[antimatter101]

Carbon monoxide is a paramagnetic compound due to unbonded electrons. Thus, in order to become more stable it should bond itself to other similar molecules to form C₂O₂. It should be more stable, but it is incredibly unstable and exists for only 0.00000001 seconds before decomposing to two molecules of carbon monoxide. It is thought of as a ketene of glyoxylic acid. It should be more stable since it satisfies the octet rule, but is less stable for some reason.

One reason could be the inert-pair effect, but due to carbon's front position in the group 14 elements the effect should be very weak and should not prevent bonding from happening. While C₂O₂ is unstable, carbon suboxide C₃O₂ is stable. In the absence of water (and maybe oxygen too) the molecule C₂O₂^2- is stable.

Why?

[antimatter101]

Also, while C₂O₂ is unstable, C₂OH₂ is stable.

Got this info from searching up Ketene.

Also ketenes are made by adding mase to an acyl chloride, which must have at least two carbons.

And yes they are useful for making polymers.

[orgopete]

I love these types of questions. However, I prefer to use a different perspective. Stability data has been given and asked is whether it matches theory.

Let me recap:
Stable
H2C=C=O
O=C=C=C=O
(-)O-CΞC-O(-)

Unstable
O=C=C=O

Because this is data, it is correct. What we may therefore ask is whether our atomic theories are sufficient to explan this data. It is apparent that a simple application of the octet rule does not predict the instability of dicarbon dioxide. I surmise that it is the addition of electrons that increases stability. This can be noted by the dianion or that of an additional carbon atom. However, I cannot explain how or why two oxygen atoms should result in a scission of the carbon-carbon bond, especially as carbon monoxide may also be written with the oxygen atom donating electrons to the carbon as a resonance form.

[antimatter101]

How about the inert-pair effect? It works on Lead. Lead oxide is more stable than lead dioxide.

[orgopete]

I had to look up the "inert-pair" effect. From the Wikipedia zarticle, it did not seem a convincing theory. Perhaps you could explain how or why it should be applied in this case.

[antimatter101]

THE inert pair effect is seen especially in groups 13 and 14 of the periodic table, in which the heavier elements tend to form compounds with a valency 2 lower than the expected group valency. It is used to account for the existence of thallium(I) compounds in group 13 and lead (II) in group 14. In forming compounds, elements in these groups promote an electron from a filled s-level state to an empty p-level. The energy required for this is more than compensated for by the extra energy gain in forming two more bonds. For the heavier elements, the bond strengths or lattice energies in the compounds are lower than those of the lighter elements. Consequently the energy compensation is less important and the lower valence states become favoured.(Copy and paste)

If carbon monoxide were wholly ionic, then C₂O₂ would not exist.

Also the hybridization of the two carbon atoms in C₂O₂ is repulsive and thus makes the molecule unstable. But how strong can the inert-pair effect be on dicarbon dioxide?

[antimatter101]

You can watch this video on youtube too.
www.youtube.com/watch?v=xJn2sYpvrc0

[orgopete]

THE inert pair effect is seen especially in groups 13 and 14 of the periodic table, in which the heavier elements tend to form compounds with a valency 2 lower than the expected group valency. It is used to account for the existence of thallium(I) compounds in group 13 and lead (II) in group 14. In forming compounds, elements in these groups promote an electron from a filled s-level state to an empty p-level. The energy required for this is more than compensated for by the extra energy gain in forming two more bonds. For the heavier elements, the bond strengths or lattice energies in the compounds are lower than those of the lighter elements. Consequently the energy compensation is less important and the lower valence states become favoured.(Copy and paste)

If carbon monoxide were wholly ionic, then C₂O₂ would not exist.

Also the hybridization of the two carbon atoms in C₂O₂ is repulsive and thus makes the molecule unstable. But how strong can the inert-pair effect be on dicarbon dioxide?

See inert pair effect, scroll down to failure.

Let me parse this out. As one may see from the wikipedia article, the inert pair effect does not seem to be a universally accepted principle, but no matter. Whether an inert pair effect is invoked or not, it is not providing any real explanation. That is, I don't find it different than simply saying that carbon monoxide is more stable than dicarbon dioxide. If there were a real bonding effect working, carbon monoxide should be part of a larger group of compounds or general principle. I do not find this to be the case, except perhaps at high temperatures in which decarbonylation reactions take place. See decomposition of pyruvic or oxalic acid. I could conclude the decomposition energies of pyruvic or oxalic acid is higher than dicarbon dioxide, but that would not explain why.

Re: ionic
There are four forces of nature, strong, weak, gravity, and electromagnetic. Chemistry is a result of electromagnetic forces, an inverse square force. Prototypical ionic bonds are simply long and weak bonds. However, you may not realize this as atomic theory suggests cations and anions are in contact. As a consequence, ions are given different radii depending on their charge.

I argue as follows. If HF (92 pm) ionizes, it ionizes to F(-) and H(+). The radius of H(+) is effectively zero, so the radius of F(-) may be thought to be ~92 pm (give or take a few pm). If two more protons were added to F(-), Na(+) would be the result and I predict a similar atomic radius (~90 pm). The bond length of NaF is 232 pm. I suggest there is a gap of ~50 pm between the ions. This gap results in a weak bond, and readily separable by water. One must be careful to not attempt to extrapolate bond strength from ionic properties. This would be similar to building a bridge out of toothpicks and using the bridge strength to measure the strength of a toothpick. The bridge will be much stronger than the sum of the toothpicks.

Furthermore, carbon monoxide is neutral. Shifting electrons or introducing formal charges does not make it ionic. Protons are positive and electrons negative. There is no net charge present. Any reaction of electrons is a measure of their availability or lack of protection by other atoms, such as a proton. If one were to compare hydronium, water, and hydroxide, the charge of oxygen is constant (+8). The difference is due to the protonation of the electrons surrounding the oxygen. Each protonation reduces the ease by which the remaining pair of electrons may react.

Re: hybridization
I am not sure I understand this argument. I know dicarbon dioxide is unstable. I can draw a perfectly fine Lewis structure for it. Since the equilibrium favors carbon monoxide, I can conclude the C=C bond is unstable. This is true whether I invoke hybridization, ionic character, or inert pair effect. Because another series of compounds has been presented with very similar characteristics, what remains is to explain why dicarbon dioxide should be different.

I think I would be more inclined to think carbon monoxide has a greater than expected stability. One can apply resonance theory or I'm sure someone has done a calculation to explain why it should be more stable. The properties that lead to its greater stability would seem to answer the instability of dicarbon dioxide. That is, I presume the stabilization of carbon monoxide would be reduced by the formation of a dicarbon dioxide molecule and reduce its stability. Then, oxalic or pyruvic acid might result in molecules of intermediate stability by being able to form a single carbon monoxide or carbon monoxide like intermediate. Again, the driving force being the formation of a carbon monoxide or oxygen stabilized carbenoid intermediate. 

[Corribus]

I argue as follows. If HF (92 pm) ionizes, it ionizes to F(-) and H(+). The radius of H(+) is effectively zero, so the radius of F(-) may be thought to be ~92 pm (give or take a few pm).

Ionic radius of fluoride is ~130 pm, much larger than the covalent radius (~71 pm).

As to the central question of the thread, stability of OCxO is complicated.  Stability when x = odd seems much higher than when x = even.  Stability appears to be related to the stability of the triplet state, as the singlet state is repulsive in nature.  The stability also appears to increase as x becomes larger, at least for the first few examples.  OCCCCO is more stable than OCCO.  Were I a betting man, I'd guess this is because the repulsive nature of the singlet state decreases as x increases (carbon is negatively charged in CO; put two of them together tail to tail and you get kind of the same effect when you try to stick two AA batteries together in the tail to tail orientation - this effect will be insulated to some degree as the carbon bridge between the two terminal CO units increases).

If you read some of the numerous theoretical works on the x = 2 case, there is almost no barrier (~3 kcal/mol) between the minimum of the metastable triplet state and the intersystem crossing point between the triplet and (repulsive) singlet surfaces.  This pretty much ensures that OCCO will rapidly dissociate into 2 CO molecules. In conjugated oligomers and polymers, as the carbon-carbon chain lengthens, the rate of intersystem crossing drops: extent of electron delocalization in singlet states is known to be far greater than in triplet states, which tend to be localized on a single repeat unit.  This is the case of excitons in excited-states, anyway, and I see no reason why it shouldn't be the case for competition between ground-state triplets and singlets, either.  This might also explain why stability increases as the number of carbons in the bridge increases.  The energy difference between cumulenic (C=C=C=C=C=C) and noncumulenic (C-C≡C-C≡C-C) forms likely plays a role here as well, especially when it comes to explaining the stability/instability of odd/even carbon numbers, though of course these factors are all interrelated.

Those are just some thoughts.  In the end its a complicated phenomenon that doesn't seem to be completely understood. 

EDIT: For the theoretical paper I mentioned, see Schroder, et al, Chem. Eur. J. 1998, 4 2550.

[orgopete]

I argue as follows. If HF (92 pm) ionizes, it ionizes to F(-) and H(+). The radius of H(+) is effectively zero, so the radius of F(-) may be thought to be ~92 pm (give or take a few pm).

Ionic radius of fluoride is ~130 pm, much larger than the covalent radius (~71 pm).

The covalent radius of fluorine is one-half of the bond length, 142 pm. However, that does not include the actual radius of the electrons surrounding the nucleus. It is simply half of the distance. If you were to recreate a shared pair of electrons, then it could be a useful measure.

How is the ionic radius of fluoride known to be 130 pm? I suggest it is because it is assumed that ions contact one another. Hence if the bond length of NaF is 232 pm, then the sum of the ions must also equal 232 pm. I do not find this a convincing argument. This implies that ionization of HF results in the electrons of fluoride to expand from 92 pm to 130 pm simply as the H-F distance increases. Why?

I am not aware of any direct measures of electron distances and I expect if there were, then the radii would be precise. However, we can vicariously estimate how electrons may move from other examples. If fluoride increases its diameter upon ionization, then we may expect water to behave similarly. The difference is a proton remains attached so we can compare the H-O distance in hydroxide and water, or water and hydronium, or ammonia and ammonium. In all cases, the H-X distance changes very little. I am assuming the nuclear force remains constant to all electrons. That is, even for HF, the fluorine-electron force and distance would be very similar for the protonated and non-protonated electrons, give or take a little. I would be surprised to find the non-bonded electrons of HF were at 130 pm while the protonated pair were at 92 pm.

I find it much easier to think that if atomic forces reflect the electromagnetic elements making up the bonds, then attraction of a nucleus for its electrons would not change drastically should the net charge change due to ionization, especially by a proton and even less so by a cation. I recognize my position is different than the premise of different radii for atom and their ions from the 1930's

The alkali metal cations hold their electrons notoriously weakly. The loss of their electrons is very facile. It seems counter intuitive to think that the shell of electrons of a sodium ion should have a strong attraction for the electrons of a fluoride ion. To the contrary, I find it plausible that if the radius of the electrons of fluorine extend to ~92 pm, then the electrons of a sodium cation should be found at a similar distance. Because sodium has an excess proton, then it should be attracted to an addition electron or pair of electrons. However, the shielding of this inner electrons should reduce the attraction as one might expect due to the inverse square law. Consequently, it seems reasonable that the bond length of disodium should be much larger, 308 pm. That distance implies a gap between the outer electrons. It would also be consistent with a gap in NaF.

Re: the remaining arguments
Could be, I don't know.

[Corribus]

The covalent radius of fluorine is one-half of the bond length, 142 pm. However, that does not include the actual radius of the electrons surrounding the nucleus. It is simply half of the distance. If you were to recreate a shared pair of electrons, then it could be a useful measure.

Using covalent radius is an especially inaccurate way to predict bond lengths for compounds containing fluorine.  The F₂ bond is especially weak due to electron-electron replusion.  Bond enthalpy between two fluorines is almost the same as that in iodine!  The actual atomic size of fluorine is about 60 pm, much smaller than predicted by the rather arbitrary "covalent radius", a metric which assumes that the element in question bonds the same way all the time.  Fluorine defies trends in many ways because of its incredibly high electronegativity.

You may find these websites helpful:

https://en.wikipedia.org/wiki/Covalent_radius_of_fluorine (ionic radius = 133 pm)
http://www.chemguide.co.uk/atoms/properties/atradius.html (see the table which lists ionic radius of fluoride = 133 pm)
http://chemistry.about.com/od/elementfacts/a/fluorine.htm (ionic radius = 133 pm)
http://en.wikipedia.org/wiki/Ionic_radius (crystal ionic radius = 119 pm, effective radius = 133 pm)

[antimatter101]

I cam so confused.

[Corribus]

I'm not sure how the thread got side-tracked on fluorine, but what exactly are you confused about?

Molecular physics of the title compound appear to be quite complicated, probably beyond what you can easily understand at the undergraduate level.  The theoretical papers are hard to digest and would take many readings to fully understand.  I think the simplest "undergraduate chemistry" explanation, without invoking spin states and potential energy surfaces, is probably this:

In cumulenic C=C=C=C (etc) compounds there is almost always a resonance form with the oligoyne: -C≡C-C≡C-.  (The respective energies of these two states can be pretty different, with the latter form often favored because of a Peierls distortion.)  Therefore O=C=C=O will have a resonance form with O≡C-C≡O.  In the latter form the oxygens will have partial positive charge and the carbons will have partial negative charge.  You can see easily that the partial negative charges will be on two adjacent carbons.  This will form a repulsive interaction which will weaken the C-C bond and make dissociation to form two stable carbon monoxide molecules fairly favorable. 

(I will reiterate my prior analogy of trying to put two batteries together in tail-to-tail orientation.  They are repulsive and easily resist any attempt to put them together.  The dipole moment of CO isn't large, but it is enough to make the ground state repulsive, which ensures thermodynamic favorability - with no activation energy - of dissociation.)

[orgopete]

I'm not sure how the thread got side-tracked on fluorine, but what exactly are you confused about?

This is how fluorine was introduced. The poster was seeking reasons for carbon monoxide to not form dicarbon dioxide, including inert pair effect, resonance theory, and ionic character. I argued that a simple application of any of these theories may not lead to a satisfactory reason for the instability.

Since the poster had brought up ionic bonds as an example, I chose to argue ionic bonds followed the inverse square law and that ionic bonds were simply weak bonds. I concede that my explanations were incomplete in total on this point and I tried to simply insert ion gaps as consistent with the inverse square law and result in weak bonds. I argue that there are only four forces of nature, strong, weak, gravity, and electromagnetic. If that is so, then ionic, covalent, hydrogen, London, dipole-dipole, etc, are different strengths of electromagnetic forces.

An additional problem also occurs on bond lengths and atomic radii, I had not explained what radii I was measuring. If one has a carbon or a fluorine atom, the atom's radius is the distance the electrons extend from the nucleus. Obviously the radius of fluorine cannot be one-half of difluorine's bond length (142 pm) as that must truncate the shared (and presumably unshared) electrons.

Further exacerbating this problem are a host of atomic radii, but I consider this to exemplify the weakness of some of our more fundamental atomic models. I find that application of a very simplistic inverse square model reveals little additional forces to alter radii. No change in nuclear charge or force occurs upon ionization to justify a significant increase in an atom's radius. Although I had not mentioned to this point, I had used this model [ACS National Meeting, Philadephia, 2012] to explain why a neutral ammonia molecule should be a stronger base than a fluoride ion. The dominant basic character is a balance between a proton's attraction to an electron pair and the Coulombic repulsion to the nucleus and that attraction can be greater than a net imbalance in protons and electrons.

My objective in trying to answer the poster's question was not to provide a bona fide solution, but more to agree with the paradox inherent in the question. The resonance structures of dicarbon dioxide implies a greater stability than carbon monoxide. I thought this an important point raised by the poster.

My net conclusion (which did evolve since my first post), was that there must have been additional and perhaps unexplained stability in carbon monoxide. I surmise a rapid decomposition of dicarbon dioxide should be placed in the context of other decarbonylation reactions. Anhydrous formic acid notoriously decomposes to carbon monoxide and water. Other compounds can also suffer loss of carbon monoxide, though none with the ease of dicarbon monoxide.

I would argue the following proves the poster's question was a well thought out question. 

Molecular physics of the title compound appear to be quite complicated, probably beyond what you can easily understand at the undergraduate level.  The theoretical papers are hard to digest and would take many readings to fully understand.  I think the simplest "undergraduate chemistry" explanation, without invoking spin states and potential energy surfaces, is probably this:

In cumulenic C=C=C=C (etc) compounds there is almost always a resonance form with the oligoyne: -C≡C-C≡C-.  (The respective energies of these two states can be pretty different, with the latter form often favored because of a Peierls distortion.)  Therefore O=C=C=O will have a resonance form with O≡C-C≡O.  In the latter form the oxygens will have partial positive charge and the carbons will have partial negative charge.  You can see easily that the partial negative charges will be on two adjacent carbons.  This will form a repulsive interaction which will weaken the C-C bond and make dissociation to form two stable carbon monoxide molecules fairly favorable. 

(I will reiterate my prior analogy of trying to put two batteries together in tail-to-tail orientation.  They are repulsive and easily resist any attempt to put them together.  The dipole moment of CO isn't large, but it is enough to make the ground state repulsive, which ensures thermodynamic favorability - with no activation energy - of dissociation.)

Note: although alternate resonance forms have been invoked, I do not agree the additional structures are resonance forms. They are oxidation products of a cumulene or dicarbon dioxide.

[Corribus]

Note: although alternate resonance forms have been invoked, I do not agree the additional structures are resonance forms. They are oxidation products of a cumulene or dicarbon dioxide.

I'm not sure what you are referring to here, particularly the bold part (bolded by myself).