Why doesn't oxygen form stable cyclic pentagonal compounds with itself?

Consider a molecule where five oxygen atoms are bonded in a pentagonal ring structure. The bond angle in an oxygen molecule (104.5 degrees) is quite similar to the interior angle of a regular pentagon (think benzene) which is 108 degrees. This would cause a fairly low amount of ring strain.

In addition, the formal charges on the oxygens would be zero. There aren't any dipoles, weak bonds, or areas of high electron density so surely it would be somewhat stable and unreactive?

Why don't they form?


1 Answer 1


Sulfur does form ring compounds, so why not oxygen?

You're not wrong to wonder, but I think people are reacting to the implication that this species has:

[no] weak bonds

Consider hydrogen peroxide H-O-O-H which has a central O-O bond. $\ce{H2O2}$ is fairly well known to decompose easily and generate radicals.

In fact, there are a variety of peroxides with that R-O-O-R motif. The peroxide refers to the O-O single bond.

In general, if you look up bond dissociation energies for R-O-O-R bonds, they're ~150-160 kJ/mol, about half the dissociation energy for $\ce{H3C}-\ce{CH3}$ and far less than the 498 kJ/mol for the $\ce{O2}$ double bond.

So it's perhaps possible to form such a ring, but with five O-O single bonds, it's likely to decompose readily, presumably into $\ce{O2}$ and $\ce{O3}$ first, which would both have greater bond strengths.


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