What's the reason for the instability of ethylene dione?

Question

Ethylene dione, with a chemical formula $\ce{C2O2}$ and an IUPAC name of ethene-1,2-dione, is an oxocarbon.

_{(source: wikimedia.org)}

The Wikipedia page says:

The existence of $\ce{OCCO}$ was first suggested in 1913. However, despite its deceptively "simple" structure, for over a century the compound had eluded all attempts to synthesize and observe it. Such elusive nature had earned $\ce{OCCO}$ the reputation of a hypothetical compound and a mysterious, "exceedingly coy molecule".

I cannot understand what the rest of the article is telling me. So, why is ethylene dione unstable?

Answer 1

Expanding what Tan Yong Boon said, there is a full scientific article detailing why ethylene dione is unstable Chem. Eur. J., 1998, 4, 2550-2557.

Long story short: Two CO molecules are far more stable than OCCO. The ground state of ethylene dione is a "diradical". This is, OCCO has two unpaired electrons instead of forming the intuitive chain of double bonds (O=C=C=O). However, if you add one or two additional electrons, the resulting anion becomes stable and adopts the shape of glyoxal (but without hydrogen atoms).

Answer 2

The strength of a bond can be determined by the amount of electron density accumulated in the bonding region. A general rule is that the more electron density, the stronger the bond. This is due to a stronger attraction of the nuclei to the bonding electrons. (Of course, if there is too much, it may in turn become overall repulsive, like in the fluorine molecule.)

With an oxygen atom attached to each carbon, there is a strong inductive effect pulling electrons away from the carbon-carbon bonding region of the ethylene dione molecule. Thus, the strength of the carbon-carbon double bond in ethylene dione is greatly reduced, resulting in its instability. This is one plausible explanation.

Answer 3

If we count $\pi$ electrons, we discover that $\ce{C2O2}$ has an unfavorable count for forming a stable molecule whereas both $\ce{CO2}$ and $\ce{C3O2}$ are much more favorable. This effect is discussed in the context of whether there can be aromaticity without rings.

When you have a linear chain of $sp$ hybridized atoms, as in $\ce{C_mO2}$ oxides, the $\pi$ orbitals are all doubly degenerate, with no nondegenerate low-energy orbital like conventional aromatic rings. Therefore the favored $\pi$ electron count for maximum stabilization is $4n$, not $4n+2$. Well, in $\ce{C_mO2}$ we find there are $2m+6$ $\pi$ electrons, so with both $m=0$ and $m=2$ we get an unfavorable count. With the double degeneracy we then get diradical species, and when we draw out molecular orbitals we find that the unpaired electrons are in $\pi$ antibonding orbitals effectively weakening the bonds holding the molecule together.

With $m=0$ ($\ce{O2}$) the only alternative is to break up a lot of bonding orbitals leaving the individual atoms, but with $m=2$ the poorly stabilized $\ce{C2O2}$ molecule can and does break into two relatively stable $\ce{CO}$ monomers.

Comparing the $\pi$ electron structures for $m=0,1,2,3$:

$m=0 (\ce{O2}): \color{red}{6}$ $\pi$ electrons, diradical, but too small to decompose into alternative stable species.

$m=1 (\ce{CO2}): \color{blue}{8}$ $\pi$ electrons, nonradical species with strong resonance stabilization.

$m=2 (\ce{C2O2}): \color{red}{10}$ $\pi$ electrons, diradical; can and will decompose to stable smaller molecules.

$m=3 (\ce{C3O2}): \color{blue}{12}$ $\pi$ electrons, nonradical, resonance stabilized; can undergo nucleophilic/electrophilic addition (polymerization) but not decompose; storable under controlled conditions.