A General method of reducing a Reducible
Representation.
The Group has h operations (called the
Order of the Group).
The Group has N Classes.
Class a has na operations. (Called the Order
of the Class.)
The Reducible Representation, Γ, has
Character χRa under Class a.
An Irreducible Representation, ΓI, has
Character χIa under Class a.
Then the number of times that Γ contains
ΓI is:
(Σa= 1 N na χRa χIa ) / h
(Far easier to do than to write down!!)
Reduce the following Reducible
Representation, Γ, of the C3v point Group.
C3v
Γ
E
12
2C3
0
h=6
Number of A1 reps.
C3v
E
2C3
Γ
12
0
A1
1
1
na χRa χIa 12
0
3σ v
2
1
6
Sum/h = 18/6 = 3
Number of A2 reps.
C3v
E
2C3
Γ
12
0
A2
1
1
12
0
3σ v
2
-1
-6
Sum/h = 6/6 = 1
Number of E reps.
C3v
E
2C3
Γ
12
0
E
2
-1
24
0
3σ v
2
0
0
3σ v
2
Sum/h = 24/6
=4
Therefore: Γ = 3A1 + A2 + 4E
Check:
3A1
A2
4E
Γ
3
1
8
12
3
1
-4
0
Some simple rules for multiplying
Representations:
Let Γ1 be the totally symmetric
representation of the group.
Then:
1) Γ x Γ1 = Γ
2) Γ x Γ Э Γ1
AxΓ=Γ
B x Γ = Γ , except B x B = A
1x1 = 1 2x2=1 1x2 = 2x1 = 2
uxu=g gxg=g uxg=gxu=u
3
-1
0
2
Determining the symmetry of bonding in
molecules.
Consider PtCl42- , square planar molecule,
D4h point group.
1) How do the Pt s, p, d orbitals
transform?
2) How do the Pt-Cl σ bonds transform?
3) What does this say about Pt – Cl
bonding?
D4h E 2C4 C2 2C'2 2C''2 i 2S4 h 2v 2d
x2+y2;
z2
A1g 1 1
1
1
1
1 1 1 1
1
A2g
B1g
B2g
Eg
A1u
A2u
B1u
B2u
Eu
1
1
1
-2
1
1
1
1
-2
-1
1
-1
0
1
-1
1
-1
0
-1
-1
1
0
1
-1
-1
1
0
1
1
1
2
-1
-1
-1
-1
-2
-1
Rz
-1
x2-y2
1
xy
0 (Rx;Ry) (xz;yz)
-1
1
z
1
-1
0 (x;y)
1
1
1
2
1
1
1
1
2
1
-1
-1
0
1
1
-1
-1
0
1
-1
-1
0
-1
-1
1
1
0
1
1
1
-2
-1
-1
-1
-1
2
-1
1
-1
0
-1
1
-1
1
0
1)
Read the s, p, d orbital info off the
Character Table:
s a1g; pz a2u; px, py eu
z2 a1g; x2 – y2 b1g; xy b2g ; xz, yz eg
If that section not provided then consider
the effect of the set of operations on the
orbital:
e.g.
D4h E 2C4 C2 2C'2 2C''2 i 2S4 h 2v 2d
xy 1 -1 1 -1 1 1 -1 1 -1 1
which is b2g . (etc…)
2) There are 4 Pt – Cl σ bonds. Consider
the effect of the set of operations on
them, and count +1 if they
(individually) do not change, 0 if they
move, -1 if they simply change sign
(which will not happen).
D4h E 2C4 C2 2C'2 2C''2 i 2S4 h 2v 2d
σ
4 0 0 2
0 0 0 4 2 0
bonds
a Reducible Representation.
Reduce the Representation:
A1g + B1g + Eu
D4h
A1g
B1g
Eu
E
1
1
2
4
2C4
1
-1
0
0
C2
1
1
-2
0
2C'2 2C''2 i
1
1 1
1 -1 1
0
0 -2
2
0 0
2S4
1
-1
0
0
h
1
1
2
4
2v
1
1
0
2
2d
1
-1
0
0
These will be the Representations of the
molecular orbitals (4 of them) that result in
the Pt – Cl σ bonds.
3) The atomic orbitals involved in those
molecular orbitals must have the same
symmetry.
Therefore s and dz2 can contribute to the a1g
mo. dx2 - y2 to the b1g m.o. and px , py to the eu
mo.
The other Pt valence mos do not contribute.
[Note: It is obvious that dz2 cannot
effectively contribute, but this is not due to
symmetry.]
Determining the symmetry of the motion of
nuclei a molecule.
Nuclear motion – translation, rotation and
vibration.
N nuclei, 3N nuclear coordinates (x, y, z) for
each atom.
Attach a (x, y, z) set to each atom and
consider the effect of the symmetry
operations of the group.
Count +1 if they (individually) do not
change, 0 if they move, -1 if they change
direction to form the Reducible
Representation.
e.g. ethane, D2h point group.
x
H
x
y
x
x C
x
H
C y
y
x
H
y
H
y
and a z perpendicular to page on each atom
D2h E C2(z) C2(y) C2(x) i (xy) (xz) (yz)
Γ 18 0
-2
0 0 


h=8
How many ag?
D2h E C2(z) C2(y) C2(x) i (xy) (xz) (yz)
Ag 1 1
1
1 1 1
1
1
Γ 18 0
-2
18 0
-2
n (ag) = 24 / 8 = 3
0 0
0 0






How many b1g?
D2h E C2(z) C2(y) C2(x) i (xy) (xz) (yz)
B1g 1 1
-1
-1 1 1
-1
-1
Γ 18 0
-2
0 0
18 0
2
0 0
n (b1g) = 24 / 8 = 3






How many b2g?
D2h E C2(z) C2(y) C2(x) i (xy) (xz) (yz)
B2g 1 -1
1
-1 1 -1
1
-1
Γ 18 0
-2
0 0 


18 0
-2
0 0   
n (b2g) = 8/ 8 = 1
Similarly: n (b3g) = 2, n (au) = 1, n (b1u) = 2,
n (b2u) = 3, n (b3u) = 3
Γ = 3ag + 3b1g + b2g + 2b3g + au + 2b1u + 3b2u
+ 3b3u
But:
Translation transforms as b1u + b2u + b3u
Rotation transforms as
b1g + b2g + b3g
Therefore
Vibrations transforms as
3ag + 2b1g + b3g + au + b1u + 2b2u + 2b3u
These are the symmetries of the (3N-6)
vibrations of ethane.