Ch 5 Lecture 3 Group Theory and MO’s
I.
The Group Theory Approach to Molecular Orbitals
A.
Steps of the Group Theory Approach
1) Find the point group (substitute D2h for D∞h, and C2v for C∞v)
2) Assign xyz coordinates (Cn = z, y points toward central atom)
3)
Find characters and representations for groups of outer atoms (2s, 2p…)
a) An orbital that changes position = 0
b) An orbital in the same position = +1
c) An orbital with reversed sign = -1
4)
Reduce each representation to its irreducible representations
a) This identifies the symmetries of the various group orbitals
b) Group orbitals = combinations of AO’s with matching symmetry
5)
Find atomic orbitals of the central atom with the same symmetry as the
group orbitals (ie, same irreducible representations) and combine those of
similar energies to form MO’s
6)
The total number of MO’s formed = number of AO’s used from all atoms
B.
H2O
1) Water is in the C2v point group
2)
The C2 axis = z-axis; xz plane is the plane of the molecule
a) Since the molecule is planar, we don’t need to assign axes to H’s
b) H(1s) orbitals have no directionality
3)
Use the H(1s) orbitals to find the reducible representation
4)
Use the reduction formula to identify the irreducible representations
a) A1 = caH(1s)a + cb(H1s)b
b) B1 = caH(1s)a - cb(H1s)b
5)
Find O AO’s with the
same symmetry
a)
1s, 2s orbitals are
of A1 symmetry (s
is always most
symmetric)
b) 2pz orbital is of A1
symmetry (use z in
character table)
c)
2px orbital is of B1
symmetry (use x in
character table)
6)
There are 6 total MO’s
a. The A1 group + O(2s) + O(pz)
1) Y1 = Bonding
2) Y3 = Nonbonding
3) Y5 = Antibonding
b. The B1 group + O(px)
1) Y2 = Bonding
2) Y6 = Antibonding
c. The oxygen py = Y4
d.
7)
8 valence electrons fill Y1-Y4
1) BO = 2
2) Lone pair = 2
Compare to Lewis Structure
a. 1 lone pair nonbonding HOMO (2py)
b. 1 lone pair LOMO (Y1 = 2s)
c. 2 bonding pairs in the middle
d. All four orbitals are different
C.
NH3
1) Ammonia is in C3v point group
2) C3 axis is assigned as the z-axis,
the yz plane passes through N
and one H
3) Use character table to find
irreducible representations
4) Identify orbital and group orbital
symmetries
a) A1 = Ha + Hb + Hc
b) E = Hb – Hc
= 2Ha – Hb – Hc
c) A1 = s, pz
d) E = px, py
5)
Combine Orbitals to give 7 MO’s
a. A1 group + 2s + 2pz = a1
i. Bonding 2a1
ii. Antibonding 4a1
iii. Nonbonding 3a1
b.
6)
E group + 2px + 2py = e
i. 2 bonding (degenerate)
1e
ii. 2 antibonding
(degenerate) 2e
Put in 8 valence electrons
a. BO = 3 (2a1, 1e)
b. Lone pair = 3a1
D.
BF3 (Lewis Acid)
1) Point Group = D3h
2) C3 axis = z-axis, F y-axis points
to B, F x-axis is in the plane of
the molecule
3) Find Reducible, Irreducible
Representations
4) Find suitable orbitals
5) Results:
a)
B—F should have
some double bond
character
i. 4a1’, 3e’ are
bonding BO = 3
ii. 1a2” is somewhat
bonding too
b) The LUMO is an
empty p orbital
i. 2a2” is LUMO
ii. Lewis acidity
explained
6) More complex
molecules by
computer program
H
III. Hybrid Orbitals
A.
Utility
1) Developed in Valence Bond Theory
2) Not needed anymore, but emphasizes symmetry
B.
Hybrid Orbitals
1) Orbitals on a central atom can be combined 1s + 3p orbitals = 4sp3 orbitals
2) Equivalent, directional orbitals result
C.
Methane
1) Reducible representation found from what happens to four vectors
2) Irreducible representations found as usual
H
C
H
H
3)
D.
A1 = 2s, T2 = px, py, pz (or dxy, dxz, dyz) = sp3 hybrid orbitals
Example: Find hybridization and orbital involved for BF3