I am following the video lectures by Neeraj Saini sir and in this video$^{\dagger}$ he said that $\ce{CO2}$ has infinite planes of symmetry.

But I think that if we take into account the orientation of $π$ bonds (which I don't know if we have to or not) then $\ce {CO2}$ molecule must have $2$ planes of symmetry, as per the following argument:

A $\ce {CO2}$ molecule has similar $\pi$ bond orientation as that of an allene. Structure of allene

Now if we consider the orientation of pi bonds then $\ce {CO2}$ must have only 2 planes of symmetry.

So, to check whether infinite planes of symmetry statement is correct or not, I have the following question:

  • Do we consider the spatial orientation of molecular orbitals while assigning plane of symmetry? And if not then why not?

$\dagger$ Note that he uses Hindi language for the purpose of conversation.

  • 1
    $\begingroup$ We don't.$\mathstrut$ $\endgroup$ – Ivan Neretin Jun 16 '20 at 8:35
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    $\begingroup$ @IvanNeretin: I agree that we don't. But why don't we? We learn about symmetry of molecular orbitals and molecules as a whole. But in a molecule, we don't consider the orientation of orbitals. Is this simply a convention? $\endgroup$ – user14250 Jun 16 '20 at 9:43
  • $\begingroup$ In a way, orbitals themselves are a convention. $\endgroup$ – Ivan Neretin Jun 16 '20 at 10:06
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    $\begingroup$ You can use symmetry to examine orbitals, for example those in benzene or butadiene, but to determine a point group the atoms are all that is needed. The allene belongs to d$_{2d}$ for example and the orbitals / bonds can be remove and the conclusion is the same. $\endgroup$ – porphyrin Jun 16 '20 at 15:06

The reason that orbitals are not considered in the symmetry of a molecule is that the orbitals cannot decrease the symmetry of the molecule. If we look at the canonical delocalized molecular orbitals of any molecule, everyone of them is symmetric or antisymmetric with respect to a symmetry operation of the molecule, with those symmetry operations determined only by the positions of the atoms, not the electron density.

Looking specifically at carbon dioxide, the simplistic picture of $\ce{CO2}$ as being like allene with lone pairs instead of the C-H bonds is not an accurate view of the electron density. If one looks at the delocalized pi-type orbitals, both span all three atoms in either the xz or yz planes (using the standard convention of the z axis being aligned with the molecular axis). Thus, both orbitals are symmetric with respect to the xy plane of symmetry of the molecule (which passes through the C).

Similarly, the lone pairs do not occupy separate sp2 orbitals that have specific alignment in a plane perpendicular to the p orbital involved in pi bonding. Instead, the create a combined electron density that can be deconvoluted into an sp-type orbital along the z-axis and partial contributions from the px and py orbitals, such that the total has no bias along the x or y axis.

If we were to look at the total electron density of $\ce{CO2}$, we would therefore see only a pill shape that does not vary as one rotates around the z-axis.

To visualize this, I recommend looking at all of the calculated delocalized molecular orbitals of $\ce{CO2}$. You can do this on the molcalc.org website if you do not have other software for that purpose.


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