# Bonding in diatomic C2, a carbon-carbon quadruple bond?

Carbon is well known to form single, double, and triple $\ce{C-C}$ bonds in compounds. There is a recent report (2012) that carbon forms a quadruple bond in diatomic carbon, $\ce{C2}$. The excerpt below is taken from that report. The fourth bond seems pretty odd to me.

$\ce{C2}$ and its isoelectronic molecules $\ce{CN+}$, BN and $\ce{CB-}$ (each having eight valence electrons) are bound by a quadruple bond. The bonding comprises not only one σ- and two π-bonds, but also one weak ‘inverted’ bond, which can be characterized by the interaction of electrons in two outwardly pointing sp hybrid orbitals.

According to Shaik, the existence of the fourth bond in $\ce{C2}$ suggests that it is not really diradical...
If $\ce{C2}$ were a diradical it would immediately form higher clusters. I think the fact that you can isolate $\ce{C2}$ tells you it has a barrier, small as it may be, to prevent that.

Molecular orbital theory for dicarbon, on the other hand, predicts a C-C double bond in $\ce{C2}$ with 2 pairs of electrons in $\pi$ bonding orbitals and a bond order of two. "The bond dissociation energies (BDE) of $\ce{B2, C2}$, and $\ce{N2}$ show increasing BDE consistent with single, double, and triple bonds." (Ref) So this model of the $\ce{C2}$ molecule seems quite reasonable.

My questions, since this is most definitely not my area of expertise:

• Is dicarbon found naturally in any quantity and how stable is it? Is it easy to make in the lab? (The Wikipedia article reports it in stellar atmospheres, electric arcs, etc.)
• Is there good evidence for the presence of a quadruple bond in $\ce{C2}$ that wouldn't be equally well explained by double bonding?
• You may be interested in this blog post by Rzepa on the $\ce{CN+}$ cation, which putatively contains a $\ce{CN}$ quadruple bond and is isoelectronic with $\ce{C2}$ Jun 5 '12 at 2:56
• @Richard Terrett Thanks for the reference...it's one I hadn't found. So, the quadruple bond is plausible from a calculation stand point (if I'm reading that right). Is there experimental evidence that could/would support one view or the other? As I said, I'm "a bit" out of my field here. Jun 5 '12 at 5:39
• There is an example that C might have quadruple bonds with U
– user378
Jun 29 '12 at 21:06
• @JaniceDelMar There is no evidence, and there never will be. The C2 molecule looks like any other homodiatomic: two fluffy balls of electron density pushed together. Where are the four ropes in that picture? May 4 '13 at 6:31
• It would not necessarily form higher clusters, because maybe 2 C-C -> C-C-C-C is an endothermic reaction. The product, too, is a diradical! It's a non-explanation. May 4 '13 at 6:37

Okay, this is not so much of an answer as it is a summary of my own progress on this topic after giving it some thought. I don't think it's a settled debate in the community yet, so I don't feel so much ashamed about it :)

A few of the things worthy of note are:

• The bond energy found by the authors for this fourth bond is $$\pu{13.2 kcal/mol}$$, i.e. about $$\pu{55 kJ/mol}$$. This is very weak for a covalent bond. You can compare it to other values here, or to the energies of the first three bonds in triple-bonded carbon, which are respectively $$348, 266$$, and $$\pu{225 kJ/mol}$$. This fourth bond is actually even weaker than the strongest of hydrogen bonds ($$\ce{F\bond{...}H–F}$$, at $$\pu{160 kJ/mol}$$). Another point of view on this article could thus be: “valence bond necessarily predicts a quadruple bond, and it was now precisely calculated and found to be quite weak.”

• The findings of this article are consistent with earlier calculations using other quantum chemistry methods (e.g. the DFT calculations in ref. 48 of the Nature Chemistry paper) which have found a bond order between 3 and 4 for molecular dicarbon.

• However, the existence of this quadruple bonds is somewhat at odds with the cohesive energy of gas-phase dicarbon, which according to Wikipedia is $$\pu{6.32 eV}$$, i.e. $$\pu{609 kJ/mol}$$. This latter value is much more in line with typical double bonds, reported at an average of $$\pu{614 kJ/mol}$$. This is still a bit of a misery to me…

The real issue is that no one has ever taken a picture (i.e. electron density) of genuine, unambigious, cases of a single, double, triple, quadruple??? bonds. And they never will, because these concepts are not based on quantum mechanics.

Two atoms reside next to each other, and if they have a favorable electrostatic interaction, then a certain type of topology arises in their electron density. (q.v. Quantum Theory of Atoms in Molecules)

You might as well say that every "bond" is a single-bond, or, equivalently, and infinity-bond.

These types of articles are bogus, as they can not be confirmed experimentally. They got lucky with the reviewers, and/or an editor who knows their readership is just dying to hear news of a quadruple bond, having heard for so many years that triple is the highest you can go.

I mean, what are people looking for? Four "ropes" that link between the two carbon atoms? Where is the unambiguous, unbiased dividing line between bond energies of a single/double/triple/quadruple bond?

• 1) Electron density can be observed. 2) Quadruple bonds are pretty obvious in metals complexes 3) Even hextuple bonds are theoretically possible in certain molecules... Aug 21 '17 at 14:33
• 1) Yes, that's why I cited electron density as an example of an observable which might be used to confirm. 2) citation needed 3) example needed. Hexatuple bonds? Aug 22 '17 at 12:13
• Aug 22 '17 at 12:30

A recent publication by Chen & Manz (2019) on the bond order of diatomic molecules did touch on this controversy in their discussion. In their study, they have employed a novel method of analysis, which they called Bond Order Component Analysis (BOCA), to quantum mechanically calculate the bond orders of 288 diatomic molecules and ions. They discussed the diatomics $$\ce {C2}$$, $$\ce {Mo2}$$ and $$\ce {O2}$$ in greater detail. The following edited text taken from the publication focuses on their discussion on the diatomic bonding in $$\ce {C2}$$:

Fig. 2A displays the natural spin orbitals, their occupancies, and their bond order components for the carbon dimer, $$\ce {C2}$$. (Here, we adopted the convention of labeling the first valence orbital (rather than the core orbital) as $$1\sigma_g$$.) The $$1\sigma_g$$, $$1\sigma_u$$, and $$2\sigma_g$$ orbital shapes show strong $$\mathrm{s–p}$$ mixing. Four natural spin orbitals had significantly positive bond order components. The $$1\pi_{u,x}$$, and $$1\pi_{u,y}$$ components ($$0.786$$) were larger than the $$1\sigma_g$$ component ($$0.667$$) which was larger than the $$1\sigma_u$$ component $$(0.407)$$. The $$2\sigma_g$$ orbital had negligible BO contribution $$(0.045)$$. These bond order components sum to $$2.727$$. Our findings are in agreement with a carbon dimer bond order of $$2–3$$ caused by four bonding components,(i.e., two smaller $$\sigma$$-bonding components and two larger $$\pi$$-bonding components) reported by several research groups as discussed by Hermann and Frenking.

So yes, there seems to be indeed, four bonding components contributing to the overall bond order of the dimer but each component contributes to less than a bond. Hence, the overall bond order is much less than the value of four.

Reference

Taoyi Chen, Thomas A. Manz, "Bond orders of the diatomic molecules," RSC Adv. 2019, 9, 17072-17092 (DOI: 10.1039/C9RA00974D).

Carbon 100% could form a quadruple bond in C2. Each and every non-noble gas atom is trying to get an outer shell of 8. Diatomic elements are elements that can bond with themselves (the three most common being Oxygen, Nitrogen, and Hydrogen). An acronym to remember the Diatomic elements is H O Br F I N Cl. As you can see, "C" isn't one of them. But, these elements can only bond with themselves because if they do, they get a full outer shell. 2 Nitrogens share 3 electrons with each other, and have 2 left over on each one. 3+3=6+2=8. Carbon has 4 valence electrons and needs to gain 4. It is light enough to be able to form a quadruple bond, and if it does, it will share all four of its electrons with the other carbon! So each carbon will have a total of 8 valence electrons.