Molecular Orbital Theory Chapter 8 and a half

Molecular Orbital Theory Chapter 8 and a half

1. 1

2. 2
MO Electron Correlation Diagram
for Ground-State H
2
H1s
σ
1s
σ*
1s
1s
H
2
H
Energy
Ground-State MO electron configuration of H
2
:
5q
6
MOT Bond order:
5
6
MOT predicts a single bond
Node

3. 3
Using MOT, predict whether or not the excited state
of H
2
shown on the board is stable and explain why.
A) It is stable because the number of bonding electrons =
the number of antibondingelectrons.
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
antibondingorbital.
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 H
2
in an Excited State
H1s
σ
1s
σ*
1s
1s
H
2
H
Energy
Excited state MO electron configuration of H
2
:
5q
5
5q
∗5
MOT Bond order:
5
6
MOT predicts no bond

5. 5
MO Electron Correlation Diagram
for H
2
−
(Ground State)
H1s
σ
1s
σ*
1s
1s
H
2
H
Energy
First: Form H
2
, then add an electron.

6. 6
MO Electron Correlation Diagram
for H
2
−
(Ground State)
H1s
σ
1s
σ*
1s
1s
H
2
−
H
Energy
Ground state MO electron configuration of H
2
−
:
σ
5q
6
:σ
5q
∗
;
5
MOT Bond order:
5
6
2F1L
5
6
MOT predicts a half-bond

7. 7
MO Electron Correlation Diagram
for He
2
(Ground State)
He1s
σ
1s
σ*
1s
1s
He
2
He
Energy
Ground-state MO electron configuration of He
2
:
5q
6
5q
∗6
MOT Bond order:
5
6
MOT predicts no bond

8. MO Valence Electron Correlation Diagrams for
2
nd
Row HomonuclearDiatomics
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
E
A
-E
B
+
–
+
–
2s + 2s
+
–
–
+
2s – 2s
E
σ
2s
σ*
2s
Constructive
Interference
Destructive Interference
node
Radial Nodes
Radial Nodes
σ
2s
Please practice drawing these molecular orbitals!

9. +
−
Molecular Orbitals for 2
nd
Row
HomonuclearDiatomics
9
Atomic Orbital Overlap
Molecular Orbital
+
−
Planar Nodes
p
p
fp
Planar Nodes
z
x
y
E
A
-E
B

10. Molecular Orbitals for 2
nd
Row
HomonuclearDiatomics
10
Atomic Orbital Overlap
Molecular Orbital
+ −
+ −
Planar Node
n
n
fn
Planar
Node
z
x
y
E
A
-E
B

11. Molecular Orbitals for 2
nd
Row
HomonuclearDiatomics
11
Atomic Orbital Overlap
Molecular Orbital
+ −
+−
Planar Node
n
n
fn
Planar Node
Planar
Node
z
x
y
E
A
-E
B

12. π
2py
π
2px
12
O
2
Valence MO Correlation Diag.
O2s
2p
π*
2pyσ*
2pz
σ
2pz
σ
2s
σ*
2s
2s
2p
O
2
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: O
2
Lewis structure:
OO
πframework:
++
––
LEM predicts all electrons
are paired, but this is not
true experimentally.
In reality, O
2
has 2
unpaired electrons.
Hybridization: σframework:
+
●
+
●
●●
●● ●●
––OO
●●
+
+
+
+

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
E
A
-E
B

15. EMO Correlation Diagrams for HomonuclearDiatomics
Second-order
(Li
2
-N
2
)
First-order
(O
2
-Ne
2
)
2p 2p

2pz

2px

2py
2p 2p

2pz

2px

2py
z
x
y
E
A
-E
B

16. i-Clicker Question
16
What molecular orbital can be associated
with the lowest-unoccupied molecular orbital
(LUMO) for ground-state C
2
?
A) B)
C)
D) E)
z
x
y

17. 17
C
2
Valence MO Correlation Diag.
C2s
2p
π*
2py
π
2py
σ*
2pz
σ
2pz
σ
2s
σ*
2s
2s
2p
C
2
C
Energy
π
2px
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 C
2
?
A) B)
C)
D) E)
z
x
y
σ*
2pz
σ
2pz
σ*
2s
π
2px
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 O
2
.
Assume the bond axis lies along the zaxis.
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
CO
Energy
2s
2s
2p
2p
π*
2p(x,y)
π
2p(x,y)σ*
2p(z)
σ
2p(z)
σ
2s
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
AB
Energy
2s
2s
2p
2p
π*
2p(x,y)
π
2p(x,y)σ*
2p(z)
σ
2p(z)
σ
2s
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
AB
Energy
2s
2s
2p
2p
π*
2p(x,y)
π
2p(x,y)σ*
2p(z)
σ
2p(z)
σ
2s
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. HeteronuclearDiatomics:
NO (neutral molecule) second-ordervalence orbital correlation
diagram
NO
Energy
2s
2s
2p
2p
π*
2p(x,y)
π
2p(x,y)
σ*
2p(z)
σ
2p(z)
σ
2s
σ*
2s
z
x
y
N-O

26. HeteronuclearDiatomics:
NO
+
(molecular cation) second-ordervalence orbital correlation
diagram
NO
Energy
2s
2s
2p
2p
π*
2p(x,y)
π
2p(x,y)
σ*
2p(z)
σ
2p(z)
σ
2s
σ*
2s
z
x
y
N-O

27. Example: NO
2
–
27
Resonance structures: Localized electrons anddelocalized electrons:

28. Describe the localized electrons
with LEMin a
framework:
28
Considering ONLY the localized
electrons, each atom in NO
2
–
is sp
2
hybridized.

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

30. Describe the delocalizedelectrons
with MOTin 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 NO
2
–
• 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
OO

B

NB
*
AB

32. Chem105 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
Chem105 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.sresult 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: NO
2
‒
, 3 p a.o.sresult 3 MOs
(1 bonding MO, 1 antibonding MO, and
1 nonbonding MO)
34