Consider the molecule $\ce{Os(CO)4Cl2}$ with a drawing given below via Chemtube3d:

Symmetry elements of Os(CO)4Cl2

I thought in this molecule, if I drew a horizontal plane, then a $σ_\mathrm{h}$ would be present i.e. A ‘horizontal’ mirror plane is perpendicular to the principle axes seeing as the mirror would, in my view, cut the atoms in half so the top would be the reflection of the bottom.

However, my professor said in a lecture that this molecule shown here does not have $σ_\mathrm{h}$ and I'm slightly puzzled as to why.

Why does this molecule not have $σ_\mathrm{h}$ plane?


Your compound has only one proper axis of rotation and it is $C_2$ as shown in the diagram. Therefore, it is the principle axis of rotation. By definition, $\sigma_\mathrm{h}$ is a plane of symmetry perpendicular to the principle axis of rotation. However, the plain you were talking about is paralleled to the principle axis of rotation. Thus, it is not a $\sigma_\mathrm{h}$ plane. Nonetheless, it is a $\sigma_\mathrm{v}$ plane, by definition.


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