Hybridization

Transcription

Hybridization

Covalent Bonding: Orbitals
Chapter 09
Copyright©2000 by Houghton
Mifflin Company. All rights reserved.
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
1s2 2s2 sp2
Ground state of C
Promote electron at n=2
Hybridize at n=2
s
px
py
pz
s
px
py
pz
sp3
sp3
2s12p3
sp3 sp3 sp3 sp3
sp3 sp3
Four sp3 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
three 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
Copyright©2000
bybetween
Houghton
24
Sigma bond (s) – electron
density
the
2
atoms
Mifflin Company.
All rights reserved.
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 orbital.

All single bonds are hybridized and Sigma bond

Double bond contains one Sigma and one Pi bonds while triple bond
contains one sigma and two Pi bonds.

Total number of electron pairs (bonded + unpaired electrons) decides
the hybridization of the atom. Double and triple bonds are COUNTED
as One Electron Pair.

2 electron pairs=SP, 3 electron pairs=SP2, 4 electron pairs=SP3,5
electron pairs=dSP3, 6 electron pairs=d2SP3
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
diagram for the He2 molecule.
55
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
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.Mifflin
Company. All rights reserved.
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
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

Hybridization

Transcription

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