I'd like to describe the symmetry of the ground state of $N_2$ in the higher symmetry groups like $D_{3h}$ and $D_{\infty h}$.

In this answer the topic is covered very good, but only for $D_{2h}$ point group, where $N_2$ orbitals are described by the irreducible representations like this:

$$\begin{array}{cccc} \hline \mathrm{s} & \mathrm{p}_x & \mathrm{p}_y & \mathrm{p}_z \\ \hline \mathrm{A_{g}} & \mathrm{B_{3u}} & \mathrm{B_{2u}} & \mathrm{B_{1u}} \\ \hline \end{array}$$

But, when I have a look at character tables for $D_{3h}$ and $D_{\infty h}$, there are no irreducible representations with basis functions $x$ and $y$ as can be seen below.

So, how am I supposed to describe the symmetry of the $p_x$ and $p_y$ orbitals to be able to represent the symmetry of the whole system subsequently?

My attempt

I've tried to describe the symmetry of $p_x$ in $D_{3h}$ like this:

$$ \begin{array}{cccccc} \hline E & 2C_3(z) & 3C^{'}_{2} & \sigma_h(xy) & 2S_3 & 3\sigma_v\\ \hline 1 & 0 & -1 & 1 & 0 & 0\\ \hline \end{array} $$

Having this reducible representation, I've tried to identify needed irreducible representation using the reduction formula described here:

$$ a_i = \frac{1}{h}\sum_Q N\cdot \chi(R)_Q \cdot \chi_i(R)_Q $$

For the first IR I'm getting the following result

$$ a_{A^{'}_1} = \frac{1}{12}\left[ 1\cdot1\cdot1 + 0 + 3\cdot(-1)\cdot1 + 1\cdot1\cdot1 + 0 + 0 \right] = -\frac{1}{12}, $$

which doesn't maky any sense, as it should be an integer, if I understand it well.

What am I doing wrong here? Is my reducible representation incorrect or does the formula not apply in this situation?

enter image description here enter image description here

  • $\begingroup$ Will $p_x$ not transform as $x$ in the point group? In $D_{3h}$ this would be $E'$. $\endgroup$
    – porphyrin
    Sep 28, 2018 at 12:34
  • $\begingroup$ @porphyrin $E'$ is two dimensional, $x$ is one-dimensional only. Otherwise, $E$ operation couldn't be +2. $\endgroup$
    – Eenoku
    Sep 28, 2018 at 12:37
  • 2
    $\begingroup$ Yes, that’s why you need to look at p_x and p_y together. If you sum up the characters for both orbitals under each symmetry operation, it should exactly match the E’ row. You can’t look at one part without the other, which is why you’re running into issues with the reduction formula. $\endgroup$
    – orthocresol
    Sep 28, 2018 at 12:48
  • $\begingroup$ @orthocresol Thank you! I'm trying that, but $C_3(x)$ is 0 for both $p_x$ and $p_y$. Then it doesn't correspond with -1 in $E'$. $\endgroup$
    – Eenoku
    Sep 28, 2018 at 12:55
  • 1
    $\begingroup$ Not at all; its the projection of the unit vector representing the p orbital on to the axis. $\endgroup$
    – porphyrin
    Sep 28, 2018 at 16:10


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.