Group theory in action: molecular vibrations
We will follow the following steps:
1. Decide on a basis to describe our molecule
2. Assign the point group of the molecule in question
3. Generate a reducible representation of our basis
4. Generate irreducible representations form the reducible representation
5. Examine the irreducible representations in terms of the molecular properties
Molecular vibrations
A basis to describe vibrations
Molecular vibrations: the relative motion of atoms with respect to one another
 These vibrations are an energy state of the molecule
 We should be able to use group theory to figure out which vibrations the molecule can actually
have.
1
Step 1. We use displacement coordinates.
Step 2. The water molecule is C2v point group.
Step 3. Generating a reducible representation
Two steps are required to generate a reducible
representation for the sets of orthogonal Cartesian coordinates
1: Figure out the number of unshifted atoms on for each symmetry operation
E
C2v
＃of unshifted atoms
C2
 v (xZ )
 V' ( yz )
1
1
3
3
2: Use the following equation to calculate the contribution to the character per unshifted atom of the
representations
360
E = 3, i  3 , S =-1+2 cos( n ) ,
n
C2v
Contribution to character
 1,
C n =1+ 2 cos(
360
)
n
E
C2
V
 V' ( yz)
3
-1
1
1
2
3: The reducible representation for the displacement coordinates is the multiplication of the number of
unshifted atoms and the contribution to character
E
C2v
C2
V
 V' ( yz )
＃of unshift atoms
3
1
1
3
Contributions to character
3
-1
1
1
9
-1
1
3
Γ
3N
Γ : The symbol of the reducible representation
3N
Step 4. Reducing the representation
1
aA1= 4 ((9 ×1×1)+(-1×1×1)+(1×1×1)+(3×1×1)) = 3, aA2= 1 , aB1= 2 , aB2= 3
Γ 3A + A2 + 2B1 +3 B2
3N
1
Exercise Use the reduction formula to work out how many times each of the irreducible representations
appears in Γ
3N
of H2O
3
What have we just done?
．Generate a reducible representation of our description of molecule.
．The reducible representation has then be reduced into its individual irreducible representation
components
．These irreducible representation components are of interest
．They describe the possible energy states of the molecule according to our basis
Step 5. Examining the irreducible representations
Displacement coordinates tell us how each atom moves relative to the others in molecule
 atoms move in the same direction at once  translation along an axis  A , B , B
1
 rotation  A , B , B
2
1
1
2
2
4
In the 3 N representation, six of the irreducible representations correspond to translations and rotations
of the molecule.
 every non-linear molecule has 3N-6 vibrations , where N is the number of atoms.
The irreducible representations of vibrations vib
vib = 3 N - T - R =(3A 1 +A 2 +2B 1 +3B 2 ) - (A 1 +A 2 +2B 1 +2B 2 ) =2A 1 + B 2
．each irreducible representation corresponds to a single vibration
．The water molecule have three distinct vibration
．all other fundamental vibrations do not exist, because they are not solution to the Schrödinger equation
NH3
Step 1, Use displacement coordinates.
Step 2, What is the point group of NH3
Step 3, Generate a reducible representation
Exercise Using displacement coordinates of each atom come out reducible representation for each
5
symmetry operation of NH3 molecule.
E
2C
# of unshifted atoms
4
1
2
Contribution to character
3
0
1
3 N
12
0
2
C
3V
3
3
V
Step 4, Reducing the representation
Exercise Use the reduction formula to work out how many times each of the irreducible representations
appears in Γ
3N
of NH3
3 N =3A 1 +A 2 +4E
6
Step 5, Examining the irreducible representations (Determine vib )
T =A 1 +E, R =A 2 +E  RT =A 1 +A 2 +2E
*Each E representation corresponds to a doublely degenerate state, 2E represents four energy states
vib = 3 N - T  R =(3A 1 +A 2 +4E 3 )-(A 1 +A 2 +2E) = 2A 1 +2E
(3N - 6= 3  4 – 6 = 6)
2A  Two energy states; 2E  four energy states
1
How can we use these results to help explain the properties of molecules?
 What do the vibrations actually look like?
 Can group theory help us to the examine the vibration in more detail?[Yes]
7
Internal Coordinates
Vibration of water molecule: vib =2A +B
1
2
 Which are bond-stretching vibration & which are bending vibrations?
Internal coordinate
r : the displacement of one atom with respect to another along the line of the bond
# of unshifted coordinates

E
C2
V
 V'
2
0
0
2
2
0
0
2
8
1
1
aA1 = 4 [(2  1  1) + (2  1  1)] = 1; aA2 = 4 [(2  1  1) + (2  (-1)  1)] = 0;
1
1
aB1 = 4 [(2  1  1) + (2  (-1)  1)] = 1; aB2 = 4 [(2  1  1) + (2  1  1)] = 1
str =A1+B2  vib = str + bend  2A1+B1=A1+B1+ bend  bend =A1
NH3
# of unshifted coordinate
E
2C3
3 v
3
0
1
1
aA1= 6 [(3  1  1)+(1  1  3)]=1;
1
aA2= 6 [(3  1  1)+(1  (-1)  3)]=0
1
aE = 6 (3  2  1) =1
str =A1+E  vib = str + bend  2A1+2E = A1+E+ bend  bend =A1+ E
9
 Can we use internal coordinates to determine bend
# of unshifted angles
E
2C3
3 v
3
0
1
bend = A1 + E
The same answer as we got from the difference method ( bend = vib - str )
＊Sometime, generating bend from the angles of the molecule, does NOT give the same answer as
difference method. A redundant coordinate is found in the bend representation, which is often A1.
When this occurs it can be simply removed from the bend representation.
10
Exercise What is the point group of the PCl5 molecule. [D3h]
Exercise Using displacement coordinates as a basis, generate the reducible representation of 3N in PCl5
molecule. [E(18), 2C3(0), 3C2(-2), h(4), 2S3(-2), 3v(4)]
Exercise How many times does each of the irreducible representations appear in 3N.
[2A1’+A2’+4E’+3A2”+2E”]
Exercise Calculate vib for the PCl5 molecule. [2A1’ + 3E’ + 2A2” + E”]
Exercise Using internal coordinate as a basis, derive a reducible representation for stretching vibrations
in PCl5 molecule. [E(5), 2C3(2), 3C2(1), h(3), 2S3(0), 3v(3)]
Exercise Reduce the representation for stretching vibrations in PCl5 molecule. [2A1’ + E’ + A2”]
Exercise Determine bend. [2E’ + A2” + E”]
Exercise Using internal coordinate as a basis, derive a reducible representation for bending vibrations in
PCl5 molecule. [2A1’ + 2E’ + A2” + E”]
11
Projection Operators
Q: Can we use group theory to analise the vibrations further, and also provide a link with experiment?
A: We can use these results and use group theory to let us “see” what the vibrations look like
 Water molecule has two possible stretching vibration, A1 and B2
 The linear combination of the coordinates also have A1 and B2 symmetry
Projection operators
Px  [  ( R ) R ]x   ( R1 )x   ( R2 )x  ...   ( Rh )x
R
P: the projection operator
x: the generating function, coordinate or vector
(R): the character of the irreducible representation
R: the symmetry operator
12
Consider the integral coordinates,


r1 and r2 , and choose one of these to be generating coordinates

r1
E

r1
C2
v
v’
C2v
E
C2
v
v’
A1
1
1
1
1
C2v
E
C2
v
v’
1× r1
1× r2
1× r2
1× r1
C2v



r1

r2





r2


P r1 =1× r1 ＋1× r2 +1× r2 +1× r1 = 2 r1 +2 r2

 normalized
1
2
 
( r1 + r2 )


 A1 representation has r1 + r2 as a basis


 A stretch has r1 and r2 increasing or decreasing at the same time.
1
13
 
r1 + r2 is know as a symmetry adopted linear combination (SALC)

 
Exercise Please use r2 as generating coordinate and find out the SALC for A1. [ r1 + r2 ]
Show the motion of the atoms in the B2 vibration in water.

r1
E

r1
C2
v
v’
C2v
E
C2
v
v’
B2
1
-1
-1
1
C2v

r2

r2

r1
14
C2v
E

- r2

r1
   
v
C2


v’

- r2

r1

P r1 = r1 - r2 - r2 + r1 = 2 r2 - 2 r2  normalize
1
2
 
( r1 - r2 )


B2 vibration in water is the one where the r1 coordinates is reducing whilst the r2 coordinates is
increase, and vice versa.

 
Exercise Please use r2 as generating coordinate and find out the SALC for A1. [ r1 - r2 ]
15
NH
3
C3v
(C3v)
E

r1
C32
C3

r1

r2

r3
v
v’

r2

r3
v”

r2
C3v
E
C3
C32
v
v’
v”
A1
1
1
1
1
1
1
E
2
-1
-1
0
0
0

  

  
For A1 P r1 =( r1 + r2 + r3 ); For E P r1 =2 r1 - r2 - r3
 The letters A and B are used for the symmetry species of
one-dimensional irreducible representation E is used for 2-D T is
used for 3-D. The E vibration is doubly degenerate and should have
different types of vibrational motions.


Exercise Please use r2 , r3 as generating coordinate and find out the
SALC for E.
16


Use r 2 and r3


 

 

P r2 =2 r 2 - r1 - r3 ; P r3 =2 r3 - r 2 - r1
 These three linear combinations are not linearly independent. In
fact, their vibrations are the same.
 Try to find a linear combinations of the three combination





 





P( r2 - r3 ) = P r2 - P r3 = 2 r 2 - r1 - r3 -2 r3 + r 2 + r1 = 3 r 2 -3 r3
 
r2 - r3 is orthogonal to the first one. It is the second E vibration.
Use the projection operator method to determine the SALC for
the bending vibration of NH
3
C3v
E
C3
C32
v
v’
v”
1
1
2
3
2
1
3
17
C3v
E
C3
C32
v
v’
v”
A1
1
1
1
1
1
1
E
2
-1
-1
0
0
0
For A1 1 +  2 +  3 +  2 + 1 +  3 =2 1 +2  2 +2  3
For E 2 1 -  2 -  3
P(  2 +  3 ) = P  2 +P  3 =  2 -  3 [與(2 1 -  2 -  3 )正交]
Exercise Please use, 2 and 3 as generating coordinate and find out the SALC for A1 and E.
18
AuBr (D )

4
D
4h
4h
# of unshifted coordinates
E
2C
4
0
4
C
0
2
2C
2
'
2
2C
"
2
0
i
2S
0
0
4
h
2
4
2
2
v
d
0
19
aA1g = 1 ; aA2g = 0 ; aB1g = 1; aB2g = 0 ; aEg = 0; aA1u = 0; aA2u = 0; aB1u = 0; aB2u = 0; aEu =1
st = A1g + B1g + Eu
E
C41
C43
C2’(x) C2’(y) C2”(1) C2”(2) i
C2
S41
S43
h
v
v’ d
d’

r1
A1g
B1g
E

   

   

 
For A1g, P r1 = r1 + r2 + r3 + r4 ; For B1g, P r1 = r1 - r2 + r3 - r4 ; For Eu, P r1 = r1 - r3

For the remaining Eu SALC, use r2 as a generating coordinate.

 
 
P r2 = r2 - r4 [與( r1 - r3 )正交]


Exercise Please use r2 , r3 as generating coordinate and find out the SALC for Eu.
20
Exercise
For PCl5, determine the SALCs for the stretching vibrations of the molecule. Use the
projection operation method to “view” the SALCs, sketch the results.
21
Spectroscopy and symmetry selection rules
Whether a transition is allowed or not can be determined by solving the following equation.

*
i
T  f d
 i :wave function of the ground state
 f :wave function of the excited state
T: transition moment operator
d  : small space
A transition is allowed if the equation is not zero.
We can use the results of group theory to simplify the use of this daunting equation greatly.
Infra-red spectroscopy
In infra-red spectroscopy, a molecule can be excited from its ground vibration state to a fundamental
vibration using infra-red radiation. This all occurs if there is a dipole moment change in the molecule
upon absorption.
22
For infra-red spectroscopy,

*
i
*
T f d   i  f d
qr : dipole moment of the molecule
For H2O, vib =2A +B .
1
1.
f : A1 or B2
2.
i : A1
2
3. : character table, xB1, yB2, zA1
 B1 
 
 i T f d   i  f d   A1  B2  f d
A 
 1
*
*
All we have to decide is whether the symmetry of the appropriate parts of the equation all multiply
together to give a non-zero result, without having to do the integration.
In group theory language all we need to decide is whether the multiplication of the symmetries gives the
totally symmetric representation (e.g. A1, A1g) or not.
23
For f = A1, we need to figure out is the results of the following:
A1  B1  A1
A1  B2  A1
A1  A1  A1
To multiply irreducible representations together, we need to use the Direct Product Rules.
A1  B1  A1  B1
A1  B2  A1  B 2
A1  A1  A1  A1
A1 is included in one of the answer, and we can conclude that absorbance due to transition from the
ground state to A1 vibrations in H2O will be observed in the infra-red spectrum.
Exercise Will the B2 vibration in H2O be observable by infra-red spectroscopy?
24
Raman spectroscopy
2
2
2
T : x , y , z , xy, yz, xz
Consider H 2O
x 2 , y 2 , z 2  A1 , xy  A2 , xz  B1 , yz  B2
If  f = A1
 B1 
 B1 
 
 
 B2 
 B2 
 
A1  A1  A1 =  A1 
A 
A 
 2
 2
One equation contains the A1 representation, hence we expect to see bands in the Raman spectrum
corresponding to the A1 vibrations.
Exercise Will the B2 vibration in H2O be observable by Raman spectroscopy?
25
If a band due to a vibration appears in the infra-red spectrum, the vibration is known as infra-red active
If a band due to a vibration appears in the Raman spectrum, the vibration is known as Raman active.
In H 2 O , we expect to see three fundamental vibration bands, all three of which are Raman and infra-red
active.
To assign vibration spectra of molecule, we need one rule and one rule-of-thumb.
Rule : polarized bands can only come from A1 vibrations
Rule-of-thumb : stretching vibration usually appear at higher frequency.
 Any molecule which possesses a centre of symmetry, i, is subject to something know as the “mutual
exclusion rule”. This rule states that any vibration in such a molecule cannot be both IR-active and
Raman-active
26
Tetra chloromethane, CCl 4
1
Infra-red spectrum ( cm 1 )
Raman spectrum( cm )
____
218 depolarized
305
324 depolarized
____
458 polarized
768
762 depolarized
1. point group Td
2. consider displacement coordinates
3.
27
1
[(15  1  1)  (1  1  3)  (1  1  6)  (3  1  6)]  1
24
a A = 0, aE  1 aT  1 aT  3
a A1 =
2
1
2
3N = A1  E  T1  3T2 ; vib  3v  T  R  A1  E  2T2
Use internal coordinates,
8 C3
3 C2
6 S4
6 d
# of unshift coordinate 4
1
0
0
2
str
1
0
0
2
Td
E
4
28
a A1 =1, a A2 =0, aE  0 aT1  0 aT2  1
str = A1  T2

bend = E  T2
Use internal coordinates,
E
8C
3
3C
2
6S
4
6
# of unshifted coordinate 6
0
2
0
2
bend
0
2
0
2
6
d
a A1 =1, a A2 =0, aE  1 aT1  0 aT2  1
bend = A1  E  T2
 The extra A1 is redundant coordinate and can be ignored  str  A1  E  T2
bend  E  T2
 x,y,z  T2
 for  f  A1
A1  T2  T1  T2
 f  T2
A1  T2  T2  A1  E  T1  T1
f E
A1  T2  E  T1  T2
 Two IR bands, one T2 stretch, one T2 bend.
29
2
2
2
xy,yz,zx  T2 ; x , y , z  E & A1
For  f  A1
A1  A1  A1  A1 (Raman active); A1  E  A1  E , A1  T2  A1  T2
For  f  T2
A1  A1  T2  T2 ; A1  E  A1  T2 , A1  T2  T2  A1  E  T2 (Raman active)
For  f  E
A1  A1  E  E ; A1  E  E  A1  E (Raman active), A1  T2  E  T1  T2
Four bands (two stretches and two bends) to be Raman active.
 Polarized band in the Raman is A1 stretch
IR-spectrum
Raman spectrum
____
218 depolarized E bend
305
324 depolarized T2 bend
_____
458 polarized A1 stretch
768
762 depolarized T2 strrtch
30