
assume that orbital
order is the same as
that for N2. The
bond order is 2.5.
71
Figure 9.42: The
molecular orbital
energy-level
diagram for both
+
the NO and CN
ions.
72
Heteronuclear Molecules and similar?
A partial
molecular
orbital
energy-level
diagram for
the HF
molecule.
73
The electron probability distribution in the
bonding molecular orbital of the HF molecule.
Note the greater electron density close to the
fluorine atom.
74
Combining LE and MO Models
s bonds can be described as being
localized.
p bonding must be treated as being
delocalized.
75
Delocalized molecular orbitals are not confined between
two adjacent bonding atoms, but actually extend over three
or more atoms.
76
(a) The benzene molecule consists of a ring of six carbon atoms
with one hydrogen atom bound to each carbon. (b) Two of the
resonance structures for the benzene molecule.
77
The s bonding system in the
benzene molecule.
78
(a) The p molecular orbital system in benzene is formed by
2
combining the six p orbitals from the six sp hybridized carbon
atoms. (b) The electrons in the resulting p molecular orbitals are
delocalized over the entire ring of carbon atoms, giving six
equivalent bonds. A composite of these orbitals is represented
here.
79
Electron density above and below the plane of the
benzene molecule.
80
-
The resonance structures for O3 and NO3 . Note
that it is the double bond that occupies various
positions in the resonance structures.
81
-
(a) The p orbitals used to form the π bonding system in the NO3
ion. (b) A representation of the delocalization of the electrons in
the π molecular orbital system of the NO3- ion.
82
QUESTION
The hybridization of the phosphorus atom in the
+
cation PH2 is:
2
1) sp .
3
2) sp .
3) dsp.
4) sp.
5) none of these.
83
ANSWER
1)
2
sp .
Section 9.1 Hybridization (p. 413)
The two of the three hybridized orbitals are
bonding orbitals, the third hybridized orbital
contains the lone pair.
84
QUESTION
2
Atoms which are sp hybridized form ____ pi
bond(s).
1) 0
2) 1
3) 2
4) 3
5) 4
85
ANSWER
2)
1
Section 9.1 Hybridization (p. 416)
2
An atom with sp hybridization has a p orbital
that remains unhybridized and can overlap with
another p orbital on a neighboring atom to
create a pi bond.
86
QUESTION
The hybridization of the central atom in
1) sp
2
2) sp
3
3) sp
3
4) dsp
2
3
5) d sp
–
I3
is:
87
ANSWER
4)
dsp
3
Section 9.1 Hybridization (p. 421)
Three of
–
I3
orbitals contain lone pairs.
88
QUESTION
Which of these statements about benzene is true?
3
1) All carbon atoms in benzene are sp
hybridized.
2) Benzene contains only p bonds between C
atoms.
3) The bond order of each C—C bond in benzene
is 1.5.
4) Benzene is an example of a molecule that
displays ionic bonding.
5) All of these statements are false.
89
ANSWER
3)
The bond order of each C—C bond in
benzene is 1.5.
Section 9.5 Combining the Localized Electron
and Molecular Orbital Models (p. 439)
This unusual bond order is due to the
delocalization of the bonding electrons. On
average three electrons are shared between two
atoms.
90
QUESTION
Which statement about N2 is false?
1) It is a gas at room temperature.
2) The oxidation state is +3 on one N and –3
on the other.
3) It has one sigma and two pi bonds between
the two atoms.
4) It can combine with H2 to form NH3.
5) It has two pairs of nonbonding electrons.
91
ANSWER
2)
The oxidation state is +3 on one N and –3
on the other.
Section 9.1 Hybridization (p. 418)
For a molecular form of an element, all atoms
share electrons equally and thus should have
oxidation states of zero.
92
QUESTION
The hybridization of the central atom, Al, in AlBr3
is:
1) sp
2
2) sp
3
3) sp
3
4) dsp
2
3
5) d sp
93
ANSWER
2)
sp
2
Section 9.1 Hybridization (p. 416)
Aluminum has three valence electrons, each of
which forms a bond with a Br atom.
94
QUESTION
The hybridization of Br in BrF3 is:
1) sp
2
2) sp
3
3) sp
3
4) dsp
2
3
5) d sp
95
ANSWER
4)
dsp
3
Section 9.1 Hybridization (p. 421)
Bromine has two lone pairs as well as three
bonding pairs.
96
QUESTION
What is the bond order of Ne2?
1) 0
2) 1/2
3) 1
4) 1 1/2
5) 2
97
ANSWER
1)
0
Section 9.2 The Molecular Orbital Model (p. 429)
Ne atoms have a full octet and have no halffilled orbitals to overlap for bonding.
98
QUESTION
What is the bond order of
1) 0
2) 1/2
3) 1
4) 1 1/2
5) 2
+
C2 ?
99
ANSWER
4)
1 1/2
Section 9.2 The Molecular Orbital Model (p. 429)
Remember when using MO diagrams that
Hund’s Rule still applies when filling the orbitals
and that forgetting this rule can give the wrong
bond order.
100
QUESTION
How many electrons are involved in pi bonding
in benzene, C6H6?
1) 12
2) 30
3) 3
4) 6
5) 18
101
ANSWER
4)
6
Section 9.5 Combining the Localized Electron
and Molecular Orbital Models (p. 439)
Each carbon atom in the benzene ring donates
an electron to the pi bonding.
102
QUESTION
Which of the following substances contains two
pi bonds?
1) C2H4
2) C3H8
3) C2H2
4) C2H6
5) CH4
103
ANSWER
3)
C2H2
Section 9.1 Hybridization (p. 418)
Each carbon in C2H2 has two unhybridized p
orbitals for pi bonding.
104
QUESTION
Consider the skeletal structure shown below:
N—C—C—N
Draw the Lewis structure and answer the
following: How many of the atoms are sp
hybridized?
1) 0
2) 1
3) 2
4) 3
5) 4
105
ANSWER
5)
4
Section 9.1 Hybridization (p. 418)
Each nitrogen and carbon has an unhybridized p
orbital for pi bonding.
106
QUESTION
Consider the skeletal structure shown below:
N—C—C—N
Draw the Lewis structure and answer the
following: How many pi bonds does the
molecule contain?
1) 0
2) 2
3) 4
4) 6
5) 7
107
ANSWER
3)
4
Section 9.1 Hybridization (p. 418)
The pi bonds create a triple bond between each
CN pair.
108
QUESTION
Which of the following statements about the
molecule BN is false?
1) It is paramagnetic.
2) Its bond order is 2.
3) The total number of electrons is 12.
4) It has two pi bonds.
5) All of these are true.
109
ANSWER
1)
It is paramagnetic.
Section 9.3 Bonding in Homonuclear Diatomic
Molecules (p. 432)
BN has 8 valence electrons, all of which are
paired so that there is no paramagnetism. This
pairing can be seen in its Lewis structure and
using MO Theory.
110
QUESTION
Which of the following molecules contains the
shortest C — C bond?
1) C2H2
2) C2H4
3) C2H6
4) C2Cl4
5) 2 and 4
111
ANSWER
1)
C2H2
Section 9.3 Bonding in Homonuclear Diatomic
Molecules (p. 430)
Triple bonds are shorter than double bonds,
which are shorter than single bonds. The
strength of the bonds increases with decreasing
length.
112
QUESTION
If four orbitals on one atom overlap four orbitals
on a second atom, how many molecular orbitals
will form?
1) 1
2) 4
3) 8
4) 16
5) None of these
113
ANSWER
3)
8
Section 9.1 Hybridization (p. 414)
Molecular orbitals form during the combination
of atomic orbitals, but orbitals cannot be created
or destroyed. If eight atomic orbitals mix, eight
molecular orbitals are formed.
114
QUESTION
Which of the following molecules or ions is not
paramagnetic in its ground state?
1) O2
+
2) O2
3) B2
4) NO
5) F2
115
ANSWER
5)
F2
Section 9.3 Bonding in Homonuclear Diatomic
Molecules (p. 432)
All species can be made paramagnetic by the
promotion of an electron from a filled orbital. In
all cases covered here, we will assume that the
molecules and ions are in their electronic ground
state.
116
QUESTION
The fact that O2 is paramagnetic can be
explained by:
1) the Lewis structure of O2.
2) resonance.
3) a violation of the octet rule.
4) the molecular orbital diagram for O2.
5) hybridization of atomic orbitals in O2.
117
ANSWER
4)
the molecular orbital diagram for O2.
Section 9.3 Bonding in Homonuclear Diatomic
Molecules (p. 434)
Application of Hund’s Rule will leave two
electrons unpaired in the two p bonding orbitals.
118
QUESTION
As the bond order of a bond increases, the bond
energy ______ and the bond length ______.
1) increases; increases
2) decreases; decreases
3) increases; decreases
4) decreases; increases
5) More information is needed to answer this
question.
119
ANSWER
3)
increases; decreases
Section 9.2 The Molecular Orbital Model (p. 426)
Multiple bonds between atoms increases the
electron density, shielding the nuclei from one
another, allowing the atoms to draw closer
together.
120