N

1
Lewis structures indicate the bonding that
occurs, but they do not indicate the ACTUAL
shapes of molecules.
N
H
H
H
2
The actual shape of a molecule is determined by
the of atoms bonded to the central atom, the
length of the bonds, and the angles between each
atom .
ICE
3
Bond Angles The angles made by lines joining the
nuclei of each atom in the molecule.
In this case, the Lewis structure geometry and
the actual molecular geometry are the same. In
many molecules, the bond angles of the actual
molecule are slightly different from the angles
portrayed by the Lewis structure.
4
So what determines the bond angles?
(Electron Repulsion)
-
-
5
VSEPR
Valence
Shell
Electron
Pair
Repulsion
6
We can start the explanation of the VSEPR model
by looking at the electrons in the Lewis
Structures.
7
Electrons can be found in a region of space as a
nonbonding pair.
N
H
H
H
Electrons can be found in a region of space as a
bonding pair.
8
These regions of space in which electrons are
found are called
Electron Domains
of electron domains equals
(
)
of bonding pairs on the central atom

of nonbonding pairs on the central atom
9
of electron domains equals
1
2
1
4
N
H
H
3
Bonding Pairs
Nonbonding pairs
H
10
Two things to note
1. Only consider electron domains around the
central atom.

• It doesnt matter whether the bond is a single,
  double, or a triple bond. Each type only counts
  as 1 electron domain.

Electron Domains?
1
N
3
1
O
O
1
11
For all of the following Draw the correct Lewis
structures. Identify the total of electron
domains. Identify the of nonbonding and
bonding domains.
CO2 BCl NH4 BCl3 H2O O3 PCl5 SO2
SF4
Other Homework 9.3 - 9.6
12
Bonded Electron Domains Indicate the
orientation of the atoms in space around the
central atom.
Nonbonding Electron Pair Domains Influence the
bond angles.
13
VSEPR
The main idea of this model is that the electron
domains in a molecule spread themselves apart as
far as possible to minimize electron-electron
repulsion.
14
It will be necessary for you to be able to
determine the correct positions of the electrons
in a molecule. This will be determined by the
of electron domains in the molecule. (Electron
Domain Geometry) ---------------------------------
-------------------------------- However, this
is not the actual geometry of the molecule. In
order to determine the actual molecular geometry,
you must consider the effect of nonbonding pairs
on the bond angles. (Molecular Geometry)
15
5 Possible Electron Domain Geometries
Electron Domains
2
Picture
Electron Geometry
Linear
16
Electron Domains
3
Picture
Electron Geometry
Trigonal Planar
17
Electron Domains
4
Picture
Electron Geometry
Tetrahedral
18
Electron Domains
5
Picture
Electron Geometry
Trigonal bipyramidal
19
Electron Domains
6
Picture
Electron Geometry
Octahedral
20
Linear
Trigonal Planar
Tetrahedral
Trigonal Bipyramidal
Octahedral
21
Again, the pictures we have seen so far are
simply geometries of the electron domains. The
actual molecular geometries are affected by the
nonbonding pairs of electrons, which changes the
angles a bit between atoms.
Lets compare our pictures to real models. (We
will look at CH4, NH3, and H2S)
22
Methane
CH4
Molecular Geometry
Tetrahedral
23
(No Transcript)
24
Ammonia
NH3
Molecular Geometry
Trigonal Pyramidal
25
(No Transcript)
26
H2S
Dihydrogen Sulfide
Molecular Geometry
Bent
27
(No Transcript)
28
(No Transcript)
29
Based on what you saw with these 3 models, which
electron repulsion is stronger?
unshared electrons versus unshared electrons
Or
unshared electrons versus shared electrons
Repulsion Strength U U gt U S gt S S
30
Summary of why electron domain geometry is
different from molecular geometry
The electron clouds in a molecule take on a
specific shape based on the of electron pairs,
but the overall shape of the molecule can be
slightly altered because nonbonding electrons
take up more space around the central atom, and
they also have stronger repulsive forces.
Note Angles that are less than 109.5 as a result
of unshared electron repulsion are usually
reported as lt 109.5
31
What molecular shape is the following?
Trigonal Pyramidal
32
What molecular shape is the following?
Octahedral
33
What molecular shape and angle is the following?
Bent
lt 109.5
34
What molecular shape and angle is the following?
Trigonal Planar
120 ?
35
What molecular shape and angle is the following?
Linear
180 ?
36
(No Transcript)
37
(No Transcript)
38
At this point, its best if you actually look
over some of the other arrangements in your
textbook. I will be passing out a small flip
book of the molecules portrayed in your textbook
for you to study.
39
Dipole Moment (Polarity) and Molecular Geometry

• We discussed the polarity of diatomic molecules
  in chapter 8.
• We will now discuss the polarity of polyatomic
  molecules.

H Cl
40

• We can view the polarity of individual bonds
  within a molecule as vector quantities.
• (magnitude and direction)

Opposite directions so polarity cancels.
41

• However, molecules that exhibit any asymmetry in
  the arrangement of electron pairs would have a
  nonzero dipole moment. These molecules are
  considered polar.

42
Determine whether each of the following are polar
or nonpolar molecules. Carbon tetrachloride CCl4
Chloroform CHCl3
43
The dipole moment of the molecule is zero because
the charges cancel each other.
The charges do not cancel, and therefore this
molecule has a nonzero dipole moment.
44
One thing to consider Nonbonding pairs can cause
a decrease in polarity because of the excess
negative charge created by the nonbonding
pair. Lets consider the molecule NF3.
Fluorine is more electronegative than
nitrogen. Results in dipole moment. However,
size of dipole moment is small. Excess neg.
charge at the top cancels some polarity.
F
F
F
45

• Question commonly asked on the AP Exam
• Of the following molecules, which has the largest
  dipole moment?
• CN-
• CO2
• N2
• HF
• F2

46

• Of the following molecules, which has a zero
  dipole moment?
• HCN
• H2S
• SO2
• NO
• PF5

47
We previously learned about covalent bonding and
the sharing of electrons.
H 1S _____
H2
H 1S _____
This doesnt explain how a molecule like CH4
forms.
48

• One might expect that the number of bonds formed
  by an atom would equal its unpaired electrons.

• Chlorine, for example, generally forms one bond
  and has one unpaired electron.
• Oxygen, with two unpaired electrons, usually
  forms two bonds.

O
O

O
O
49
(No Transcript)
50
How would the Hydrogen atoms bond? Hybridization
2p
C atom (ground state)
51
Valence Bond Theory

• Valence bond theory is an approximate theory to
  explain the covalent bond from a quantum
  mechanical view.

• According to this theory, a bond forms between
  two atoms when the following conditions are met.
• Two atomic orbitals overlap.
• The total number of electrons in both orbitals is
  no more than two. (one from each)

52
Hybrid Orbitals

• The bonding in carbon might be explained as
  follows

• Four unpaired electrons are formed as an electron
  from the 2s orbital is promoted (excited) to the
  vacant 2p orbital.
• The following slide illustrates this excitation.

53
2p
2p
2s
Energy
C atom (ground state)
C atom (promoted)
54
Hybrid Orbitals

• One bond on carbon would form using the 2s
  orbital while the other three bonds would use the
  2p orbitals.

• This does not explain the fact that the four
  bonds in CH4 appear to be identical.
• Valence bond theory assumes that the four
  available atomic orbitals in carbon combine to
  make four equivalent hybrid orbitals.

55
Hybrid Orbitals

• Hybrid orbitals are orbitals used to describe
  bonding that are obtained by taking combinations
  of atomic orbitals of an isolated atom.

• In this case, a set of hybrids are constructed
  from one s orbital and three p orbitals, so
  they are called sp3 hybrid orbitals.
• The four sp3 hybrid orbitals take the shape of a
  tetrahedron.

56
You can represent the hybridization of carbon in
CH4 as follows.
2p
C-H bonds
1s
C atom (ground state)
C atom (hybridized state)
C atom (in CH4)
57
.5
58
Hybrid Orbitals

• Note that there is a relationship between the
  type of hybrid orbitals and the geometric
  arrangement of those orbitals.

• Thus, if you know the geometric arrangement, you
  know what hybrid orbitals to use in the bonding
  description.
• Also, whenever we mix a certain number of
  orbitals, we get the same number of hybrid
  orbitals.

59
Hybrid Orbitals
60
(No Transcript)
61
Hybrid Orbitals

• To obtain the bonding description of any atom in
  a molecule, you proceed as follows

• Write the Lewis electron-dot formula for the
  molecule.

2. From the Lewis formula, use the VSEPR theory
to determine the arrangement of electron pairs
around the atom.
62
Hybrid Orbitals

• From the geometric arrangement of the electron
  pairs, obtain the hybridization type.

• Assign valence electrons to the hybrid orbitals
  of this atom one at a time, pairing only when
  necessary.

• Form bonds to this atom by overlapping singly
  occupied orbitals of other atoms with the singly
  occupied hybrid orbitals of this atom.

63
A Problem to Consider

• Describe the bonding in H2O according to valence
  bond theory. Assume that the molecular geometry
  is the same as given by the VSEPR model.

• From the Lewis formula for a molecule, determine
  its geometry about the central atom using the
  VSEPR model.

64
A Problem to Consider

• Describe the bonding in H2O according to valence
  bond theory. Assume that the molecular geometry
  is the same as given by the VSEPR model.

65
A Problem to Consider

• Describe the bonding in H2O according to valence
  bond theory. Assume that the molecular geometry
  is the same as given by the VSEPR model.

• From this geometry, determine the hybrid orbitals
  on this atom, assigning its valence electrons to
  these orbitals one at a time.

66
A Problem to Consider

• Describe the bonding in H2O according to valence
  bond theory. Assume that the molecular geometry
  is the same as given by the VSEPR model.

• Note that there are four pairs of electrons about
  the oxygen atom.

• According to the VSEPR model, these are directed
  tetrahedrally, and from the previous table you
  see that you should use sp3 hybrid orbitals.

67
A Problem to Consider

• Describe the bonding in H2O according to valence
  bond theory. Assume that the molecular geometry
  is the same as given by the VSEPR model.

• Each O-H bond is formed by the overlap of a 1s
  orbital of a hydrogen atom with one of the singly
  occupied sp3 hybrid orbitals of the oxygen atom.

68
You can represent the bonding to the oxygen atom
in H2O as follows
O-H bonds
lone pairs
1s
O atom (ground state)
O atom (hybridized state)
O atom (in H2O)
69
A Problem to Consider

• Describe the bonding in XeF4 using hybrid
  orbitals.

• Write the Lewis formula for the molecule and
  determine its geometry about the central atom
  using the VSEPR model.

70
A Problem to Consider

• The Lewis formula of XeF4 is

Octahedral
71
A Problem to Consider

• Describe the bonding in XeF4 using hybrid
  orbitals.

• From this geometry (octahedral), determine the
  hybrid orbitals on this atom, assigning its
  valence electrons to these orbitals one at a
  time.

72
A Problem to Consider

• Describe the bonding in XeF4 using hybrid
  orbitals.

• The xenon atom has four single bonds and two lone
  pairs. It will require six orbitals to describe
  the bonding.

73
A Problem to Consider

• Describe the bonding in XeF4 using hybrid
  orbitals.

• Each Xe-F bond is formed by the overlap of a
  xenon sp3d2 hybrid orbital with a singly occupied
  fluorine 2p orbital.

• You can summarize this as follows

74
5d
5p
5s
Xe atom (ground state)
75
5d
sp3d2
Xe atom (hybridized state)
76
5d
sp3d2
lone pairs
Xe-F bonds
Xe atom (in XeF4)
77
Determine the of hybrid orbitals and type of
hybridization of BeH2 Linear, requires 2 sp
orbitals
78
Determine the of hybrid orbitals and type of
hybridization of BF3 Trigonal planar, requires 3
sp2 orbitals
79
Determine the of hybrid orbitals and type of
hybridization of NH4 Tetrahedral, requires 4 sp3
orbitals
80
Determine the of hybrid orbitals and type of
hybridization of PCl5 and SF6 PCl5 5 sp3d
orbitals SF6 octahedral 6 sp3d2
PCl5
SF6
81
Multiple Bonding

• According to valence bond theory, one hybrid
  orbital is needed for each bond (whether a single
  or multiple) and for each lone pair.

82
Multiple Bonding

• Each carbon atom is bonded to three other atoms
  and no lone pairs, which indicates the need for
  three hybrid orbitals.

• This implies sp2 hybridization.
• The third 2p orbital is left unhybridized and
  lies perpendicular to the plane of the trigonal
  sp2 hybrids.
• The following slide represents the sp2
  hybridization of the carbon atoms.

83
(unhybridized)
2p
2p
sp2
Energy
C atom (ground state)
C atom (hybridized)
84
Multiple Bonding

• To describe the multiple bonding in ethene, we
  must first distinguish between two kinds of bonds.

• A s (sigma) bond is a head-to-head overlap of
  orbitals with a cylindrical shape about the bond
  axis. This occurs when two s orbitals overlap
  or p orbitals overlap along their axis.
• A p (pi) bond is a side-to-side overlap of
  parallel p orbitals, creating an electron
  distribution above and below the bond axis.

85
Inaccurate?
86
Multiple Bonding

• Now imagine that the atoms of ethene move into
  position.

• Two of the sp2 hybrid orbitals of each carbon
  overlap with the 1s orbitals of the hydrogens.

87
Multiple Bonding

• The remaining unhybridized 2p orbitals on each
  of the carbon atoms overlap side-to-side forming
  a p bond.

• You therefore describe the carbon-carbon double
  bond as one s bond and one p bond.

88
(No Transcript)
89
Single bonds consist of one s (sigma)
bond. Double bonds consist of one s (sigma)
bond and one p (pi) bond. Triple bonds consist
of one s (sigma) bond and two p (pi) bonds.
90
How many sigma bonds, and how many pi bonds are
in the following structure?
H
s
s
H
C
C
N
s
s
s
p
H
p
Acetone 5 sigma 2 pi
91
Molecular Orbital Theory

• Molecular orbital theory is a theory of the
  electronic structure of molecules in terms of
  molecular orbitals, which may spread over several
  atoms or the entire molecule.

• As atoms approach each other and their atomic
  orbitals overlap, molecular orbitals are formed.
• In the quantum mechanical view, both a bonding
  and an antibonding molecular orbital are formed.

92
Electronic Configurations of Diatomic Molecules

• In heteronuclear diatomic molecules, such as CO
  or NO, we must have additional molecular orbitals.

• The overlap of p orbitals results in two sets
  of s orbitals (two bonding and two antibonding)
  and one set of p orbitals (one bonding and one
  antibonding).
• The next slide illustrates the relative energies
  of these molecular orbitals.

93
Return to Slide 27