# Mode Analysis

ΓTotal = ΓStretch + ΓBend + ΓTranslation + ΓRotation

and

ΓVibration = ΓStretch + ΓBend

For easier understanding, there are steps immediately followed by examples using the D3h.

### Finding Γ Total

First, add all the x, y and z rows on the Character Tables of Symmetry groups. If x, y or z are in () on the far right then only count them once, otherwise count the row a second time (Keep the column separated). This is called Γx,y,z. Next, move the molecule with the designated column symbols and if an atom does not move then it is counted. Finally, multiply Γx,y,z and UMA and it will equal Γtotal.

D3h E 2C2 3C2 σh 2S3 v  IR  Raman
A1'  1  1  1  1  1  1    x2+y2, z2
A2'  1  1  -1  1  1  -1  Rz
E'  2  -1  0  2  -1  0  (x,y)  (xy, x2-y2)
A1"  1  1  1  -1  -1  -1
A2"  1  1  -1  -1  -1  1  z
E"  2  -1  0  -2  1  0  (Rx,Ry)  (xz, yz)
Γx,y,z  2+1=3  -1+1=0  0-1=-1  2-1=1  -1-1=-2  0+1=1
UMA (unmoved atoms)

6

3

2

4

1

4

Γtotal  (3)(6)=18  (0)(3)= 0  (-1)(2)=-2  (1)(4)=4  (-2)(1)=-2  (1)(4)=4

### D3h molecule

-need to find the irreducible representation of Gamma total.

Take Γtotal multiply by the number in front of the symbols (the order) and multiply by each number inside of the character table. Add up each row, and divide each row by the total order. For the D3h the order is 12.

D3h 1E 2C2 3C2 σh 2S3 v add the up and divide by 12
A1' (1x1x18)=18 (2x1x0)=0 (3x1x-2)=-6 (1x1x4)=4 (2x1x-2)=-4 (3x1x4)=12  (24/12)=2
A2' (1x1x18)=18 (2x1x0)=0 (3x-1x-2)=-6 (1x1x4)=4 (2x1x-2)=-4 (3x-1x4)=-12  (0/12)=0
E' (1x2x18)=36 (2x-1x0)=0 (3x0x-2)=0 (1x2x4)=8 (2x-1x-2)=4 (3x0x4)=0  (48/12)=4
A1" (1x1x18)=18 (2x1x0)=0 (3x1x-2)=-6 (1x-1x4)=-4 (2x-1x-2)=4 (3x-1x4)=-12  (0/12)=0
A2" (1x1x18)=18 (2x1x0)=0 (3x-1x-2)=6 (1x-1x4)=-4 (2x-1x-2)=4 (3x1x4)=12  (36/12)=3
E" (1x2x18)=36 (2x-1x0)=0 (3x0x-2)=0 (1x-2x4)=-8 (2x1x-2)=-4 (3x0x4)=0  (24/12)=2

Γtotal= 2A1' + A2' + 4E' + 3A2" + 2E"

### Γ Translation

The irreducible form of Γtrans, one needs to look at the second to last column. look at the row that the x, y and z and take the irreducible representation. For instance the D3h would be Γtrans= E'+A2"

### Γ Rotation

One can find Γrot the same way as Γtrans. Instead of looking at x,y,z, one would look at the Rx, Ry, Rz. For the D3h. The Γrot = A2'+E"

### Γ Vibration

To find ΓVibration, just take Γtottransrot= Γvibration.

D3h example

 Γtot 2A1' + A2' + 4E' + 3A2" + 2E" Γtrans - E'   - A2" Γrot - A2'                       -E" Γvibration 2A1'           + 3E' + 2A2" + E"

#### Number of Vibrational Active IR Bands

Only Rx, Ry, Rz, x, y, and z can be ir active. which means only A2', E', A2", and E" can be IR active bands for the D3h. Next add up the number in front of the irreducible representation and that is how many IR active bonds. For instance for the same problem there are 3E'+2A2". There are 5 bands, three of them (meaning the E) are two fold degenerate.

#### Number of Vibrational Active Raman bands

Only x2+y2, z2, xy, xz, yz, x2-y2 can be Raman active. which means only A1', E', and E" can be raman active for the D3h. Next add up the number in front of the irreducible representation, and that is how many Raman active bonds there are. For instance for the same problem there are 3E' + E". There are four bands, 4 of which are two fold degenerate.

## Finding ΓStretch

Looking at the molecules point group, do each of the symmetry representation and count the number of unmoved bonds. Γstretch = Γσrad . next multiply unmoved bonds by the symmetry operations and then the numbers inside the character tables. Then add the rows up and divide by the order of the point group.

D3h E 2C2 3C2 σh 2S3 v  IR  Raman
A1'  1  1  1  1  1  1    x2+y2, z2
A2'  1  1  -1  1  1  -1  Rz
E'  2  -1  0  2  -1  0  (x,y)  (xy, x2-y2)
A1"  1  1  1  -1  -1  -1
A2"  1  1  -1  -1  -1  1  z
E"  2  -1  0  -2  1  0  (Rx,Ry)  (xz, yz)
Γx,y,z  2+1=3  -1+1=0  0-1=-1  2-1=1  -1-1=-2  0+1=1
UMB (unmoved bonds)

5

2

1

3

0

3

D3h 1E 2C2 3C2 σh 2S3 v add the up and divide by 12
A1' (1x1x5)=5 (2x1x2)=4 (3x1x1)=3 (1x1x3)=3 (2x1x0)=0 (3x1x3)=9  (24/12)=2
A2' (1x1x5)=5 (2x1x2)=4 (3x-1x1)=-3 (1x1x3)=3 (2x1x0)=0 (3x-1x3)=-9  (0/12)=0
E' (1x2x5)=10 (2x-1x2)=-4 (3x0x1)=0 (1x2x3)=6 (2x-1x0)=0 (3x0x3)=0  (12/12)=1
A1" (1x1x5)=5 (2x1x2)=4 (3x1x1)=3 (1x-1x3)=-3 (2x-1x0=0 (3x-1x3)=-9  (0/12)=0
A2" (1x1x5)=5 (2x1x2)=4 (3x-1x1)=-3 (1x-1x3)=-3 (2x-1x0)=0 (3x1x3)=9  (12/12)=1
E" (1x2x5)=10 (2x-1x2)=-4 (3x0x1)=0 (1x-2x3)=-6 (2x1x0)=0 (3x0x3)=0  (24/12)=0

Γstretch = Γσ = Γrad = 2A1' + 1E'+ 1A2"

Γπ= Γtan =