Download Plan for Wed, 12 Aug 09

What Do Molecules Look Like?
The Lewis Dot Structure approach provides
some insight into molecular structure in terms
of bonding, but what about 3D geometry?
Recall that we have two types of electron
pairs: bonding and lone.
Valence-Shell Electron-Pair Repulsion
(VSEPR). 3D structure is determined by
minimizing repulsion of electron pairs.
Electron pairs
(both bonding
and lone) are
distributed
around a central
atom such that
electron-electron
repulsions are
minimized.
Electron pairs (both bonding and lone) are
distributed around a central atom such that
electron-electron repulsions are minimized.
2 electron pairs
3 electron pairs
4 electron pairs
Period 1, 2
5 electron pairs
6 electron pairs
Period 3 & beyond
Arranging Electron Pairs
• Must consider both bonding and lone pairs when
minimizing repulsion.
• Example: CH4 (bonding pairs only)
H
H C H
H
Lewis Structure
VSEPR Structure
Arranging Electron Pairs (cont.)
Example: NH3 (both bonding and lone pairs).
H
H N
H
Lewis Structure
VSEPR Structure
Note:
“electron pair geometry” vs.
“molecular shape”
VSEPR Structure Guidelines
The previous examples illustrate the strategy for applying
VSEPR to predict molecular structure:
1. Construct the Lewis Dot Structure
2. Arrange bonding/lone electron pairs in space such that
repulsions are minimized (electron pair geometry).
3. Name the molecular shape from the position of the
atoms.
VSEPR Shorthand:
1. Refer to central atom as “A”
2. Attached atoms are referred to as “X”
3. Lone pair are referred to as “E”
Examples:
CH4: AX4
NH3: AX3E
H2O: AX2E2
BF3: AX3
VSEPR: 2 electron pairs
Linear (AX2): angle
between bonds is 180°
Example: BeF2
F Be F
Experiments show that molecules with
multiple bonds can also be linear.
Multiple bonds are treated as a
single effective electron group.
F
F Be F
Be
F
180°
More than one central atom?
Determine shape around each.
VSEPR: 3 electron pairs
Trigonal Planar (AX3):
angle between bonds
is 120°
Multiple bond is
treated as a single
effective electron
group.
Example: BF3
F
F
F
B
120°
F
B
F
F
VSEPR: 4 electron pairs (cont.)
Tetrahedral (AX4): angle
between bonds is ~109.5°
Example: CH4
H
109.5°
H C H
H
tetrahedral e- pair geometry AND tetrahedral molecular shape
Bonding vs. Lone pairs
Bond angle in a tetrahedral arrangement of electron pairs
may vary from 109.5° due to size differences between
bonding and lone pair electron densities.
bonding pair is constrained by
two nuclear potentials; more
localized in space.
lone pair is constrained by
only one nuclear potential;
less localized (needs more
room).
VSEPR: 4 electron pairs
Trigonal pyramidal (AX3E): Bond angles are <109.5°,
and structure is nonplanar due to repulsion of lone pair.
Example: NH3
H
H N
H
107°
tetrahedral e- pair geometry; trigonal pyramidal molecular shape
VSEPR: 4 electron pairs (cont.)
Classic example of tetrahedral angle
shift from 109.5° is water (AX2E2):
104.5o
“bent”
tetrahedral e- pair geometry; bent molecular shape
VSEPR: 4 electron pairs (cont.)
Comparison of CH4 (AX4), NH3 (AX3E), and H2O
(AX2E2):
AX2E
AX3E
AX2E2
1. Refer to central atom as “A”
2. Attached atoms are referred to as “X”
3. Lone pair are referred to as “E”
Molecular vs. Electron-Pair Geometry
H
O
N
C
F
H
H
H
Central Atom
Compound
Electron-Pair Geometry
Molecular Shape
Carbon, C
CH4
tetrahedral
tetrahedral
Nitrogen, N
NH3
tetrahedral
trigonal pyramidal
Oxygen, O
H2O
tetrahedral
bent
Fluorine, F
HF
tetrahedral
linear
What is the electron-pair geometry and the
molecular shape for HCFS?
S
a) trigonal planar, bent
H
b) trigonal planar, trigonal planar
c) tetrahedral, trigonal planar
d) tetrahedral, tetrahedral
C
F
VSEPR: Beyond the Octet
Systems with expanded valence shells will have
five or six electron pairs around a central atom.
F
Cl
Cl
Cl
F
P
Cl
S
F
F
Cl
F
90°
120°
F
90°
F
F
F
S
F
F
F
90°
VSEPR: 5 electron pairs
• Consider the structure of SF4 (34 e-, AX4E)
• What is the optimum arrangement of electron pairs around S?
F
F
F
F
F
??
S
F
S
F
F
F
S
F
F
F
Compare e– pair angles
lone-pair / bond-pair: two at 90o, two at 120o
bond-pair / bond-pair: four at 90o, one at 120o
three at 90o
three at 90o, three at 120o
Repulsive forces (strongest to weakest):
lone-pair/lone-pair > lone-pair/bond-pair > bond-pair/bond-pair
VSEPR: 5 electron pairs
The optimum structure maximizes the angular
separation of the lone pairs.
I3- (AX2E3):
5-electron-pair geometries
AX4E
AX3E2
AX2E3
our
previous
example
VSEPR: 6 electron pairs
Which of these is the more likely structure?
See-saw
Square Planar
6-electron-pair geometries
AX5E
AX4E2
our
previous
example
Molecular Dipole Moments
We can use VSEPR to
determine the polarity of a
whole molecule.
1. Draw Lewis structures to
determine 3D arrangement
of atoms.
2. If one “side” of the molecule has
more EN atoms than the other,
the molecule has a net dipole.
Shortcut: completely
symmetric molecules will not
have a dipole regardless of
the polarity of the bonds.
Molecular Dipoles
The C=O bonds have dipoles
of equal magnitude but
opposite direction, so there is
no net dipole moment.
The O-H bonds have dipoles of
equal magnitude that do not
cancel each other, so water has
a net dipole moment.
Molecular Dipoles (cont.)
symmetric
asymmetric
symmetric
Molecular Dipole Example
• Write the Lewis dot and VESPR structures for
CF2Cl2. Does it have a dipole moment?
F
F
32 e-
Cl
C
Cl
F
Cl
F
Cl
Tetrahedral
Advanced VSEPR Application
Molecules with more than one central atom…
methanol (CH3OH)
H
H C O
H
tetrahedral e- pairs
tetrahedral shape
tetrahedral e- pairs
bent shape
H
The VSEPR Table
# e- pairs
e- Geom.
Molec. Geom.
2
AX2
BeF2
linear
linear
3
AX3
BF3
trigonal planar
trigonal planar
AX2E
O3
trigonal planar
bent
AX4
CH4
tetrahedral
tetrahedral
AX3E
NH3
tetrahedral
pyramidal
AX2E2
H2O
tetrahedral
bent
4
The VSEPR Table
# e- pairs
5
6
e- Geom.
Molec. Geom.
AX5
PF5
trigonal
bipyramidal
trigonal
bipyramidal
AX4E
SF4
trigonal
bipyramidal
see saw
AX3E2
ClF3
trigonal
bipyramidal
T-shaped
AX2E3
I3-
trigonal
bipyramidal
linear
AX6
SF6
octahedral
octahedral
AX4E2
XeF4
octahedral
square planar
What is the expected shape of ICl2+?
+
20 e-
Cl
I
Cl
AX2E2
A. linear
C. tetrahedral
B. bent
D. square planar
Valence Bond Theory
Basic Principle of Localized Electron Model:
A covalent bond forms when the orbitals from two
atoms overlap and a pair of electrons occupies the
region between the two nuclei.
Rule 1: Maximum overlap. The bond strength depends on
the attraction of nuclei to the shared electrons, so:
The greater the orbital overlap, the stronger the bond.
Valence Bond Theory
Basic Principle of Localized Electron Model:
A covalent bond forms when the orbitals from two
atoms overlap and a pair of electrons occupies the
region between the two nuclei.
Rule 2: Spins pair. The two electrons in the overlap region
occupy the same space and therefore must have opposite spins.
There may be no more than 2 electrons in a molecular orbital.
Valence Bond Theory
Basic Principle of Localized Electron Model:
A covalent bond forms when the orbitals from two
atoms overlap and a pair of electrons occupies the
region between the two nuclei.
Rule 3: Hybridization. To explain experimental observations,
Pauling proposed that the valence atomic orbitals in a
molecule are different from those in the isolated atoms. We
call this concept
Hybridization
What is hybridization?
•
Atoms
adjust to meet the “needs” of the
molecule.
• In a molecule, electrons rearrange in an attempt
to give each atom a noble gas configuration and
to minimize electron repulsion.
• Atoms in a molecule adjust their orbitals through
hybridization in order for the molecule to have a
structure with minimum energy.
• The source of the valence electrons is not as
important as where they are needed in the
molecule to achieve a maximum stability.
Example: Methane
• 4 equivalent C-H covalent bonds
• VSEPR predicts a tetrahedral geometry
The Valence Orbitals of a Carbon Atom
Carbon: 2s22p2
How do we explain formation of 4 equivalent C-H bonds?
Hybridization: Mixing of Atomic Orbitals to
form New Orbitals for Bonding
+
–
+
–
+
–
+
+
– +
+
–
–
Other Representations of Hybridization:
y1 = 1/2[(2s) + (2px) + (2py) + (2pz)]
y2 = 1/2[(2s) + (2px) - (2py) - (2pz)]
y3 = 1/2[(2s) - (2px) + (2py) - (2pz)]
y4 = 1/2[(2s) - (2px) - (2py) + (2pz)]
Hybridization is related to the number of
valence electron pairs determined from VSEPR:
Methane (CH4)
VSEPR: AB4
 tetrahedral
 sp3 hybridized
Electron pair
geometry
determines
hybridization, not
vice versa!!
109.47 º
Hybridization is related to the number of
valence electron pairs determined from VSEPR:
Ammonia (NH3)
VSEPR: AB3E
 tetrahedral
 sp3 hybridized
H
N
108.1 º
H
H
Hybridization is related to the number of
valence electron pairs determined from VSEPR:
Water (H2O)
VSEPR: AB2E2
 tetrahedral
 sp3 hybridized
105.6 º
s bonding and p bonding
• Two modes of bonding are important for
1st and 2nd row elements: s bonding and p bonding
• These two differ in their relationship to the
internuclear axis:
s bonds have electron density ALONG the axis
p bonds have electron density
ABOVE AND BELOW the axis
Problem: Describe the hybridization and bonding of
the carbon orbitals in ethylene (C2H4)
VSEPR: AB3
 trigonal planar
 sp2 hybridized orbitals for s bonding
sp2 hybridized orbitals used for s bonding
remaining p orbital used for p bonding
Bonding in ethylene (C2H4)
Problem: Describe the hybridization and bonding of
the carbon orbitals in Carbon Dioxide (CO2)
VSEPR: AB2
 linear
 sp hybridized orbitals for s bonding
Bonding in Carbon Dioxide (CO2)
Atoms of the same kind can have different hybridizations
CH3
C
N:
Acetonitrile (important solvent and industrial chemical)
H
H
C2
C1
N
Bonds
H
s
C2: AB4
C1: AB2
2s2 2px2py
sp3
N:
sp sp
p p
s
p
sp
p p
p
ABE
2s2 2px2py2pz
sp
lone pair
sp
What have we learned so far?
• Molecular orbitals are combinations of
atomic orbitals
• Atomic orbitals are “hybridized” to satisfy
bonding in molecules
• Hybridization follows simple rules that can
be deduced from the number of chemical
bonds in the molecule and the VSEPR model
for electron pair geometry
Hybridization
• sp3 Hybridization (CH4)
– This is the sum of one s and three p orbitals on the
carbon atom
– We use just the valence orbitals to make bonds
– sp3 hybridization gives rise to the tetrahedral
nature of the carbon atom
Hybridization
• sp2 Hybridization (H2C=CH2)
– This is the sum of one s and two p orbitals on the
carbon atom
– Leaves one p orbital uninvolved – this is free to
form a p bond (the second bond in a double
bond)
Hybridization
• sp Hybridization (O=C=O)
– This is the sum of one s and one p orbital on the
carbon atom
– Leaves two p orbitals free to bond with other
atoms (such a O in CO2), or with each other as in
HC≡CH
General Notes
• This is a model and only goes so far, but it is
especially helpful in understanding geometry and
expanding Lewis dot structures.
• Orbitals are waves. Hybridized orbitals are just the
sums of waves – constructive and destructive
interference.
What is important to know
about hybridization?
• You should be able to give the hybridization of an
atom in a molecule based on the formula given.
• Example: CH3-CH2-CHO
• Step 1: Draw the Lewis Dot Structure
What is important to know
about hybridization?
• Step 2: What is the electron pair geometry and
molecular shape?
AXE2
Trigonal Planar
AX3
AX4
Tetrahedral
AX4
Tetrahedral
Trigonal Planar
What is important to know
about hybridization?
• Step 3: Use the molecular shape to determine the
hybridization.
AXE2
Trigonal Planar
sp2
AX3
AX4
Tetrahedral
sp3
AX4
Tetrahedral
sp3
Trigonal Planar
sp2
The Localized Electron Model is very powerful
for explaining geometries and basic features of
bonding in molecules, but it is just a model.
Major limitations of the LE model:
•
•
•
Assumes electrons are highly localized between the nuclei (sometimes
requires resonance structures)
Doesn’t easily deal with unpaired electrons (incorrectly predicts physical
properties in some cases)
Doesn’t provide direct information about bond energies
Example: O2
.. ..
Lewis dot structure O=O
.. ..
All electrons are paired
Contradicts experiment!
The Molecular Orbital Model
Basic premise: When atomic orbitals interact
to form a bond, the result is the formation of
new molecular orbitals
HY = EY
Important features of molecular orbitals:
1. Atomic Orbitals are solutions of the Schrödinger
equation for atoms.
Molecular orbitals are the solutions of the same
Schrödinger equation applied to the molecule.
Molecular Orbital Theory
2. Atomic Orbitals can hold 2 electrons with opposite spins.
Molecular Orbitals can hold 2 electrons with opposite
spins.
3. The electron probability for the Atomic Orbital is given
by Y2.
The electron probability for the Molecular Orbital is given
by Y2.
4. Orbitals are conserved - in bringing together 2 atomic
orbitals, we have to end up with 2 molecular orbitals!
How does this work?
Molecular Orbitals are simply
Linear Combinations of Atomic Orbitals
Example: H2
s anti-bonding (s*)
-
+
Molecular Orbitals have phases (+ or -)
+
s bonding
Next Question: Why does this work?
Constructive and Destructive
Interference
Constructive interference between
two overlapping orbitals leads to a
bonding orbital.
Destructive interference
between two orbitals of opposite
sign leads to an anti-bonding
orbital.
Bonding is driven by stabilization of electrons
• Electrons are negatively charged
• Nuclei are positively charged
= + = nucleus
The bonding combination puts electron density between the two nuclei stabilization
The anti-bonding combination moves electron density away from region
between the nuclei - destabilization
MO Diagrams
• We can depict the relative energies of molecular
orbitals with a molecular orbital diagram:
The new molecular orbital is lower in energy than the atomic orbitals
s* M.O. is raised in energy
s M.O. is lowered in energy
H atom: (1s)1 electron configuration
H2 molecule: (s1s)2 electron configuration
Same as previous description of bonding
s*
s
Review of Orbital Filling
• Pauli Exclusion Principle:
– No more than 2 e- in an orbital, spins must be
paired (↑↓)
• Aufbau Principle (a.k.a. “Building-Up”):
– Fill the lowest energy levels with electrons first
• 1s 2s 2p 3s 3p 4s 3d 4p …
• Hund’s Rule:
– When more than one orbital has the same energy,
electrons occupy separate orbitals with parallel
spins:
Yes
No
No
Filling Molecular Orbitals with Electrons
1) Orbitals are filled in order of increasing Energy (Aufbau
principle)
H2
Filling Molecular Orbitals with Electrons
He2
2) An orbital has a maximum capacity of two electrons with
opposite spins (Pauli exclusion principle)
Filling Molecular Orbitals with Electrons
3) Orbitals of equal energy (degenerate orbitals) are half filled,
with spins parallel, before any is filled completely (Hund’s rule)
Bond Order
Bond Order =
# bonding
electrons
#anti-bonding
electrons
2
The bond order is an indication of bond strength:
Greater bond order
Greater bond strength
(Shorter bond length)
Bond Order: Examples
Bond order = (2-0)/2 = 1
Single bond
H2
Stable molecule (436 kJ/mol
bond)
Bond order = (2-2)/2 = 0
He2
No bond!
Unstable molecule (0 kJ/mol
bond)
He2+
Bond order = (2-1)/2 = 1/2
Half of a single bond
Can be made, but its not very
stable (250 kJ/mol bond)
Fractional bond orders are okay!
H2+
Bond order = (1-0)/2 = 1/2
Half of a single bond
Can be made, but its not very
stable (255 kJ/mol bond)
Forming Bonds
• A s bond can be formed a number of ways:
– s, s overlap
– s, p overlap
– p, p overlap
Only orbitals of the same phase (+, +) can form bonds
Anti-bonding Orbitals
• For every bonding orbital we form, we also form an antibonding orbital:
MO Theory in Bonding
• Homonuclear atoms (H2, O2, F2, N2)
H2
(Only 1s
orbitals
available for
bonding)
Covalent Bonding in Homonuclear
Diatomics
• Atomic orbitals must overlap in space in order
to participate in molecular orbitals
• Covalent bonding is dominated by the valence
orbitals (only valence orbitals are shown in the
MO diagrams)
Covalent Bonding in Homonuclear
Diatomics
Region of
shared edensity
–
+
+
Valence configurations of the 2nd row atoms:
Li
2s1
Be
2s2
B
C
2s22p1 2s22p2
N
2s22p3
O
2s22p4
F
2s22p5
So far we have focused on bonding involving the
s orbitals.
What happens when we have to consider the p
orbitals?
For diatomic molecules containing atoms with valence electrons in
the p orbitals, we must consider three possible bonding
interactions:
= nucleus
p-type
p-type
s-type
(–)
destructive
mixing
(+)
constructive
mixing
Major limitations of the LE model:
2) Doesn’t easily deal with unpaired electrons
(incorrectly predicts physical properties in some cases)
Example: O2
.. ..
- Lewis dot structure ..
O=O
..
- All electrons are paired
Contradicts experiment!
Experiments show O2 is paramagnetic
A quick note on magnetism…
Paramagnetic
The molecule contains unpaired electrons and is
attracted to (has a positive susceptibility to) an applied
magnetic field
Diamagnetic
The molecule contains only paired electrons and is
not attracted to (has a negative susceptibility to) an
applied magnetic field
Example: the O2 Diatomic
Oxygen atom has a 2s22p4 valence
configuration
O atom
O atom
____ s2p*
___ ___ p2p*
Bond Order = (8-4)/2 = 2
O2 is stable
(498 kJ/mol bond strength)
M.O.
O2
___ ___ ___2p
___ ___ ___ 2p
___ ___ p2p
____ s2p
Energy
____ s2s*
___ 2s
___ 2s
____ s2s
(s2s)2(s2s*)2(s2p)2(p2p)4(p2p*)2
Both have degenerate orbitals
A prediction from the M.O. diagram of O2
.. ..
O=O
.. ..
The Lewis dot structure predicts O2
should be diamagnetic-all electrons
are paired.
The unpaired electrons predicted by
the M.O. diagram should behave as
small magnetsO2 should be magnetic!
N2 Video
O2 Video
What have we learned so far?
1. Molecular orbitals (MO) are linear combinations of atomic orbitals
2. Both s and p atomic orbitals can be mixed to form MOs
3. Molecular orbitals are bonding and anti-bonding
4. Bonding and anti-bonding MOs lead to the definition of the bond order
5. Bond order is related to the bond strength (bond dissociation energy)
MO Diagram for H2 vs. N2
N2
sp*
H2
p2p*
s2p
p2p
Atomic orbital overlap sometimes forms both s
and p bonds. Examples: N2, O2, F2
s2s*
s2s
M.O. Diagram for N2
s*(2p)
p*
Electron energy (kJ mol-1)
p*
-1,155
s(2p)
-1,240
-1,240
p
p
-1,479
s*(2s)
Valence
Valence
-2,965
Core
s(2s)
Core
-37,875
-37,871
1s(N) + 1s(N)
1s(N) –1s(N)
A Complication…
M.O. Diagram for B2
(similar for C2 and N2)
M.O. Diagram for O2
(similar for F2 and Xe2)
O
O2
O
A Complication…
M.O. Diagram for B2
(similar for C2 and N2)
M.O. Diagram for O2
(similar for F2 and Ne2)
No s-p mixing
s-p mixing
Why does s-p mixing occur?
Electron repulsion!!
s2s and s2p both have significant e- probability
between the nuclei, so e- in s2s will repel e- in s2p
Effect will decrease as you move across the Periodic
Table
 increased nuclear charge pulls the s2s e- closer,
making the s2s orbital smaller and decreasing
the s2s and s2p interaction
Molecular Orbitals of X2
Molecules
sp orbital mixing (a little hybridization)
• lowers the energy of the s2s orbitals and
• raises the energy of the s2p orbitals.
• As a result, E(s2p) > E(p 2p) for B2, C2, and N2.
• As one moves right in Row 2, 2s and 2p get further
apart in energy, decreasing s–p mixing  E(s2p) <
E(p2p) for O2, F2, and Ne2. See text pages 680-681.
• Note that s–p mixing does not affect bond order or
magnetism in the common diatomics (N2, O2, and
F2). Hence it is not of much practical importance.
s-p mixing
No s-p mixing
When does s-p mixing occur?
B, C, and N all have  1/2 filled 2p orbitals
O, F, and Xe all have > 1/2 filled 2p orbitals
• If 2 electrons are forced to be in the same orbital,
their energies go up.
• Electrons repel each other because they are
negatively charged.
• Having > 1/2 filled 2p orbitals raises the energies of
these orbitals due to e- - e- repulsion
s-p mixing only occurs when the s and p atomic orbitals are close in energy (
1/2 filled 2p orbitals)
Relating the M.O. Diagrams to Physical
Properties
Sample Problem:
Using MO Theory to Explain Bond Properties
Problem: Consider the following data for these homonuclear diatomic species:
Bond energy (kJ/mol)
Bond length (pm)
No. of valence electrons
N2
945
110
10
N2+
841
112
9
O2
498
121
12
O2 +
623
112
11
Removing an electron from N2 decreases the bond energy of the resulting ion,
whereas removing an electron from O2 increases the bond energy of the resulting
ion. Explain these facts using M.O. diagrams.
Sample Problem:
Using MO Theory to Explain Bond Properties
Problem: Consider the following data for these homonuclear diatomic species:
Bond energy (kJ/mol)
Bond length (pm)
No. of valence electrons
N2
945
110
10
N2+
841
112
9
O2
498
121
12
O2 +
623
112
11
Plan: We first draw the MO energy levels for the four species, recalling that they
differ for N2 and O2. Then we determine the bond orders and compare them with
the data: bond order is related directly to bond energy and inversely to bond
length.
Sample Problem - Continued
Solution: The MO energy levels are:
N2
N2+
O2
O2 +
sp*
sp*
p2p*
p2p*
s2p
p2p
p2p
s2p
s2s*
s2s*
s2s
s2s
Bond Orders:
(8-2)/2 = 3
(7-2)/2 = 2.5
(8-4)/2 = 2
(8-3)/2 = 2.5
Sample Problem:
Using MO Theory to Explain Bond Properties
Problem: Consider the following data for these homonuclear diatomic species:
Bond energy (kJ/mol)
Bond length (pm)
No. of valence electrons
Bond Order
N2
945
110
10
N2+
841
112
9
3
O2
498
121
12
2.5
2
O2 +
623
112
11
2.5
What have we learned so far?
1. Molecular orbitals (MO) explain the properties of valence electrons in molecules
(Example: O2)
2. s and p atomic orbitals can be mixed to form s, s*, p, and p* molecular orbitals
3. Electrons in p or p* molecular orbitals can have the same energies: Degenerate
orbitals
4. The ordering of s2p and p2p molecular orbitals depends on the electron
occupancy: s-p mixing
Bonding in Diatomic Molecules
Covalent
Ionic
Ionic
Covalent
Homonuclear:
H2
Heteronuclear:
HF
Electronegativity
Nonpolar covalent bond
(450 kJ/mol bond)
Polar covalent bond
(565 kJ/mol bond)
Electrons are not equally shared
in heteronuclear bonds
HF
Electronegativity
Because F (EN = 4.0) is more
electronegative than H (EN = 2.2), the
electrons move closer to F.
This gives rise to a polar bond:
H
Figure 14.45
F
M.O.s of a Polar Covalent Bond: HF
s Antibonding (s*)
Mostly H(1s)
This approach simplifies model and only
considers electrons involved in bond.
H
F
H
F
s Bonding
Mostly F(2p)
MOs OF XY MOLECULES
Equal or unequal e sharing between 2 atoms is
reflected in the composition of the MOs:
When 2 atoms X and Y have the same electronegativity (purely covalent
bond), their overlapping AOs have the same energy, and the bonding and antibonding MOs are each half X and half Y AO. All electrons spend equal time
near X and Y. Examples: N2, O2, F2.
If EN(Y) > EN(X) (polar covalent X+Y), the Y AO has lower energy than
the X AO. The bonding MO is more like the Y AO and the anti-bonding MO
more like the X AO. Bonding e spend more time near Y than X; vice versa for
anti-bonding e. Example: CO.
MOs OF XY____
MOLECULES
s*
___ ___ p*
↑
Energy
___ ___ ___ 2p
____ s
___ ___ p
___ ___ ___ 2p
____ s*
___ 2s
___ 2s
____ s
C Atom (4e–)
Electronegativity
___ ___ ___2p
___ 2s
Cδ+Oδ– (10e–) O Atom (6e–)
 CO Bond Order = 3.0 (same as N2).
 CO Bond Energy = 1,076 kJ/mol (N2 = 945 kJ/mol).
 Isoelectronic to CO and N2: CN–, NO+.
 NO has 1e– in p*  bond order = 2.5; this e– is more on N than O; NO 
NO+ easy…
Bonding in NO
• Two possible Lewis dot structures for
NO
• The simplest structure minimizes formal
charges and places the lone (unpaired)
electron on the nitrogen.
• The Lewis structure predicts a bond
order of 2, but experimental evidence
suggests a bond order between 2 and 3.
• How does MO theory help us
understand bonding in NO?
. ..
N=O
.. ..
.. .
N=O
.. ..
-1
+1
When the electronegativities of the 2 atoms are
more similar, the bonding becomes less polar.
2p
2p
2s
NO
Electronegativity
2s
N
. ..
N=O
.. ..
EN(N) = 3.0
EN(O) = 3.4
O
Bond order = 2.5, unpaired electron is in a N-like orbital
NO is easily oxidized to form NO+. Why? What changes can we predict in the bonding
and magnetism of the molecule?
NO
NO+
oxidation
Bond Order = (8-3)/2 = 2.5
Paramagnetic
Bond Order = (8-2)/2 = 3
Diamagnetic
M.O. diagram for
NO
-597
p2p *
p2p * (empty)
s2p
-1307
-1444 p2p
p2p
s2s*
s2s
-1835
-3320
-1374
Key Points of MO Theory –
Heteronuclear Molecules
• The more electronegative atom has orbitals lower in
energy than the more positive atom.
• Electrons in bonding orbitals are closer to the more
electronegative atom, anti-bonding electrons are
closer to the more positive atom.
• For most diatomic molecules, s-p mixing changes the
orbital energy levels, but since these orbitals are
almost always fully occupied, their order is less
important to us.
Combining the Localized Electron and Molecular Orbital
Models (into a convenient working model)
Figure 14.47
Only the p bonding changes between these
resonance structures - The M.O. model
describes this p bonding more effectively.
Atomic Orbitals
Molecular Orbitals
Figure 14.51
Another example:
Benzene
s bonding:
p bonding:
p atomic orbitals
p molecular orbital
MO Theory Expectations
• You should be able to:
– predict which atomic orbitals are higher or lower
in energy (based on electronegativity differences).
– correctly fill a molecular orbital diagram.
– correctly calculate bond order.
– predict molecular magnetic properties based on
orbital occupation.
– understand how molecular properties change
upon ionization (oxidation or reduction) of
molecules.