Editing Character table (section)

== Finding the vibrational modes of the ethylene molecule using character table ==
Ethylene is a member of the D2h point group, which has eight Mulliken symbols in the first column. Besides, the ethylene molecule contains six atoms, each with an x, y, and z axis. So, the molecule has a total of 18 axes.

For vibrational modes of the molecule, it is necessary to calculate the irreducible representation Γ<sub>irreducible</sub>. Also, the irreducible representation is related with the reduible representation.

Here is another method to calculate the representation calculation. It is necessary to find the change of x, y and z axes. If the atom changes the place after the operation, there is no contribution to the Γ<sub>reducible</sub>. If the atom keeps the same place after the operation, then check the axis, if the axis keeps same direction, the contribution to the Γ<sub>reducible</sub>.is 1; if the axis reverses to the opposite direction, the contribution to the Γ<sub>reducible</sub>.is -1; if the axis rotates at a certain angle ''θ'', the contribution is cos ''θ''. After calculating all axes of all atoms, there is the value of the reducible representation Γ<sub>reducible</sub> for this operation.
In this case, ethylene is the D<sub>''2h''</sub> point group with eight symmetry operations in the first line, each operation provides the different Γ<sub>reducible</sub>.

E: Identity Symmetry. All atoms remain in their original positions, so they all have the same x, y, and z axes. The 18 axes remain in the same position, each contributing one to the reducible. The reducible number for E is 18.

C<sub>2</sub>(x), C<sub>2</sub>(y): As the molecule rotates along the x or y axis, each atom moves and contributes zero to the reducible. The overall Γ<sub>reducible</sub> for C2(x) and C2(y) are 0.

C<sub>2</sub>(z): The molecule rotates along the z axis, with only two carbon atoms remaining in the same position. The x and y axes of each carbon atom reverse to the opposite place, but z axis keeps the same direction, contributing negative one of each atom. The overall Γ<sub>reducible</sub> is -2.

i: The molecule is inverse through the center. Since all atoms move places, the overall Γ<sub>reducible</sub> for i is 0.

σ(xy): The molecule flips across the xy plane. The overall Γ<sub>reducible</sub> for σ(xy) is 0, as all atoms move places.

σ(xz): The molecule flips across the xz plane, but two carbon atoms remain in the same place. The x and z axes remain unchanged, each contributing to a single reducible number. However, the y axis reverses and contributes to negative one Γ<sub>reducible</sub>. So, each carbon contributes one Γ<sub>reducible</sub>, the overall Γ<sub>reducible</sub> is 2.

σ(yz): It is different from other operations. All six atoms maintain their original positions. The y and z axes remain the same, but the x axis reverses, resulting in one Γ<sub>reducible</sub> for each atom. The total Γ<sub>reducible</sub> is 6.

'''New character table for ethylene <math>\Gamma_{\text{red}}</math>'''
{| class="wikitable"
|
|'''E'''
|'''C<sub>2</sub>(x)'''
|'''C<sub>2</sub>(y)'''
|'''C<sub>2</sub>(z)'''
|'''i'''
|'''σ(xy)'''
|'''σ(xz)'''
|'''σ(yz)'''
|-
|'''Γ<sub>reducible</sub>'''
|'''18'''
|'''0'''
|'''0'''
|'''-2'''
|'''0'''
|'''0'''
|'''2'''
|'''6'''
|}
The next step is to calculate the irreducible presentation based on the reducible presentation. Here is the calculation.

 <math>N_{A_g} = \frac{1}{8}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times 1)+(6\times 1\times 1)] = 3</math><math>N_{B_1g} = \frac{1}{8}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times (-1))+(6\times 1\times (-1))] = 1</math><math>N_{B_2g} = \frac{1}{8}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times 1)+(6\times 1\times (-1))] = 2</math><math>N_{B_3g} = \frac{1}{8}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times (-1))+(6\times 1\times 1)] = 3</math>

 <math>N_{A_u} = \frac{1}{8}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times (-1))+(6\times 1\times (-1))] = 1</math>

 <math>N_{B_1u} = \frac{1}{8}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times 1)+(6\times 1\times 1)] = 3</math>

 <math>N_{B_2u} = \frac{1}{8}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times (-1))+(6\times 1\times 1)] = 3</math>

 <math>N_{B_3u} = \frac{1}{8}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times 1)+(6\times 1\times (-1))] = 2</math>

Γ<sub>irreducible</sub> = 3''A''<sub>g</sub>+1''B''<sub>1g</sub>+2''B''<sub>2g</sub>+3''B''<sub>3g</sub>+1''A''<sub>u</sub>+3''B''<sub>1u</sub>+3''B''<sub>2u</sub>+2''B''<sub>3u</sub>


Translational motion has x, y and z functions in “linear functions, roatations”. So, Γ<sub>trans</sub> = 1''B''<sub>1u</sub>+1''B''<sub>2u</sub>+1''B''<sub>3u</sub>

Rotational motion has R<sub>x</sub>, R<sub>y</sub> and R<sub>z</sub> functions in “linear functions, roatations”. So, Γ<sub>rot</sub> = 1''B''<sub>1g</sub>+1''B''<sub>2g</sub>+1''B''<sub>3g</sub>

Vibrational motio: Γ<sub>vib</sub> = Γ<sub>irreducible</sub>-Γ<sub>trans</sub>-Γ<sub>rot</sub> = 3''A''<sub>g</sub>+1''B''<sub>2g</sub>+2''B''<sub>3g</sub>+1''A''<sub>u</sub>+2''B''<sub>1u</sub>+2''B''<sub>2u</sub>+1''B''<sub>3u</sub>


The final step is to determine which vibrations are IR or Raman active.  This means that the symmetry operation can be detected using the infrared or Raman spectrum.

First, for IR to work, they must have x, y, and z functions in "linear functions, rotations". In Γ<sub>vib</sub>, only 2''B''<sub>1u</sub>+2''B''<sub>2u</sub>+1''B''<sub>3u</sub> are IR active.

To be Raman active, "quadratic functions" must include x<sup>2</sup>, y<sup>2</sup>, z<sup>2</sup>,xy, xz, yz, x<sup>2</sup>+y<sup>2</sup> or x<sup>2</sup>-y<sup>2</sup> functions. In Γ<sub>vib</sub>, only 3''A''<sub>g</sub>+1''B''<sub>2g</sub>+2''B''<sub>3g</sub> are Raman active.