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An exercise asks me to find $\Gamma_{vib}$ for that molecule. I know that: $$\Gamma_{cart}=\Gamma_{vib}+\Gamma_{translations}+\Gamma_{rot}$$

Since the molecule I'm considering belongs to the $D_{3d}$ group (in the staggered configuration), from the character table you get: $$\Gamma_{translations}=A_{2u}+2E_{u}$$ $$\Gamma_{rot}=A_{2g}+2E_{g}$$ and then: $$\Gamma_{vib}=\Gamma_{cart}-2E_{u}-2E_{g}-A_{2g}-A_{2u}$$ My question is: does exist a fast way to find out $\Gamma_{cart}$?

The only method that I know is to express each operation of the group in terms of matrices using the Cartesian basis, so I should write down 12 matrices 24x24, and then find out the corresponding reducible representation, but I'd take ages. I'm pretty sure that there must exist a faster method.

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To transcribe what is given in levineds's linked answer and remove the parts only relevant to linear molecules, here are the steps you can apply to determine the reducible representation of the vibrations.

  1. Determine how the molecule is changed by all symmetry operations. If an atom is moved by a symmetry operation, it contributes it doesn't contribute to $\Gamma_\text{atoms}$. If the atom remains in place, add +1 to the column of that symmetry operation.

For hexafluoroethane, this gives: $$\small \begin{array}{c|cccccc} \hline D_{3\mathrm{d}} & E & 2C_3 & 3C_2' & i & 2S_6 & 3\sigma_d & \\ \hline \Gamma_\text{atoms} & 8 & 2 & 0 & 0 & 0 & 4 &\\ \end{array}$$

  1. Multiply each column of $\Gamma_\text{atoms}$ by the trace of that operation's matrix representation. $$\begin{array}{c|c}E&3\\\hline C_2&-1\\\hline \sigma&1\\\hline i&-3\\\hline C_n&1+2\cos(\frac{2\pi}{n})=1+2\cos\theta\\\hline S_n&-1+2\cos(\frac{2\pi}{n})=-1+2\cos\theta\end{array}$$

$$\small \begin{array}{c|cccccc} \hline D_{3\mathrm{d}} & E & 2C_3 & 3C_2' & i & 2S_6 & 3\sigma_d & \\ \hline \Gamma_\text{cart} & 24 & 0 & 0 & 0 & 0 & 4 &\\ \end{array}$$

  1. Use reduction formula to determine irreps. $n(i)=\frac{1}{h}\sum_R N\cdot\chi_r(R)\cdot\chi_i(R)$ where $n(i)$ is the number of the $i^{\text{th}}$ irrep, $R$ is a symmetry operation, $h$ is the order of the group, $N$ is the coefficient in front of the operation, and $\chi(R)$ is the character of the operation in the reducible/irreducible representation.

I won't go through the calculation (there are online calculators that you can use to do this), but this gives. $$\Gamma_{xyz}=3A_{1g}+1A_{2g}+4E_g+1A_{1u}+3A_{2u}+4E_u$$ which if we subtract off the translational and rotational irreps you have already obtained gives: $$\Gamma_{vib}=3A_{1g}+3E_g+1A_{1u}+2A_{2u}+3E_u$$ (You had small error with your rot and trans, the $E$ irreps are degenerate and so count for 2 degrees of freedom. This means in each case, you would only have a single $E$ irrep.)

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