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I am attempting to use the reduction formula to find the irreducible representation of $\ce{XeF4}$ to determine its IR stretching vibrations. I know the point group of $\ce{XeF4}$ is $D_{4h}$ and have the character table but I am having a hard time understanding how to get the characters for the reducible representation. I was taught to do it by considering moved/unmoved vectors emanating from the point of the molecule giving values of 1, 0 and -1 for each vector after the symmetry operations have transformed them. Using this method I to find 2$E_u$ irreps, but I found in a paper (http://sces.phys.utk.edu/~moreo/mm08/penchoff.pdf) that $\ce{XeF4}$ also has other irreps including an Au irrep involved in IR activity. How do I find the irreps that I missed? I noticed the paper uses different reducible representations ($\Gamma_{xyz}$, $\Gamma_{unmoved}$ and $\Gamma_{Tot}$) and I don't understand why.

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A way to do it is to put 3 axes ($x$, $y$ and $z$) on each atom (to represent the 3N degrees of freedom) and determine the characters for each axis.

After that, if you need the representations for the 3N-6 (3N-5 for a linear molecule) vibrational modes, you need to remove the representations for the 3 translational modes (what they call $\Gamma_{xyz}$), as well as the representations for the 3 (2 for a linear molecule) rotational modes. Those are usually stated in the character table for the space group, but you can also determine them by checking the characters for a vector.

Example with $\ce{H2O}$

Character table for point group $C_2v$

C2v | E  | C2 (z) | v(xz) | v(yz) |  lin  |    quad
-------------------------------------------------------
A1  | +1 |     +1 |    +1 |    +1 |   z   | x2, y2, z2
A2  | +1 |     +1 |    -1 |    -1 |   Rz  |     xy  
B1  | +1 |     -1 |    +1 |    -1 | x, Ry |     xz  
B2  | +1 |     -1 |    -1 |    +1 | y, Rx |     yz  

Character table for all 3N = 9 axes

C2v | E  | C2 (z) | v(xz) | v(yz) 
----------------------------------
Ox  | +1 |     -1 |    +1 |    -1 
Oy  | +1 |     -1 |    -1 |    +1 
Oz  | +1 |     +1 |    +1 |    +1 
H1x | +1 |      0 |     0 |    -1 
H1y | +1 |      0 |     0 |    +1 
H1z | +1 |      0 |     0 |    +1 
H2x | +1 |      0 |     0 |    -1 
H2y | +1 |      0 |     0 |    +1 
H2z | +1 |      0 |     0 |    +1
---------------------------------- 
Γtot| +9 |     -1 |    +1 |    +3

Reduction of $\Gamma_{tot}$ to irreducible representations

Using the reduction formula $n_i = \frac{1}{h}\sum_R \chi_{\Gamma_{tot}}(R) \chi_i(R)$ for each irreducible representation $i$ of the point group, we get :

  • A1 : $\frac{1}{4} \times (9 + (-1) + 1 + 3) = 3$
  • A2 : $\frac{1}{4} \times (9 + (-1) + (-1) + (-3)) = 1$
  • B1 : $\frac{1}{4} \times (9 + 1 + 1 + (-3)) = 2$
  • B2 : $\frac{1}{4} \times (9 + 1 + (-1) + 3) = 3$

Therefore, $\Gamma_{tot}$ = 3 A1 + A2 + 2 B1 + 3 B2.

We can check that there are the 3N = 9 representations that are associated with the 3N degrees of freedom of the molecule.

For degenerate representations ($\chi(E) > 1$), the representations correspond to $\chi(E)$ modes.

Isolation of the vibrational representation

Determine $\Gamma_{trans}$ and $\Gamma_{rot}$ (from the character table or by hand).

$\Gamma_{trans}$ corresponds to the linear terms in the table, $\Gamma_{rot}$ corresponds to the rotational linear terms in the table.

For $C_2v$ :

  • $\Gamma_{trans}$ = A1 + B1 + B2
  • $\Gamma_{rot}$ = A2 + B1 + B2

You can check that there are 3 modes for $\Gamma_{trans}$ and 3 (2 for a linear molecule) modes for $\Gamma_{rot}$

For degenerate representations, you will see things like $x + y$ in the character table. This means that this representation corresponds to both x and y and should be included only once

Finally, $\Gamma_{vib}$ = $\Gamma_{tot} - \Gamma_{trans} - \Gamma_{rot}$ = 2 A1 + B2.

Activity of the vibrational modes in IR and Raman

IR activity is determined by the dipole moment and a mode will be active if the irreducible representation contains a linear part (x, y or z).

Raman activity is determined by the polarisability and a mode will be active if the irreducible representation contains a quadratic part ($x^2$, $xy$, ...).

For $C_2v$, all the representations correspond to active modes in Raman and all except A2 correspond to active modes in IR.

Therefore, the 3 vibrational modes of $\ce{H2O}$ (2 A1 and B2) are active in both Raman and IR.

Two modes have the same irreducible representation: A1. That does not mean that they have the same wavenumber. For $\ce{H2O}$, there are 3 vibrational modes with distinct wavenumbers.

For degenerate representations, there will be $\chi(E)$ modes with the same wavenumber.

In groups that have an inversion center (like $D_4h$), linear and quadratic terms are mutually exclusive. Therefore, an irreducible representation will always correspond to a mode active only in IR or Raman, not both.

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