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.
1 Answer
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.
