Molecular Orbital Theory Chapter 8 and a half

1. 1

2. 2
MO Electron Correlation Diagram
for Ground-State H2
H
1s
σ1s
σ*1s
1s
H2
H
Energy
Ground-State MO electron configuration of H2:
MOT Bond order:
MOT predicts a single bond
Node

3. 3
Using MOT, predict whether or not the excited state
of H2 shown on the board is stable and explain why.
A) It is stable because the number of bonding electrons =
the number of antibonding electrons.
B) It is unstable because in this excited state, the two
electrons have opposite spins and Hund’s rule is not
obeyed.
C) It is unstable because the predicted bond order is 0.
D) It is unstable because there is an electron in an
antibonding orbital.
E) Unable to predict. MOT can only describe ground-state
MO electron configurations. Thus, MO cannot be used
to predict the stability of excited states.

4. 4
MO Electron Correlation Diagram
for H2 in an Excited State
H
1s
σ1s
σ*1s
1s
H2
H
Energy
Excited state MO electron configuration of H2:
∗
MOT Bond order:
MOT predicts no bond

5. 5
MO Electron Correlation Diagram
for H2
− (Ground State)
H
1s
σ1s
σ*1s
1s
H2
H
Energy
First: Form H2, then add an electron.

6. 6
MO Electron Correlation Diagram
for H2
− (Ground State)
H
1s
σ1s
σ*1s
1s
H2
−
H
Energy
Ground state MO electron configuration of H2
−: σ σ∗
MOT Bond order: 2 1
MOT predicts a half-bond

7. 7
MO Electron Correlation Diagram
for He2 (Ground State)
He
1s
σ1s
σ*1s
1s
He2
He
Energy
Ground-state MO electron configuration of He2:
∗
MOT Bond order:
MOT predicts no bond

8. MO Valence Electron Correlation Diagrams for
2nd Row Homonuclear Diatomics
Second-order First-order
1s
2s
2p
1s
2s
2p
1s
*1s
2s
*2s
2pz
*2pz
1s
2s
2p
1s
2s
2p
1s
*1s
2s
*2s
2pz
2px 2py
*2px *2py
*2pz
8
2px 2py
*2px *2py
z
x
y
EA-EB
+
–
+
–
2s + 2s
+
–
–
+
2s – 2s
E
σ2s
σ*2s
Constructive
Interference
Destructive
Interference
node
Radial Nodes
Radial Nodes
σ2s
Please practice drawing these molecular orbitals!

9. +
−
Molecular Orbitals for 2nd Row
Homonuclear Diatomics
9
Atomic Orbital Overlap Molecular Orbital
+
−
Planar Nodes Planar Nodes
z
x
y
EA-EB

10. Molecular Orbitals for 2nd Row
Homonuclear Diatomics
10
Atomic Orbital Overlap Molecular Orbital
+
−
+
−
Planar Node
Planar
Node
z
x
y
EA-EB

11. Molecular Orbitals for 2nd Row
Homonuclear Diatomics
11
Atomic Orbital Overlap Molecular Orbital
+
−
+
−
Planar Node
Planar Node
Planar
Node
z
x
y
EA-EB

12. π2py
π2px
12
O2 Valence MO Correlation Diag.
O
2s
2p
π*2py
σ*2pz
σ2pz
σ2s
σ*2s
2s
2p
O2
O
Energy
π*2px
z
x
y
O-O

13. Recall from L#30: Failures of LEM
13
Failure #3: LEM fails to predict certain molecular
properties.
Example: O2
Lewis structure: O O
π framework:
+ +
– –
LEM predicts all electrons
are paired, but this is not
true experimentally.
In reality, O2 has 2
unpaired electrons.
Hybridization:
σ framework:
+ ●
+
●
●●
●●
●●
– –
O O
●●
+
+
+
+

14. E
MO Correlation Diagrams for Homonuclear Diatomics
Second-order First-order
1s
2s
2p
1s
2s
2p
1s
*1s
2s
*2s
2pz
*2pz
1s
2s
2p
1s
2s
2p
1s
*1s
2s
*2s
2pz
2px 2py
*2px *2py
*2pz
14
2px 2py
*2px *2py
z
x
y
EA-EB

15. E
MO Correlation Diagrams for Homonuclear Diatomics
Second-order
(Li2-N2)
First-order
(O2-Ne2)
2p 2p
2pz
2px 2py
2p 2p
2pz
2px 2py
z
x
y
EA-EB

16. i-Clicker Question
16
What molecular orbital can be associated
with the lowest-unoccupied molecular orbital
(LUMO) for ground-state C2?
A) B)
C)
D) E)
z
x
y

17. 17
C2 Valence MO Correlation Diag.
C
2s
2p
π*2py
π2py
σ*2pz
σ2pz
σ2s
σ*2s
2s
2p
C2
C
Energy
π2px
π*2px
z
x
y

18. i-Clicker Solution
18
What molecular orbital can be associated
with the lowest-unoccupied molecular orbital
(LUMO) for ground-state C2?
A) B)
C)
D) E)
z
x
y
σ*2pz
σ2pz
σ*2s
π2px
π*2px
nodes

19. 19
i-Clicker Question
A)2px
B)2py
C) * 2px
D) *2py
E) *2pz
Using MOT, determine the highest occupied
molecular orbital (HOMO) for ground-state O2.
Assume the bond axis lies along the z axis.
z
x
y
O-O

20. 1) Homonuclear, nonpolar covalent bond
- A bond in which each atom of a bonded pair
contributes one e– to form a pair of e–s (e– pair
bond)
20
Recall from L#4: Bond Types
Consider: 2 identical atoms (A) with A
 AA = A – A = 0
- each atom “pulls” equally on the shared, bonding
electrons, and there is an equal distribution of the
electron density between the atoms
A A
Symmetric
e– Density

21. 2) Polar covalent bond
- A bond formed by sharing of e–s between 2
different atoms (A and B)
- A  B
- In general, AB = |A ‒ B|  ~2
- Unequal sharing of e–s
- Molecules (molecular compounds) contain polar
and/or nonpolar covalent bonds
21
Recall from L#4: Bond Types
Consider: A bound to B with A > B
A B
Asymmetric
e– Density
‒ +
: partial charge

22. CO Valence MO Correlation Diagram
z
x
y
C-O
C O
Energy
2s
2s
2p
2p
π*2p(x,y)
π2p(x,y)
σ*2p(z)
σ2p(z)
σ2s
σ*2s

23. i-Clicker Question: Please identify atoms A, B, and the
molecular charge corresponding to the MO diagram
z
x
y
A-B
A B
Energy
2s
2s
2p
2p
π*2p(x,y)
π2p(x,y)
σ*2p(z)
σ2p(z)
σ2s
σ*2s
A B Charge
A) N N 0
B) N O +1
C) O N +1
D) C O +1
E) C N −2

24. i-Clicker Question: Please identify atoms A, B, and the
molecular charge corresponding to the MO diagram
z
x
y
A-B
A B
Energy
2s
2s
2p
2p
π*2p(x,y)
π2p(x,y)
σ*2p(z)
σ2p(z)
σ2s
σ*2s
A B Charge
A) N N 0
B) N O +1
C) O N +1
D) C O +1
E) C N −2

25. Heteronuclear Diatomics:
NO (neutral molecule) second-order valence orbital correlation
diagram
N O
Energy
2s
2s
2p
2p
π*2p(x,y)
π2p(x,y)
σ*2p(z)
σ2p(z)
σ2s
σ*2s
z
x
y
N-O

26. Heteronuclear Diatomics:
NO+ (molecular cation) second-order valence orbital correlation
diagram
N O
Energy
2s
2s
2p
2p
π*2p(x,y)
π2p(x,y)
σ*2p(z)
σ2p(z)
σ2s
σ*2s
z
x
y
N-O

27. Example: NO2
–
27
Resonance structures:
Localized electrons and delocalized electrons:

28. Describe the localized electrons
with LEM in a framework:
28
Considering ONLY the localized
electrons, each atom in NO2
– is sp2
hybridized.

29. Describe the localized electrons
with LEM in a framework:
29
• All atoms lie in the
plane of the slide
(yz plane)
• N–O bond:
sp2 + sp2
• N lone pair: sp2
• 2 O lone pairs: sp2
• Each atom has a
pure px orbital
remaining.
y
z
x
N
O O

30. Describe the delocalized electrons
with MOT in a framework:
30
There are 4 delocalized electrons over
the entire O–N–O molecule, in the
framework.

31. Notes on framework described by MOT in NO2
–
• Nonbonding MO: the orbitals
are on non-adjacent atoms
and therefore there is no
significant overlap.
• The MOs are ranked in
energy according the # of
nodes perpendicular the
plane of atoms participating
in the framework.
• The delocalized bond and
the delocalized lone pair are
now depicted without the
need for resonance.
31
N
O O
B
NB
*AB

32. Chem 105 combined LEM/MOT rules for
drawing/positioning nodes in the framework:
• Draw the nodes perpendicular to the plane of
the atoms in the molecule or molecular ion
participating in bonding.
• Arrange nodes as symmetrically as possible.
– Similar to how nodes were arranged in the violin
string standing waves.
– Planar nodes should be placed either in between
two atoms or such that the plane contains an atom’s
nucleus (or nuclei).
• Amplitude sign (phase) should change when
crossing a node. 32

33. • When atoms participating in bonding are
heteronuclear atoms:
– Lobes should be drawn larger for the more
electronegative atoms in bonding MOs.
– Lobes should be drawn larger for the less
electronegative atoms in antibonding ∗
MOs.
33
Chem 105 combined LEM/MOT rules for
drawing/positioning nodes in the framework
(Continued):

34. Notes for Conjugated Linear -Molecules
• For conjugated linear -molecules,
– An even number of contributing p atomic orbitals
results in an equal number of bonding and anti-
bonding MOs, with zero non-bonding MOs.
• Example: 1,3-Butadiene, 4 p a.o.s result 4  MOs
(2 bonding MOs and 2 antibonding MOs)
– An odd number of contributing p atomic orbitals
results in an equal number of bonding and anti-
bonding MOs plus one non-bonding MO.
• Example: NO2
‒, 3 p a.o.s result 3  MOs
(1 bonding MO, 1 antibonding MO, and
1 nonbonding MO)
34