Lecture Note
Constructing pi Orbitals Using Group Theory (Continied)
-
University:
Seattle Pacific University -
Course:
CHM 3372 | Organic Chemistry II Academic year:
2023
-
Views:
157
Pages:
30
Author:
rezancuk2j
Constructing a MO of NH3
Nitrogen AO symmetries are •
To develop a MO scheme for NH3 assume that only the 2s and2p orbitals of nitrogen interact
with the hydrogen 1s orbitals (i.e., the nitrogen 1s orbital is nonbonding).
•
All AOs match with SALCs, so there are no nonbonding levels.
•
Both s and pz AOs match the A1 SALC,
SALC so s‐p
s p mixing is likely.
likely
•
If bonding and antibonding combinations were formed for both s and pz AOs, we would end up
with eight MOs, but only seven AOs on N and H are available (disregarding the 1s AO on N). We
must make only three MOs from the s and pz AOs and the A1 SALC. For simplicity, we will assume
that the s and pz AOs each form essentially separate bonding MOs, but that together they form a
single mixed antibonding MO. The resulting MO scheme is shown below, where hashed lines
indicate lesser contributions from s‐p mixing.
•
Two AOs of nitrogen match symmetry with one of the SALCs, thereby forming three MOs. Of
these, the lowest energy MO is essentially a bonding combination formed between the hydrogen
SALC and the 2s AO on nitrogen, and the highest MO is an antibonding combination formed from
a mixture of the two AOs from nitrogen and the same symmetry SALC. Symmetry
SALCs
AOs
E'
(1 node)
A 1'
(0 nodes)
px
py
s
pz Symmetry
E'
A 1'
MOs
2e'
2e '*
e'
e '*
2a1'
2a1'*
a1 '
a1'* Lewis base character of NH3, because the σ(z) MO has considerable electron density above
the nitrogen, not unlike the customary picture of the VB model's lone‐pair sp3 hybrid. MOs of Benzene, C6H6
•
C6H6, has three pairs of electrons delocalized in a system extending around the hexagonal ring.
•
The six 2p orbitals perpendicular to the ring on the six carbon atoms combine to form three
bonding (1, 2, 3) and three antibonding (4*, 5*, 6*) MOs.
•
The symmetries and forms of these MOs can be deduced by applying the operations of the point
group D6h to a set of six vectors perpendicular to the ring, one at each carbon, to generate a
reducible representation .
= B2g + E1g + A2u + E2u MOs of Benzene, C6H6
h = 24
D 6h
A 1g
A 2g
B1g
B2g
E 1g
E 2g
A 1u
A 2u
B1u
B2u
E 1u
E 2u
E
2C 6
2C 3
C2
3C 2'
3C 2''
i
6
0
0
0
-2
2
0
0
0
6
6
6
6
12
12
6
6
6
6
12
12
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-6
6
-6
6
0
0
-6
6
-6
6
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
B2g
E1g
A2u + E2u
2S 3 2S 6
d
0
0
0
0
0
0
0
0
0
0
0
0
0
h
3
d
3
-6
6
0
2
-6
-6
6
6
12
-12
12
6
6
-6
-6
-12
12
0
0
0
0
0
0
0
0
0
0
0
0
6
-6
-6
6
0
0
-6
6
6
-6
0
0
/h
v
0
0
0
24
24
0
0
24
0
0
0
24
0
0
0
1
1
0
0
1
0
0
0
1 = B2g + E1g + A2u + E2u
MOs of Benzene, C6H6
(B2g) = (1/6) (a ‐ b + c ‐ d + e ‐ f)
(E2u) = (1/23) (2a ‐ b ‐ c + 2d ‐ e ‐ f)
(E2u) = (1/2) ((- b + c ‐ e + f)
(E1g) = (1/23) (2a + b ‐ c ‐ 2
d ‐ e ‐ f)
(E1g) = (1/2) (b + c ‐ e ‐ f)
(A2u) = (1/6) (a + b + c + d + e + f) = B2g + E1g + A2u + E2u
B2g
(6 nodes)
MOs of Benzene, C6H6
(B2g) = (1/6) (a ‐ b + c ‐ d + e ‐ f)
anti-bonding
E2u
( nodes))
(4
anti-bonding
E1g
(2 nodes)
bonding
(E2u) = (1/23) (2a ‐ b ‐ c + 2d ‐ e ‐ f)
(E2u) = (1/2) ((- b + c ‐ e + f)
(E1g) = (1/23) (2a + b ‐ c ‐ 2
d ‐ e + f)
(E1g) = (1/2) (b + c ‐ e ‐ f)
A2u
(0 nodes)
bonding
(A2u) = (1/6) (a + b + c + d + e + f) MO Energy Level Scheme for Benzene, C6H6
•
By the “shadow method” we can anticipate that the ‐MO scheme will look like the following:
•
Three pairs in bonding MOs add a total of three bond orders over six C–C linkages, or 0.5 for each.
•
When this is added to the sigma bond between each carbon pair, the C–C bond order becomes
1.5. MOs of Cyclopentadienyl, C5H5‐
•
C5H5 , has three pairs of electrons delocalized in a system extending around the
pentagonal ring.
•
The five 2p orbitals perpendicular to the ring on the five carbon atoms combine to
form three bonding (1, 2, 3) and three antibonding (4*, 5*, 6*) MOs.
•
The symmetries and forms of these MOs can be deduced by applying the operations of
the point group D5h to a set of five vectors perpendicular to the ring, one at each
carbon,
b
t generate
to
t a reducible
d ibl representation
t ti . MOs of Cyclopentadienyl, C5H5‐
D 5h
A 1'
A 2'
E 1'
E 2'
A 1''
A 2''
E 1''
E 2''
2C 52 5C 2
2S 5 2S 35 5
E
2C 5
5
0
0
-1
-5
0
0
1
5
5
10
10
5
5
10
10
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-5
5
0
0
-5
5
0
0
-5
-5
-10
-10
5
5
10
10
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
5
-5
0
0
-5
5
0
0
h
d
/h
v
0
0
0
0
0
20
20
20
0
0
0
0
0
1
1
1
h = 20 •
The five p‐orbitals on the planar Cp ring (D5h symmetry) can be combined to
produce five molecular orbitals according to the reducible representation:
•
One combination has the full symmetry of the ring (a2’’)
•
There are two doubly degenerate combinations (e1’’ and e2’’) having one and two
planar nodes at right angles to the plane of the ring.
•
The relative energies of these orbitals increase as the number of nodes increases.
increases
•
The a2’’ and e1’’ orbitals are both fully occupied in the electronic configuration of
the Cp‐ anion whereas the e2’’ orbitals are net anti
anti‐bonding
bonding and are unfilled.
unfilled E2''
(4 nodes)
anti-bonding
i b di
E1''
(2 nodes)
weakly bonding
A2''
(0 nodes)
bonding
The ‐molecular orbitals of the cyclopentadienyl ring (D5h) MOs of the Allyl anion, C3H5‐
•
The alkyl anion has a delocalized, open, three‐center system.
•
Although it is customary to assume that pz orbitals are involved in forming orbitals,
orbitals in this case
for constructing the MO diagram we will assume that px orbitals are used, in keeping with the
standard character table and conventions of defining z as the principal axis and the yz plane as
the plane of the C‐C‐C chain. B1
(2 nodes)
aanti-bonding
t bo d g
A2
(1 node)
d )
weakly bonding
B1
(0 nodes)
bonding and MOs of Boron Trifluoride, BF3
•
Although BH3 is unstable, the BX3 trihalides (X = F, Cl, Br) are stable but reactive compounds that
have been well characterized.
•
A significant advantage of the BX3 compounds is the potential for bonding between the 2pz
orbital of boron and the npz orbitals of the halogens (n = 2, 3, 4).
•
g fluorine is the most electronegative
g
of the halides this bondingg is most significant
g
for
Although
BF3 due to closely matched energies with the central boron atom. MOs of BF3
h = 12
D 3h
A 1'
A 2'
E'
A 1''
A 2''
E''
E
2C 3
3C 2
3
0
1
3
3
6
3
3
6
0
0
0
0
0
0
3
-3
0
3
-3
0
A 1'
E'
2S 3
3
3
0
1
3
3
6
-3
-3
-6
0
0
0
0
0
0
3
-3
0
-3
3
0
h
d
/h
v
12
0
12
0
0
0
1
0
1
0
0
0 A 1'
E' •
Based upon the symmetry of the 3F SALCs derived from the F 1s orbitals symmetry
allowed overlap with the B 2s 2px and 2py orbitals is predicted.
•
Due to energy considerations however this interaction is very weak and generally
ignored.
A 1'
E'
E'
(1 node)
3
1
2
2
1
1
A 1'
(0 nodes) •
Overlap of the B 2s 2px and 2py orbitals with the 3F SALCs derived from the F 2px
orbitals is more prominent.
•
The 2py orbitals of the 3F atoms are considered non‐bonding and are not
considered here.
A 1'
Symmetry
E'
SALCs
AOs
E'
(1 node)
px
A 1'
(0 nodes)
py
s •
Symmetry allowed sigma MOs derived from the 3F (2px) SALCs and the B 2s 2px and
2py orbitals.
A 1'
E'
(1 node)
E'
3
3
2
2
1
1
A 1'
(0 nodes) BF3
B
3F
2
1
The non‐bonding p orbital
is responsible for the Lewis
acid properties of BF3
2p x , p y (e')
2p z ( a2'')
n
a1' e'
2s (a1'))
2
1
2
n
e'
1
n
a1' MOs of BF3
d SALCs
Symmetry
AOs
E''
(1 node)
2
n
3
A 2''
(0 nodes)
n
MOs
n
pz Non‐bonding SALCs of BF3
•
The 2py orbitals of the 3F atoms are considered non‐bonding
non bonding and are considered separatel
separately
when deriving their irreducible representation.
h = 12
D3h
E
2C 3
3C 2
2S 3
n
3
0
-1
3
0
-1
A1'
A2'
E'
A1''
A2''
E''
3
3
6
3
3
6
0
0
0
0
0
0
-3
3
0
-3
3
0
3
3
6
-3
-3
-66
0
0
0
0
0
0
-3
3
0
3
-3
0
h
d
3
/h
v
0
12
12
0
0
0
0
1
1
0
0
0 Symmetry
y
y
E'
E
A 2'
SALCs BF3
B
3F
2
1
*
2p x , p y (e')
2p z ( a2'')
1
n
2
n
3
a2'' e''
a2' e'
a1' e'
n
2s (a1'))
2
1
2
n
e'
1
n
a1' BF3
B
3F
2
1
*
2p x , p y (e')
2p z ( a2'')
a2'' e''
a2' e''
a1' e'
1
n
2
n
3
n
2s (a1')
2
1
•
Consider the MO diagrams for both NH3 and BF3 and how the mechanism for the
Lewis acid‐base reaction may proceed.
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