Chapter 9 Molecular Structure
Under what rules do atoms form molecules?
New materials, such as the
lightweight gear used by
these climbers, allow us to
explore our world further
than thought possible
before. The development
of new materials like these
relies on the principles of
molecular structure
introduced in this chapter.
Assignment for Chapter 9
9.7, 9.12, 9.15, 9.21, 9.25, 9.28, 9.33
9.44, 9.54, 9.56
Some of the basic geometrical shapes that are
used to describe the shapes of simple molecules
Tetrahedral
Octahedral
Shape names and bond angles
(lone pairs not included)
T-shaped
VSEPR Model
(Valence-Shell Electron-Pair Repulsion model)
Bonding electrons and lone pairs take up
positions as far from one another as
possible, for then they repel each other the
least.
VSEPR Model (1):
locate the high concentrations
The positions taken up by regions of high electron concentration
(green) around a central atom.
Electron pairs or bonds are as far away from
one another as possible and so experience the
minimal repulsion from other electrons
VSEPR Model (2):
determine the shapes
Two ‘highs’
Five ‘highs’
Three ‘highs’
Six ‘highs’
Four ‘highs’
Seven ‘highs’
A summary of the positions taken up by regions of high electron
concentration (other atoms and lone pairs) around a central atom.
The locations of these regions are given by straight lines sticking out
of the central atom.
Use this chart to identify the arrangement of a given number of
atoms and lone pairs and then use Fig. 9.2 to identify the shape
of the molecule from the location of its atoms.
VSEPR Model: Example
Electron pairs or bonds are as far away from one
another as possible and so experience the minimum
repulsion from other electrons
..
..
:Cl -Be- Cl:
..
..
Shape: linear (two ‘highs’)
Shape: trigonal planar
(three ‘highs’)
VSEPR Model: Example
Electron pairs or bonds are as far away from one
another as possible and so experience the minimum
repulsion from other electrons
Shape: trigonal bipyramidal (five ‘highs’)
VSEPR Model
• In order to reduce repulsions, bonding pairs
and lone pairs take up positions around an
atom that maximize their separations. The
shape of the molecule is determined by the
location of the atoms attached to the central
atom.
Classroom Exercise
Predict the structure of SiCl4 and AsF5
Classroom Exercise
Predict the structure of SiCl4 and AsF5
Cl
|
Cl- Si -Cl
|
Cl
Si: 3s23p2
F
F
As
F
SiCl4
AsF5
F
F
As: 3d104s24p3
Molecules with Multiple Bonds
The VSEPR model does not distinguish single and multiple
bonds. A multiple bond is treated as just another region of
high electron concentration.
..
..
O =C= O
..
(Two ‘highs’)
..
(Three ‘highs’)
The VSEPR model does not distinguish single and multiple
bonds. A multiple bond is treated as just another region of
high electron concentration.
Two central atoms
Example
HC  CH
Two central atoms, no lone pairs.
The most possible VSEPR arrangement:
Formaldehyde
CH2O
H
|
..
H- C = O
..
CH2O
Classroom Exercise
Predict the structure of HCN
H-C  N :
Different resonance structures correspond
to a single shape
How about if the central atoms
have lone pairs?
Lone pairs on attached atoms have little effect on molecular shape,
but the lone pairs on central atoms may have significant effect.
• A: central atom
• X: atom bonded to the central atom
• E: lone pair on the central atom
The single nonbonding electron on radicals is treated as a ‘lone pair’.
Lone Pair on the Central Atom
AX3E
Repulsion Order:
Lone pair/lone pair > lone pair/bonding pair > bonding pair/bonding pair
H2O
Example
..
H-O-H
..
AX2E2
Shape: Angular (not tetrahedron!)
Three views of water molecular shape
NH3
H
|
H- N -H
..
AX3E
Triangular pyramid
(NOT tetrahedron!)
Classroom Exercise
Predict the shape of NO2AEX2
..  -
..
O = N  O:
..
..
..
NO2Smaller
angle
Angular (NOT planar triangle!)
AX4E
Axial lone pair
Equatorial lone pair
AX4E2
Square planar
(NOT octahedron!)
How to Use VSEPR Model
1. Write Lewis structure
and determine the number of
electron pairs
2. Maximize the separations.
3. Decide the positions
of lone pairs (on the central atom).
4a. Name the shape (without
considering the lone pair).
4b. Consider distortion using
repulsion order.
Lone pair/lone pair > lone pair/bonding pair > bonding pair/bonding pair
Example: SF4
AX4E
Equatorial lone pair
Bent seesaw
T-shaped
(NOT triangular bipyramid!)
ClF3
..
..
:F:
|
..
:F- Cl - F:
..
..
..
AX3E
T-shaped
(NOT triangular bipyramid!)
Classroom Exercise: XeF4
..
:F:
|
..
..
..
:F - .. Xe - F:
..
..
|
:F:
..
Square planar
(NOT octahedron!)
AX4E2
Quiz
• Write the VSEPR formula of water and
sulfur tetrafluoride, draw and name their
structures.
Answer
• Write the VSEPR formula of water and
sulfur pentafluoride. Draw and name their
structures.
..
H-O-H
..
AX2E2
AX4E
Angular (not tetrahedron!)
T-shaped
(not triangular bipyramid!)
詩云
VSEPR 頌
Ode to VSEPR
中心邊緣孤對通，
三位一體電斥鬆。
主要形狀心中記，
孤對不在名字中。
Central, attached plus lone pairs found
The trinity held by electrical repulsive bounds.
Memorize the major molecular shapes
and ignore the lone pair in naming the compounds.
Charge Distribution in
Molecules
Where does have high or low
electron density (concentration)?
Electron distribution is responsible
for molecular properties and
functions.
Polar Bonds
r
Partial negative charge
 r
Partial positive charge
1 Debye
r=100 pm (=1 °A)
+e
-e
μ=1 D
Polar Bonds Forming Polar Molecules
Polar Bonds Forming Nonpolar Molecules
Total dipole moment = 0
Partial negative charge
Partial negative charge
Partial positive charge
Charge distribution of CO2
The shape of a molecule governs
whether it is polar or not.
Nonpolar
Polar
Molecular shape and polarity
BF3
SF4
SF6
..
:F:
|
..
..
:F ..
..
S
:F
..
|
..
:F
..
:F
..
..
-F:
..
The strengths and lengths of bonds
(a) The three normal vibrational modes
of H2O.
(b) The four normal vibrational modes
of CO2.
Infrared spectroscopy measures
normal vibrational frequencies
The infrared spectrum of tryptophan (an amino acid).
Bond enthalpy ΔHB
H2(g)2H(g)
ΔHo=+436 kJ/mol
ΔHB(H-H)= 436 kJ/mol
ΔHB(F-CF3)= 484 kJ/mol
ΔHB(H-CH3)= 412 kJ/mol
The bond enthalpies, in kilojoules per mole (kJ/mol), of
diatomic nitrogen, oxygen and fluorine molecules. Note
how the bonds weaken as they change from a triple bond
in N2 to a single bond in F2.
The bond enthalpies, in kilojoules per mole (kJ/mol), of
hydrogen halide molecules. Note how the bonds weaken as
the halogen atom becomes larger.
F
Cl
Br
I
Bond strengths
in diatomic molecules
• The bond strength between two atoms is
measured by the bond enthalpy.
• The bond enthalpy typically increases as the
order of the bond increases, decreases as the
number of lone pairs on neighboring atoms
increases, and decreases as the atom radius
increases.
Bond strengths in polyatomic molecules
The bond strength between a given pair of atoms varies slightly.
The average bond enthalpy is a guide to the strength of a bond in
any molecule.
Multiple bond strength < bond order * single bond strength
Double bonds are shorter than single bonds, leading
Bond strengths larger than double of single-bond strengths.
The pairs of electrons in a multiple bond repel each other
and can weaken the bond. As a result, a double bond
between carbon atoms is not twice as strong as two single
bonds would be.
Multiple bond strength < bond order * single bond strength
The bond enthalpies for bonds between hydrogen and the
p-block elements. The bond strengths decrease from top to
bottom of each group as the atoms increase in size.
How to use average bond enthalpies
Average bond enthalpies can be used to estimate reaction
enthalpies and to predict the stability of a molecule.
Step 1: decide which bonds are broken
and which formed. Calculate the
change in enthalpy when the bonds
are broken in the reactants.
Step 2: calculate the bond enthalpies
for the new product bonds.
Step 3: calculate the bond formation
enthalpies for the product bonds from
the bond dissociation enthalpies
obtained in step 2 and reversing the
sign.
Step 4: add the enthalpy change required
to break the reactant bonds.
Example of using bond enthalpies
to estimate a reaction enthalpy
• Decide whether the following reaction is
exothermic or endothermic:
CH2CH3I (g) + H2O (g)CH3CH2OH (g) + HI (g)
Bonds broken: C-I (238 kJ/mol) , O-H (463 kJ/mol)
ΔHo=238 + 463 = 701 (kJ/mol)
Bond formed: H-I（299 kJ/mol）, C-O (360 kJ/mol)
ΔHo=299 + 360 = 659 (kJ/mol)
Te reaction enthalpy: ΔHr=701 - 659 = +42 (kJ/mol) endothermic
Classroom exercise
Which of the following reactions is
Tables 9.2,3
exothermic?
(a) CCl3CHCl2 (g) + HF(g)CCl3CHClF (g) + HCl (g)
(b) CCl3CHCl2 (g) + HF(g)CCl3CCl2F (g) + H2 (g)
(a) Bonds broken: Cl-C (338 kJ/mol), H-F(565 kJ/mol)
bonds formed: C-F (484 kJ/mol), H-Cl(431 kJ/mol)
(b) Bonds broken: H-C (412 kJ/mol), H-F(565 kJ/mol)
bonds formed: C-F (484 kJ/mol), H-H(436 kJ/mol)
Quiz
1. Is the dipole moment of SF4 zero or not?
2. Explain what is average bond enthalpy and
its applications.
Quiz: Is the dipole moment of SF4
zero or not?
Example
Bond Length Depends on Bond Order
Chemical Bond Theories
The limitations of Lewis Model:
Bonding electron pairs are not localized between two bonded atoms.
• Valence-Bond (VB) Theory
• Molecular Orbital (MO) Theory
Valence-Bond (VB) Theory
Chemical bonds are formed by pairs of (unpaired) valence electrons
There are three types of valence bond:
σ-bond
π-bond
hybrid bonds
VB for H2
Figure 9.25 When the electrons (depicted  and ) in two hydrogen 1sorbitals pair and the s-orbitals overlap, they form a s-bond, depicted
here by the boundary surface of the electron cloud. The cloud forms a
cylinder around the internuclear axis and spreads over both nuclei.
That is, it has cylindrical symmetry.
Notice the difference between
VB and Lewis model.
1s
The spins of the two electrons must
be opposite to each otehr
End to end paring
Overlap
(high electron density)
1s
VB for HF
Figure 9.26 A -bond can also be formed when electrons in 1sand 2pz-orbitals pair (z is the direction along the internuclear
axis). The two electrons in the bond are concentrated within
the region of space surrounded by the boundary surface.
End to end pairing
VB for N2
Figure 9.27 A -bond is formed by the pairing of electron spins
in two 2pz-orbitals on neighboring atoms. At this stage, we are
ignoring the effect of the 2px- (and 2py-) orbitals that may also
contain unpaired electrons. The electron pair is concentrated
within the boundary surface shown in the bottom diagram.
-bond in N2
End to end pairing
Figure 9.28 A -bond is formed when electrons in two 2p-orbitals
(top) pair and overlap side by side. The middle diagram shows the
overlap of the orbitals, and the bottom diagram shows the
corresponding boundary surface. Even though the bond has a
complicated shape, with two lobes, it is occupied by one pair of
electrons and counts as one bond.
π-bond in N2
Side by side pairing
Figure 9.29 The bonding pattern in a nitrogen molecule, N2. (a) The
two atoms are bonded together by one -bond (blue) and two
perpendicular -bonds (yellow). (b) When the three orbitals are put
together, the two -bonds merge to form a long donut-shaped cloud
surrounding the -bond cloud; the overall structure resembles a
cylindrical hot dog.
-bond and
π-bonds
in N2
VB Theory
A bond forms when unpaired electrons in valence shell
atomic orbitals on two atoms pair. The atomic orbitals
they occupy overlap end-to-end to form σ-bonds or
side by side to form π-bonds.
• A single bond is a σ-bond
• A double bond is a σ-bond plus a π-bond
• A triple bond is a σ-bond plus two π-bonds
For compounds containing C,H,N,
the number of  bond = 1+nC - 12 nH  12 nN
More Examples
• How many σ-bonds and how many π-bonds
in CO2 and CO?
• CO2: O=C=O
• CO: C≡O
• CO2: O=C=O: σ-bond: 2, π-bond: 2
• CO: C≡O, σ-bond: 1, π-bond: 2
Classroom Exercise
• How many σ-bonds and how many π-bonds
in NH3 and CH2O?
• NH3: σ-bond: 3, π-bond: 0
• CH2O: σ-bond: 3, π-bond: 1
How about CH4?
Only two unpaired electrons
in carbon.
How can one carbon be bonded
with four hydrogens?
It seems the two 2s electrons should also be used
for bonding we want to break up the pair.
Promotion (sp)
Well, we’ve got four unparied
electrons, but….one is very different
from the other three.
H
|
H- C -H
|
H
All experimental results show that the four
C-H bonds are equivalent.
Hybridization
sp3 hybridization:
s + 3p  4 sp
Figure 9.30 The hybrid orbitals of a carbon atom in methane.
(a) One s-orbital and three p-orbitals blend into four sp3 hybrid
orbitals that each point toward one apex of a tetrahedron. (b)
The directions of the four orbitals.
Figure 9.31 (a) Each C—H bond in methane is formed by the
pairing of an electron in a hydrogen 1s-orbital and an electron in
one of the four sp3 hybrid orbitals of carbon. (b) As a result, there
are four equivalent -bonds in a tetrahedral arrangement.
End to end
σ-bond formed by s and sp3
Promotion and Hybridization in NH3
N
N
Nonbonding orbital (lone pair)
N
Figure 9.32 Two common hybridization schemes. (a) An s- and two porbitals can blend together to give three sp2 hybrid orbitals that point
toward the corners of an equilateral triangle. (b) An s-orbital and a porbital hybridize into two sp hybrid orbitals that point in opposite
directions. Only one of the orbitals is shown in each case.
O
H
C
OH
CH2O
sp2 hybridization:
s + 2p  3 sp
Figure 9.33 The atomic orbitals that overlap to form the three
-bonds in formaldehyde, CH2O. LP, lone pair.
Figure 9.34 Unhybridized p-orbitals on the C and O atoms
overlap to form the -bond in formaldehyde.
O
H
C
OH
Figure 9.35 The atomic orbitals that overlap to form the two bonds in carbon dioxide. The lone pairs are also shown. Note
that the planes of -bonds around each C atom are
perpendicular to each other. LP, lone pair.
Figure 9.36 Each of the two unhybridized p-orbitals on the C
atom in CO2 overlaps with an unhybridized p-orbital on an O
atom to form a -bond between each pair of atoms. The two bonds are perpendicular to each other.
Hybrids Including d-Orbitals
Figure 9.37 One of the five sp3d hybrid orbitals, and their five
directions, that may be formed when d-orbitals are available.
They form a trigonal bipyramidal arrangement of electron
pairs. Each arrow represents the location of a bond or electron
pair.
F
F
As
F
AsF5
F
F
As: 3d104s24p3
sp3d
Hybrids Including d-Orbitals
Figure 9.38 One of the six sp3d2 hybrid orbitals, and their six
directions, that may be formed when d-orbitals are available.
They form an octahedral arrangement of electron pairs. Each
arrow represents the location of a bond or electron pair.
S: 3s23p4
sp3d2
How the electron arrangement of a molecule is matched to a
hybridization scheme.
1.
2.
3.
4.
Draw Lewis structure and determine the electron arrangement about the
central atom. The number of σ-bonds and lone pairs required for the
electron arrangement is the number of orbitals used by the central atom.
Construct hybrid orbitals from atomic orbitals using the same number of
atomic orbitals as hybrid orbitals required. Start with the s-orbital, then
add p- and d-orbitals as needed to create the patterns listed in Table 9.5.
Use any remaining p-orbitals to form π-bonds with the p-orbitals of the
adjacent atoms
Check Table 9.5.
Example: Water
1.
2.
3.
4.
Lewis structure
4 orbitals required for the central atomsp3
No unhybridized p-orbitals after sp3 hybridization
Result: 4 sp3 orbitals.
sp3
Example: SF6
1.
2.
3.
4.
Lewis structure
6 orbitals required for the central atomsp3d2
No unhybridized p-orbitals after sp3d hybridization
Result: 6 sp3d2 orbitals.
..
:F:
|
..
..
:F ..
..
S
:F
:F
..
..
-F:
..
..
|
..
:F
..
S: 3s23p4
sp3d2
Classroom Exercise
1.
2.
3.
4.
Lewis structure
5 orbitals required for the central atomsp3d
No unhybridized p-orbitals after sp3d hybridization
Result: 5 sp3d orbitals.
P: 3s23p3
sp3d
Multiple Carbon-Carbon Bonds
Ethane
4 bonds for the central atoms  sp3 hybridization.
C-C Single Bond
Figure 9.40 The valence-bond description of the bonding in an ethane
molecule, C2H6. The boundary surfaces of only two of the bonds are
shown. Each pair of neighboring atoms is linked by a -bond formed
by the pairing of electrons in either H1s-orbitals or Csp3 hybrid
orbitals. All the bond angles are close to 109.5° (the tetrahedral
angle).
Experimental results indicate that all six atoms are in the same plane
and the H-C-H angle is 120o, suggesting sp2 hybridization.
-Bonds in Ethene
Figure 9.41 The atomic orbitals that overlap to form the five bonds in ethene (ethylene).
Carbon-Carbon Double Bond
Figure 9.42 The formation of a -bond from the side-by-side
overlap of the unhybridized p-orbitals on the C atoms in
ethene.
Carbon-Carbon Double Bond
Figure 9.43 The bonding pattern in ethene (ethylene), showing the
framework of -bonds and the single -bond (represented by two
lines above and below the -bond) formed by side-to-side overlap of
unhybridized C2p-orbitals. The double bond is resistant to twisting
because twisting would reduce the overlap between the two C2porbitals and weaken the -bond.
sp2 Hybridization: Ring
Figure 9.44 The framework of -bonds in benzene: each
carbon atom is sp2 hybridized, with bond angles of 120° in the
hexagonal molecule. Only bonding around one carbon atom is
shown explicitly; all the others are the same.
sp2 Hybridization: Benzene
Figure 9.45 The unhybridized p-orbital on each C atom in
benzene can form a -bond with either of its immediate
neighbors. Two arrangements are possible, each one
corresponding to one Kekulé structure. One Kekulé structure
and the corresponding -bonds are shown here.
Resonance
Figure 9.46 As a result of resonance between two structures
like the one shown in the preceding illustration
(corresponding to resonance of the two Kekulé structures),
the -electrons form double donut-shaped clouds, one above
and one below the plane of the ring.
Carbon-Carbon Triple Bond
Figure 9.47 The atomic orbitals that overlap to form the three
-bonds in ethyne (acetylene).
Carbon-Carbon Triple Bond
Figure 9.48 The pattern of bonding in ethyne (acetylene). (a)
The carbon atoms are sp hybridized, and the two remaining porbitals on each ring form two perpendicular -bonds. (b) The
resulting pattern is very similar to that for nitrogen (see Fig.
9.29), but two C—H groups replace the N atoms.
Exercise
Describe the structure of a formic acid molecule, HCOOH in terms of
hybrid orbitals, bon angles, and σ- and π-bonds.
Carbon is sp2 hybridized,
OH oxygen is sp3,
CO oxygen is sp2.
Figure 9.49 The pattern of -bonds in formic acid (methanoic
acid). LP, lone pair.
Carbon is sp2 hybridized,
OH oxygen is sp3,
CO oxygen is sp2.
Exercise
Describe the structure of a hydrogen cyanide molecule, HCN in terms of
hybrid orbitals, bon angles, and σ- and π-bonds.
Carbon is sp hybridized.
Nitrogen is sp hybridized.
H-C  N :
Classroom Exercise
Describe the structure of a propane molecule, CH3-CH=CH2 in terms of
hybrid orbitals, bon angles, and σ- and π-bonds.
Importance of π-Bonds
Single bond is flexible.
Double bond prevents one part of
a molecule from rotating relative
to another part.
Importance of π-Bonds
I see it!
Molecular Orbital (MO) Theory
Electrons are distributed to entire molecule.
Case 1: VB Invalid
Only 12 valence electrons.
But at least 7 bonds
(14 electrons) are needed
to bind 8 atoms!
Electron deficient compounds
cannot be understood with VB.
Case 2: VB Invalid
Figure 9.50 The paramagnetic properties of oxygen are evident when
liquid oxygen is poured between the poles of a magnet. The liquid
sticks to the magnet instead of flowing past it.




OO
O2 is a biradical!


OO
Wrong!




O O

..

..
.
..
O
..
.. 2
.
Bonding Orbitals
Figure 9.51 When two 1s-orbitals overlap in such a way that
they have the same signs in the same regions of space, their
wavefunctions (red lines) interfere constructively and give rise
to a region of enhanced amplitude between the two nuclei
(blue line).
Anti-bonding Orbitals
Figure 9.52 When two 1s-orbitals overlap in the same region of
space with opposite signs, their wavefunctions (red lines)
interfere destructively and give rise to a region of diminished
amplitude and a node between the two nuclei (blue line).
Figure 9.53 A molecular orbital energy-level diagram for the
bonding and antibonding molecular orbitals that can be built
from two s-orbitals. The signs of the s-orbitals are depicted by
the different shades of blue.
Ground State of the MO of H2
Figure 9.54 The two electrons in an H2 molecule occupy the
lower (bonding) molecular orbital and result in a stable
molecule.

2
1s
Two MOs of H2
Lowest Unoccupied MO
Highest Occupied MO
Why are HOMO and LUMO important?
• HOMO offers information on the electrons in the “top” orbitals that
would determine the physical properties of the ground state.
• LUMO offers the information on the most probable orbitals the
electrons may go into when a reaction occurs.
• Combing HOMO and LUMO, chemical reactions and properties
related to them can be understood.
• In particular, the gap between LUMO and HOMO is the most important
factor that affect the properties of materials (semiconductors, lightemitting materials, displaying materials etc ) and biomolecules
(DNA, RNA, enzymes, proteins, ATP, sacchrides etc).
MO of He2
Figure 9.55 Two of the four electrons in a hypothetical He2
molecule occupy the bonding orbital, but the Pauli principle
forces the remaining two electrons to occupy the antibonding
orbital. As a result, the He2 molecule does not have a lower
energy than two separate He atoms and does not exist as a
stable species.
In total: 4 electrons
He 2 :  
2
1s
2*
1s
MO of Li2 through N2
Figure 9.56 A typical molecular orbital energy-level diagram for
the homonuclear diatomic molecules Li2 through N2. Each box
represents one molecular orbital and can accommodate up to
two electrons.
MO of N2
B: number of electrons
in bonding orbitals
A: number of electrons
in anti-bonding orbitals
Bond Order = (B-A)/2＝(8-2)/2=3
In total: 10 electrons
N2 :    
2
2s
2*
2s
4
2p
2
2p
Classroom Exercise
Draw the molecular orbital energy-level diagram for the
homonuclear diatomic molecules Be2.
In total: 4 “valence” electrons
Be 2 :  
2
2s
Bond order = 0
2*
2s
MO of O2 and F2
Figure 9.57 The molecular orbital energy-level diagram for the
homonuclear diatomic molecules to the right of Period 2,
specifically O2 and F2.
O2 :     
2
2s
2*
2s
2
2p
4
2p
F2 :     
2
2s
2*
2s
2
2p
4
2p
2
2p
4
2p
Why Is O2 a Biradical and Paramagnetic?
In total: 12 electrons
O2 :    
2
2s
2*
2s
4
2p
N
*2
2p
Bond order = 2
..
..
.
..
O
..
.
2
..
Many fundamentally important biochemical and physiological processes are
related to this page!
S
MO of F2
In total: 14 electrons
    
2
2s
*2
2s
2
2p
Bond order = 1
4
2p
*4
2p
Classroom Exercise
Deduce the electronic configuration
and bond order of the carbide ion ( C 22  )
   
2
2s
2*
2s
4
2p
2
2p
Bond Order = (B-A)/2 ＝(8-2)/2=3
Molecular Spectroscopy:
Ultraviolet (UV) and Visible Absorption (Vis)
Investigating Matter 9.2 (a) The absorption spectrum of
chlorophyll as a plot of percentage absorption against
wavelength. Chlorophyll a is shown in red, chlorophyll b in
blue.
That’s the reason why
living leaves look green,
(some leaves turn red in fall)
and dead leaves look yellow
or grey.
Investigating Matter 9.2 (b) In a -to- * transition, an electron
in a bonding -orbital is excited into an empty antibonding *orbital.
160 nm (UV)
Conjugated Double Bond
• Bond forming and breaking are responsible
for our vision.
I see it!
•There are many energy levels for this type of bonds
(visible lights may be absorbed) in carotene. The transitions between
these levels are responsible for the color of carrots, mangoes,
persimmons and shrimps.
CH3
CH3
MO of Polyatomic Molecules
• Delocalized Electrons:
The MOS spread over all atoms.
The electron pairs in bonding orbitals
help to bind together the whole molecule,
not just a pair of atoms.
Case Study 9 (a) Sunscreens protect our
skin from ultraviolet radiation, which can
cause burns and even skin cancer.
O
C
NH2
Tanning cream: zinc oxide, p-Aminobenzoic acid (PABA), etc
OH
Case Study 9 (b) The sensitivity of living things to
electromagnetic radiation in the ultraviolet region
(green) superimposed over the ultraviolet spectrum
of solar radiation (violet).
More Complicated Cases of MO
Only 12 valence electrons.
But at least 7 bonds
(14 electrons) are needed
to bind 8 atoms!
Electron deficient compounds
cannot be understood with VB.
Actually in this case, 2 electrons
bind 3 nuclei together!
Diborane B2H6
Assignment for Chapter 9
9.7, 9.12, 9.15, 9.21, 9.25, 9.28, 9.33
9.44, 9.54, 9.56
Quiz
• What is the most significant difference
between MO and VB theories?
• What is LUMO? What is HOMO? Why are
they important?
• Explain with MO theory that oxygen
molecule is a biradical and paramagnetic.
What is the most significant difference
between MO and VB theories?
In VB, bonding electrons are paired up and
are supposed to be in the region between the
bonding atoms. In MO, however, bonding
electrons belong to the entire molecule.
What is LUMO?What is HOMO?
Why are they important?
• LUMO: lowest unoccupied MO.
• HOMO: highest occupied MO.
• HOMO offers information on the electrons in the “top” orbitals
that would determine the physical properties of the ground state.
• LUMO offers the information on the most probable orbitals the
electrons may go into when a reaction occurs.
• Combing HOMO and LUMO, chemical reactions and properties
related to them can be understood.
• In particular, the gap between LUMO and HOMO is the most
important factor that affect the properties of materials
(semiconductors, light-emitting materials, displaying materials etc
) and biomolecules (DNA, RNA, enzymes, proteins, ATP,
sacchrides etc).
Why Is O2 a Biradical and Paramagnetic?
In total: 12 electrons
O2 :    
2
2s
2*
2s
4
2p
Bond order = 2
..
..
.
..
O
..
2
..
.
N
*2
2p
S