Chapter 9 Molecular Structure

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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.

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Assignment for Chapter 9
9.7, 9.12, 9.15, 9.21, 9.25, 9.28, 9.33 9.44,
9.54, 9.56
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Some of the basic geometrical shapes that are
used to describe the shapes of simple molecules
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Octahedral
Tetrahedral
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Shape names and bond angles(lone pairs not
included)
T-shaped
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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.

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Electron pairs or bonds are as far away from one
another as possible and so experience the minimal
repulsion from other electrons
VSEPR Model (1) locate the high concentrations
The positions taken up by regions of high
electron concentration (green) around a central
atom.
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VSEPR Model (2) determine the shapes
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.
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Electron pairs or bonds are as far away from one
another as possible and so experience the minimum
repulsion from other electrons
VSEPR Model Example
Shape linear (two highs)
Shape trigonal planar (three highs)
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Electron pairs or bonds are as far away from one
another as possible and so experience the minimum
repulsion from other electrons
VSEPR Model Example
Shape trigonal bipyramidal (five highs)
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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.

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Classroom Exercise

• Predict the structure of SiCl4 and AsF5

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Classroom Exercise

• Predict the structure of SiCl4 and AsF5

Si 3s23p2
As 3d104s24p3
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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.
(Three highs)
(Two highs)
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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
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Example
Two central atoms, no lone pairs.
The most possible VSEPR arrangement
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Formaldehyde

• CH2O

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Classroom Exercise

• Predict the structure of HCN

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Different resonance structures correspond to a
single shape
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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.
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Lone Pair on the Central Atom
AX3E
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Repulsion Order
Lone pair/lone pair gt lone pair/bonding pair gt
bonding pair/bonding pair
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Example
H2O
AX2E2
Shape Angular (not tetrahedron!)
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Three views of water molecular shape
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NH3
AX3E
Triangular pyramid (NOT tetrahedron!)
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Classroom Exercise

• Predict the shape of NO2-

AEX2
Smaller angle
Angular (NOT planar triangle!)
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AX4E
Axial lone pair
Equatorial lone pair
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AX4E2
Square planar (NOT octahedron!)
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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 gt lone pair/bonding pair gt
bonding pair/bonding pair
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Example SF4
AX4E
Equatorial lone pair Bent seesaw
T-shaped (NOT triangular bipyramid!)
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ClF3
T-shaped (NOT triangular bipyramid!)
AX3E
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Classroom Exercise XeF4
AX4E2
Square planar (NOT octahedron!)
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Quiz

• Write the VSEPR formula of water and sulfur
  tetrafluoride, draw and name their structures.

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Answer

• Write the VSEPR formula of water and sulfur
  pentafluoride. Draw and name their structures.

AX2E2
AX4E
T-shaped (not triangular bipyramid!)
Angular (not tetrahedron!)
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??
VSEPR ? Ode to VSEPR

• ???????,
• ????????
• ???????,
• ????????
• 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.

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Charge Distribution in Molecules
Where does have high or low electron density
(concentration)?
Electron distribution is responsible for
molecular properties and functions.
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Polar Bonds
r
Partial positive charge
Partial negative charge
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1 Debye
r100 pm (1 A)
-e
e
µ1 D
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Polar Bonds Forming Polar Molecules
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Polar Bonds Forming Nonpolar Molecules
Total dipole moment 0
Partial negative charge
Partial negative charge
Partial positive charge
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Charge distribution of CO2
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The shape of a molecule governs whether it is
polar or not.
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Polar
Nonpolar
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Molecular shape and polarity
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BF3
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SF4
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SF6
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The strengths and lengths of bonds
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• The three normal vibrational modes
• of H2O.

(b) The four normal vibrational modes of
CO2.
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Infrared spectroscopy measures normal vibrational
frequencies
The infrared spectrum of tryptophan (an amino
acid).
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Bond enthalpy ?HB
H2(g)?2H(g) ?Ho436 kJ/mol
?HB(H-H) 436 kJ/mol
?HB(F-CF3) 484 kJ/mol
?HB(H-CH3) 412 kJ/mol
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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.
58
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
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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.

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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.
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Multiple bond strength lt bond order single bond
strength
Double bonds are shorter than single bonds,
leading Bond strengths larger than double of
single-bond strengths.
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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 lt bond order single bond
strength
63
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.
64
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.
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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)
?Ho238 463 701 (kJ/mol)
Bond formed H-I(299 kJ/mol), C-O (360 kJ/mol)
?Ho299 360 659 (kJ/mol)
Te reaction enthalpy
?Hr701 - 659 42 (kJ/mol)
endothermic
66
Classroom exercise

• Which of the following reactions is exothermic?

Tables 9.2,3
(a) CCl3CHCl2 (g) HF(g)?CCl3CHClF (g) HCl (g)
(b) CCl3CHCl2 (g) HF(g)?CCl3CCl2F (g) H2 (g)

• 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)
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Quiz
1. Is the dipole moment of SF4 zero or not?
2. Explain what is average bond enthalpy and
its applications.
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Quiz Is the dipole moment of SF4 zero or not?
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Example
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Bond Length Depends on Bond Order
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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

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Valence-Bond (VB) Theory
Chemical bonds are formed by pairs of (unpaired)
valence electrons
There are three types of valence bond
s-bond p-bond hybrid bonds
77
Figure 9.25 When the electrons (depicted ? and ?)
in two hydrogen 1s-orbitals 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.
VB for H2
Notice the difference between VB and Lewis
model.
1s
1s
The spins of the two electrons must be opposite
to each otehr
End to end paring
Overlap (high electron density)
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VB for HF
79
Figure 9.26 A ?-bond can also be formed when
electrons in 1s- and 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
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VB for N2
81
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
82
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.
p-bond in N2
Side by side pairing
83
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 p-bonds in N2
84
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 s-bonds or side by side to
  form p-bonds.

• A single bond is a s-bond
• A double bond is a s-bond plus a p-bond
• A triple bond is a s-bond plus two p-bonds

85
More Examples

• How many s-bonds and how many p-bonds in CO2 and
  CO?

• CO2 OCO
• CO CO

• CO2 OCO s-bond 2, p-bond 2
• CO CO, s-bond 1, p-bond 2

86
Classroom Exercise

• How many s-bonds and how many p-bonds in NH3 and
  CH2O?

• NH3 s-bond 3, p-bond 0
• CH2O s-bond 3, p-bond 1

87
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.
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Promotion (s?p)
Well, weve got four unparied electrons, but.one
is very different from the other three.
All experimental results show that the four C-H
bonds are equivalent.
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Hybridization
sp3 hybridization s 3p ? 4 sp
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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.
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Figure 9.31 (a) Each CH 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
s-bond formed by s and sp3
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Promotion and Hybridization in NH3
Nonbonding orbital (lone pair)
93
Figure 9.32 Two common hybridization schemes. (a)
An s- and two p-orbitals can blend together to
give three sp2 hybrid orbitals that point toward
the corners of an equilateral triangle. (b) An
s-orbital and a p-orbital hybridize into two sp
hybrid orbitals that point in opposite
directions. Only one of the orbitals is shown in
each case.
sp2 hybridization s 2p ? 3 sp
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Figure 9.33 The atomic orbitals that overlap to
form the three ?-bonds in formaldehyde, CH2O. LP,
lone pair.
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Figure 9.34 Unhybridized p-orbitals on the C and
O atoms overlap to form the ?-bond in
formaldehyde.
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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.
97
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.
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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.
Hybrids Including d-Orbitals
As 3d104s24p3
sp3d
100
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.
Hybrids Including d-Orbitals
S 3s23p4
sp3d2
101
How the electron arrangement of a molecule is
matched to a hybridization scheme.

1. Draw Lewis structure and determine the electron
   arrangement about the central atom. The number
   of s-bonds and lone pairs required for the
   electron arrangement is the number of orbitals
   used by the central atom.
2. 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.
3. Use any remaining p-orbitals to form p-bonds with
   the p-orbitals of the adjacent atoms
4. Check Table 9.5.

102
Example Water

1. Lewis structure
2. 4 orbitals required for the central atom?sp3
3. No unhybridized p-orbitals after sp3
   hybridization
4. Result 4 sp3 orbitals.

103
Example SF6

1. Lewis structure
2. 6 orbitals required for the central atom?sp3d2
3. No unhybridized p-orbitals after sp3d
   hybridization
4. Result 6 sp3d2 orbitals.

S 3s23p4
sp3d2
104
Classroom Exercise

1. Lewis structure
2. 5 orbitals required for the central atom?sp3d
3. No unhybridized p-orbitals after sp3d
   hybridization
4. Result 5 sp3d orbitals.

P 3s23p3
sp3d
105
Multiple Carbon-Carbon Bonds
106
Ethane
4 bonds for the central atoms ? sp3 hybridization.
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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).
C-C Single Bond
108
Experimental results indicate that all six atoms
are in the same plane and the H-C-H angle is
120o, suggesting sp2 hybridization.
109
Figure 9.41 The atomic orbitals that overlap to
form the five ?-bonds in ethene (ethylene).
?-Bonds in Ethene
110
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
111
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
C2p-orbitals and weaken the ?-bond.
Carbon-Carbon Double Bond
112
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 Ring
113
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.
sp2 Hybridization Benzene
114
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.
Resonance
115
Figure 9.47 The atomic orbitals that overlap to
form the three ?-bonds in ethyne (acetylene).
Carbon-Carbon Triple Bond
116
Figure 9.48 The pattern of bonding in ethyne
(acetylene). (a) The carbon atoms are sp
hybridized, and the two remaining p-orbitals on
each ring form two perpendicular ?-bonds. (b) The
resulting pattern is very similar to that for
nitrogen (see Fig. 9.29), but two CH groups
replace the N atoms.
Carbon-Carbon Triple Bond
117
Exercise
Describe the structure of a formic acid molecule,
HCOOH in terms of hybrid orbitals, bon angles,
and s- and p-bonds.
Carbon is sp2 hybridized, OH oxygen is sp3, CO
oxygen is sp2.
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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.
119
Exercise
Describe the structure of a hydrogen cyanide
molecule, HCN in terms of hybrid orbitals, bon
angles, and s- and p-bonds.
Carbon is sp hybridized. Nitrogen is sp
hybridized.
120
Classroom Exercise
Describe the structure of a propane molecule,
CH3-CHCH2 in terms of hybrid orbitals, bon
angles, and s- and p-bonds.
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Importance of p-Bonds
Single bond is flexible.
Double bond prevents one part of a molecule from
rotating relative to another part.
122
Importance of p-Bonds
I see it!
123
Molecular Orbital (MO) Theory
Electrons are distributed to entire molecule.
124
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.
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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.
Case 2 VB Invalid
126
O2 is a biradical!
Wrong!
127
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).
Bonding Orbitals
128
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).
Anti-bonding Orbitals
129
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.
130
Figure 9.54 The two electrons in an H2 molecule
occupy the lower (bonding) molecular orbital and
result in a stable molecule.
Ground State of the MO of H2
131
Two MOs of H2
Lowest Unoccupied MO
Highest Occupied MO
132
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, light-
• emitting materials, displaying materials etc
  ) and biomolecules
• (DNA, RNA, enzymes, proteins, ATP, sacchrides
  etc).

133
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.
MO of He2
In total 4 electrons
134
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 Li2 through N2
135
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)/23
In total 10 electrons
136
Draw the molecular orbital energy-level diagram
for the homonuclear diatomic molecules Be2.
Classroom Exercise
In total 4 valence electrons
Bond order 0
137
Figure 9.57 The molecular orbital energy-level
diagram for the homonuclear diatomic molecules to
the right of Period 2, specifically O2 and F2.
MO of O2 and F2
138
Why Is O2 a Biradical and Paramagnetic?
In total 12 electrons
Bond order 2
Many fundamentally important biochemical and
physiological processes are related to this page!
139
MO of F2
In total 14 electrons
Bond order 1
140
Classroom Exercise
Deduce the electronic configuration and bond
order of the carbide ion ( )
Bond Order (B-A)/2 (8-2)/23
141
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.
Molecular SpectroscopyUltraviolet (UV) and
Visible Absorption (Vis)
Thats the reason why living leaves look
green, (some leaves turn red in fall) and dead
leaves look yellow or grey.
142
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)
143
Conjugated Double Bond

• Bond forming and breaking are responsible for our
  vision.

I see it!
144

• 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.

145
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.

146
Case Study 9 (a) Sunscreens protect our skin from
ultraviolet radiation, which can cause burns and
even skin cancer.
Tanning cream zinc oxide, p-Aminobenzoic acid
(PABA), etc
147
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).
148
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.
Diborane B2H6
Actually in this case, 2 electrons bind 3 nuclei
together!
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Assignment for Chapter 9
9.7, 9.12, 9.15, 9.21, 9.25, 9.28, 9.33 9.44,
9.54, 9.56
152
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.

153
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.

154
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).

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Why Is O2 a Biradical and Paramagnetic?
In total 12 electrons
Bond order 2