# XeF4 - why are there only 2 dihedral mirror planes?

I started studying molecular symmetry and I became somehow confused about dihedral mirror planes. I found a definition, which says, that it is a vertical mirror plane, which bisects two $C_2$ axes.

But when I look at $\ce{XeF4}$ molecule, I see, that it has 4 $C_2$ axes and 4 vertical mirror planes. All of those planes are bisecting 2 $C_2$ axes and both all of them also go through some atom.

Still, there are only 2 dihedral mirror planes $\sigma_\mathrm{d}$ and the remaining two planes are just common vertical ones $\sigma_\mathrm{v}$.

So, could you please tell me what's the real difference between vertical and dihedral mirror planes?

OK, in a link given in comments by Tyberius is said, that dihedral planes are such planes, which bisects as many bonds as possible, while "normal" vertical planes bisects as many atoms as possible. Then I don't get, why ethene (see the picture below) doesn't have one dihedral plane, as one of them clearly bisects its central bond and we can't symmetrically bisect more bonds in this molecule.

### Ethene

• chembio.uoguelph.ca/educmat/chm2060_preuss/L3-2013.pdf Dec 8, 2017 at 22:23
• @Tyberius Ok, now I see "bisects as many bonds as possible"... Could you, please, tell me then, why there is no $\rho_d$ in Ethene? I think, that one of the vertical planes should be dihedral, as it bisects the central double bond and there is no larger number of bonds to be bisected symmetrically. Dec 8, 2017 at 22:35
• It's in the definition you quoted: 'bisects two C2 axes'. Dec 8, 2017 at 23:21
• @orthocresol I thought about that, but while I didn't find it in the provided slideshow, I thought it's not correct... Dec 8, 2017 at 23:28
• I took the liberty to rearrange the structures on the pictures horizontally so that users with smaller or wider screens could compare them both without scrolling back and forth; plus I changed the notations for mirror planes from $\rho$ to more common $\sigma$ (see How are point group character tables typeset correctly?). If you are unhappy with this, please feel free to rollback to the original version:) Dec 9, 2017 at 4:05

This is a situation of coordinate systems. In the $D_\mathrm{4h}$ point group, as you've noticed, there are 4 $C_{2}$ rotation axes in the plane. The easiest way to set the coordinate system is to have one pair be coincident with the $x$ and $y$ axes and the other pair coincident with the functions $x=y$ and $x=-y$.

Herein lies the issue: which pair is which? I think most people would agree that a reasonable choice is that the primary set should be coincident with the Cartesian axes. Indeed, an example character table for this group makes this choice as well (these are the $C_{2}'$ elements). This then defines which pair of mirror planes are $\sigma_\mathrm{d}$ and which pair are $\sigma_\mathrm{v}$.

This covers $\ce{XeF4}$ only, not ethene, because ethene belongs to the $D_\mathrm{2h}$ point group.

Here's the character table for that group. There's no dihedral mirror plane because the mirror planes perpendicular to the plane of molecule are coincident with the rotation axes.

So it seems that the use of $\sigma_\mathrm{v}$ vs $\sigma_\mathrm{d}$ is basically a matter of convention when both terms can be applied. Really, even $\sigma_\mathrm{h}$ vs $\sigma_\mathrm{v}$ is somewhat arbitrary if there isn't a single clear principal axis.

By a certain convention $\sigma_\mathrm{d}$ are vertical mirror planes that bisect $2C_2$ axes. When there are multiple mirror planes that fit this criteria, those that cut through the most atoms are referred to as $\sigma_\mathrm{v}$, while the others are called $\sigma_\mathrm{d}$.

This again is somewhat arbitrary and can be ambiguous. In particular, for $D_\mathrm{2h}$ molecules like ethene any of the mirror planes could fit the description for $\sigma_\mathrm{h}$ depending on which $C_2$ you say is the principal. For this point group, you can unambiguously label the axes and planes using the Cartesian axes