For instance, consider a system with $p_x$ and $p_z$ orbitals at the vertices of a square (on xy-plane). A square by itself would have $D_4$ symmetry. However, because of the $p_x$ orbital; the $90^\circ$ rotation ($C_4$) and $270^\circ$ rotation ($C_4^{-1}$) are no longer symmetry operations. Now the rest of the $D_4$ members do preserve the symmetry, but do not form a group since the subset is not closed.

How does one go about formulating a symmetry group for cases like these?

  • $\begingroup$ Never saw a rotation denoted with $r$. Normally, 90° and 270° rotations are denoted with $C_4$ and $C_4^{-1}$ (or $3C_4),$ respectfully. This should give you a hint when you look up the point group using a character table. $r$ and $r^3$ in your context would refer to the distance and volume, respectfully. $\endgroup$
    – andselisk
    Commented Jun 28, 2020 at 17:01
  • $\begingroup$ Orbitals are not relevant to molecular symmetry. Only atoms are. Are the atoms at the four corners of the square identical? If so, then 90 degree rotation does return the original molecule. $\endgroup$
    – Andrew
    Commented Jun 28, 2020 at 17:05
  • $\begingroup$ @andselisk Sorry about the issue with notations, I've fixed it. Thanks! $\endgroup$
    – Feynfan
    Commented Jun 28, 2020 at 17:19
  • 1
    $\begingroup$ I am mildly confused: a square should be $D_\mathrm{4h}$, surely? $\endgroup$ Commented Jun 28, 2020 at 17:23
  • $\begingroup$ @Andrew Actually this is sort of a hypothetical problem from physics. It's not really a molecule, and I'm considering a system that contains hypothetical atoms with only $p_x$ and $p_z$ like orbitals at the four vertices. I'm afraid a $C_4$ rotation will turn $p_x$ into $p_y$ breaking the symmetry. $\endgroup$
    – Feynfan
    Commented Jun 28, 2020 at 17:24

1 Answer 1


Hmm, so I'm not entirely sure what notation you're using ($D_4$ sounded like Schönflies to me), but I'm sure you can figure out the equivalent.

In general I agree with Andrew's point that molecular symmetry is determined by atomic positions and not orbitals. But at least to me as of now, this seems to be a thought exercise more than anything (after all, no real atom has only two p-orbitals), so I'll play along. In Schönflies notation your system would be labelled $D_\mathrm{2h}$. The three $C_2$ axes and three mirror planes are illustrated below. I left out the identity operation $E$ and the centre of inversion $i$.

Symmetry operations

  • $\begingroup$ +1, orthocresol, What software did you use to draw these pictures? Is it a pen used PowerPoint? $\endgroup$
    – AChem
    Commented Jun 29, 2020 at 3:56
  • 2
    $\begingroup$ @M.Farooq, stylus on tablet, more specifically Apple Pencil + iPad Pro using the Notability app (starting to sound like an Apple commercial so I had better stop here...) $\endgroup$ Commented Jun 29, 2020 at 3:59
  • $\begingroup$ Figure looks nice! I feel a common stylus in PowerPoint should also be able to draw similar figures. Don't have these Apple products. $\endgroup$
    – AChem
    Commented Jun 29, 2020 at 4:05
  • $\begingroup$ Yep, I've done similar things in PowerPoint before on my iPad, so I don't think special software is necessary. As far as hardware goes, there are many on the market nowadays. $\endgroup$ Commented Jun 29, 2020 at 4:19

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