# Why can two carbon atoms not form more than triple bond with each other?

Why two carbon atoms don't form more than triple bond with each other? Please tell me simply as I am a student of 10th class.

• As you are in class 10, I believe that you haven't started learning about Molecular Orbital Theory. If you draw the MOT diagrams for $C_2$ molecule, you get a bond order of 2. This would mean that $C_2$ is stable. Mar 20, 2017 at 9:42
• Mar 20, 2017 at 9:48
• See: onlinelibrary.wiley.com/doi/10.1002/anie.201208206/abstract for a extensive discussion by Roald Hoffmann, Henry Rzepa and Sason Shaik in Angewandte Chemie about this topic. Mar 21, 2017 at 11:40

## 2 Answers

Unfortunately, an answer to this requires a quick dip into orbital theory. You may have heard of s, p and d orbitals; carbon as a main group element of group 14 has the electronic configuration of $[\ce{He}]\,\mathrm{2s^2\,2p^2}$, meaning two of its electrons are in an s orbital and two in p orbitals in its atomic ground state. The process by which these orbitals form bonds is simply overlap: If you have two orbitals that can be moved together and these two orbitals overlap, you generate a bonding and an antibonding orbital from these two. (This is a mathematical operation known as linear combination; plugging two orbitals into a linear combination means that exactly two orbitals are mathematically required to come out of the linear combination.)

The bonding (or antibonding) orbitals that are generated by this process are labelled σ or π orbitals. Note that these are the Greek letters corresponding to s and p: a σ orbital is basically an s orbital streched in one direction and a π orbital is a p orbital streched in the same manner.

Due to a number of constraints that you will learn about soon enough™, a carbon–carbon triple bond $\ce{C#C}$ will consist of one σ bond and two π bonds as shown in the image below.

Figure 1: $\ce{H-C#C-H}$ showing the σ and π orbitals required to make the respective bonds. Image taken from jahschem.wikispaces.org.

See how two of the three possible p orbitals on each carbon (figure 1’s left half) overlap to form two π bonds; the third possible p orbital takes part in both the σ bond to the other carbon and that to the hydrogen — it has been linear combined with the s orbital to form an sp hybrid orbital outside the scope of this answer.

The main point of interest here is that it is, for reasons of symmetry, not possible for more than two p orbitals of a single atom to form π bonds to a single other atom at any one time. And carbon does not have any additional accessable orbitals that it could use for further bonds. Thus, it is not possible for carbon to form quadruple bonds.

The next orbital available would be a d orbital; the transition metals are those in which the d orbitals are valence orbitals and take part in bonding. Once you have a better understanding of orbitals, it may be worthwhile to look up quadruple bonds of transition metals such as in $\ce{[(H2O)Cr({\unicode{x3bc}-}OAc)4Cr(H2O)]}$ or dirhenium octachloride. The latter is particularly interesting because the fourth — δ, in analogy to d — bond can be cleaved by irradiation causing a structural change that clearly shows there must have been a quadruple bond previously. To the best of my knowledge, quadruple bonds have not been shown to exist between main group elements though.

And finally, $\ce{C2}$ does exist; however, explaining it requires orbital theories on a much higher level than in the scope of this answer.

When you say four bonds with each other, it means 1 hybrid orbital for sure which will be the sigma bond inbetween the two carbon atoms. Now inbetween two atoms you can never have more than 1 sigma bond , the reason is for sigma bond we need head one overlapping of hybrid orbitals. If hybrid orbital of one carbon atom is already in head on overlap with the hybrid orbital of the other carbon atom, the other 3 hybrid orbitals of the carbon atom cannot do any further head on overlapping because angle between the hybrid orbitals would be 109 degrees approx. So basically they cant reach the hybrid orbital of the other C atom. So you can't have more than 1 sigma bond.

As we need one hybrid orbital for sigma bond, we would be creating only one hybrid orbital then, hence sp hybridisation.

If we talk about four bonds inbetween the carbon atom, then the other three has to be pie bonds that will be from 3 pure p orbitals, but 3 pure p orbitals is not possible. Because one of the p orbital has already used up in sp hybridisation with the s orbital creating one sp hybrid orbital for sigma bonding. So we have only two pure p orbitals left with us. So how can you form 3 pie bonds when you have only 2 pure p orbital remaining with us. Hence its not possible to have 4 bonds inbetween two carbon atoms.

• I think a picture in your answer would be helpful to illustrate why there is no more space for a third pi-bond. I only suggest this because I do not think the OP has yet covered, in class, the orientation of sigma and pi bonds in space.
– Bob
Mar 20, 2017 at 10:48