1. The problem statement, all variables and given/known data


2. The attempt at a solution

I know that the MO diagram for CN- is this: CN- MO diagram

I am unsure how to draw the MO diagram for cyanogen. I know that the higher the bond order, the shorter the bond length. I have tried to draw a MO diagram for cyanogen by combining two of the MO diagram for CN:

NCCN MO Diagram

Not sure if this is necessary but I have drawn the following resonance structures: cyanogen resonance structures

I think that structure (I) contributes the most as structures (II) and (III) have a positive charge on the nitrogen.

I think the C-C bond in cyanogen is shorter than the "usual" C-C bond lengths due to delocalisation across it but I don't know how to explain this using Molecular Orbital theory.

  • $\begingroup$ 1) is the C-C bond length in cyanogen longer or shorter than "usual" C-C bond lengths? 2) Start by listing all the possible molecular orbitals and populate them from the lowest energy up. I.e. you know there will be two C-N sigma orbitals and two C-N sigma stars; a C-C sigma and a C-C sigma star, etc. $\endgroup$
    – Dissenter
    Dec 25, 2014 at 23:12
  • $\begingroup$ 1) @Dissenter The C-C bond length in cyanogen is shorter than usual C-C lengths 2) 6 electrons are in sigma bonds, the 2 C-N sigma bonds (so two C-N sigma stars) and the one C-C sigma bond (so one C-C sigma star). Each nitrogen has a lone pair of electrons, this accounts for (2x2) 4 more electrons, so these will be non-bonding orbitals in the MO diagram? That leaves 8 electrons to fill the 4 C-N pi bonds (so 4 C-N pi stars?). How do I know the order of the molecular orbitals and which orbitals form these molecular orbitals e.g. do two sigmas bond together from CN to form a new sigma in NCCN? $\endgroup$ Dec 26, 2014 at 0:40
  • $\begingroup$ I think the answer is hidden in this question: chemistry.stackexchange.com/q/21973/4945 At least an overview of all valence orbitals is given there. $\endgroup$ Dec 26, 2014 at 2:08

2 Answers 2


See my answer to your question on methyl vinyl ether for some background on molecular orbital (MO) theory.

In this case, your molecule is analogous to the all-carbon system, butadiene. In cyanogen you have 4 p orbitals that can align, overlap and mix together to generate 4 molecular orbitals and you have 4 electrons (1 from each p orbital involved in a pi bond) to place in your molecular orbitals. Since the carbons in cyanogen are $\ce{sp}$ hybridized you actually have a second set of molecular orbitals that are identical to the first in all respects except that the 4 p orbitals in the second set are rotated 90 degrees with respect to the other set of p orbitals.

Let's count electrons. We have 4 electrons in each of these orthogonal molecular orbitals that each contains 4 p orbitals, so we have 8 electrons in our two orthogonal sets of molecular orbitals. We also have 2 lone pairs remaining on each nitrogen for a total of 4 electrons. Finally we have the sigma system; we have 2 C-N sigma bonds and 1 C-C sigma bond each containing 2 electrons for a total of 6 sigma electrons. For the whole molecule we have 8+4+6=18 electrons as expected.

As I mentioned in my other answer, we typically focus on the delocalized molecular orbitals formed by combining the p orbitals and disregard the sigma system (usually) when we create our MOs. Disregarding the sigma system usually simplifies matters.

There are a number of ways to create the MO diagram for butadiene or cyanogen, but all of them will have the common theme of mixing 4 p atomic orbitals and generating 4 molecular orbitals. Here is one way to do it, let's start by mixing 2 p orbital to create the MO's for ethylene (or nitrile).

enter image description here

image source

Now we can mix 2 ethylenes (or nitriles) to generate butadiene (or cyanogen)

enter image description here

image source

We label molecular orbitals $\ce{\Psi_1 ~- ~\Psi_4}$ in order of increasing energy (so $\ce{\Psi_1}$ is the molecular orbital at the bottom of the figure and $\ce{\Psi_4}$ is at the top). Don't worry about the "S"'s and "A"'s in the diagram, they are just telling us about the symmetry properties of the molecular orbitals which we don't need to focus on right now.

Notice how $\ce{\Psi_1}$ has overlap between the orbitals on the two middle carbons. This indicates overlap and a higher bond order between these 2 carbons, just as your resonance structures II and II indicate. Just like in butadiene, this carbon-carbon bond has some double bond character and, therefore there is a higher barrier to rotation about it then you might otherwise expect.

  • $\begingroup$ Sorry @ron, I don't understand your diagram for cyanogen. What does the S and A stand for? Cyanogen has 18 electrons but your MO diagram only has 4 electrons? Where is \ce{\Psi_1} in your diagram? How did you go from the MO diagram for CN to this?×Comments may only be edited for 5 minutes×Comments may only be edited for 5 minutes×Comments may only be edited for 5 minutes $\endgroup$ Dec 26, 2014 at 0:23
  • $\begingroup$ @StackExchangeLover I've revised my answer to address the questions in your comment. $\endgroup$
    – ron
    Dec 26, 2014 at 17:05

This feature of cyanogen was already mentioned in the comment by Geoff to the Question "How many electrons are there in the pi-system of cyanogen?". (And I was very tempted to vote to close as a duplicate.)

Your resonance structures are correct and it is noteworthy, while I might be the predominant one, the other two are still considerably influencing the electronic structure. The problem with this is, that you cannot sufficiently express delocalisation in Lewis terms. A molecular orbitals approach is necessary to understand the bonding situations. Below you can find the complete valence MO scheme of cyanogen, which I repost updated from the linked question.
MO scheme cyanogen
You can clearly see, that the eight $\pi$ orbitals are delocalised over the complete molecule. Half of them include the bond between the central carbon atoms, i.e. the degenerate $\pi_\mathrm{u}$ orbitals.
For a carbon carbon single bond you would expect, a bond length of $\mathbf{d}(\ce{C-C})=150~\mathrm{pm}$ and for a double bond $\mathbf{d}(\ce{C=C})=134~\mathrm{pm}$. The calculated bond length is about $\mathbf{d}(\ce{C\bond{~-}C})=138~\mathrm{pm}$ (DF-BP86/def2-SVP) and clearly between a single and a double bond. This has of course an effect on the neighbouring carbon nitrogen bonds. You would expect the triple bond to be $\mathbf{d}(\ce{C#N})=108~\mathrm{pm}$ and the double bond $\mathbf{d}(\ce{C=N})=120~\mathrm{pm}$. The calculated bond length is $\mathbf{d}(\ce{C\bond{~=}N})=118~\mathrm{pm}$ and only little shorter than a double bond.

Covalent radii are taken from Pekka Pyykkö and Michiko Atsumi, Chem. Eur. J., 2009, 15 (46), 12770-1779.

  • $\begingroup$ Thank you for not closing my question @Martin. Sorry I don't understand your diagram. How does the diagram tell me that the pi orbitals are delocalised over the complete molecule? You need 4 atomic orbitals to form one molecular orbital? What is the significance of the black and non-black parts? What do the subscript u and g mean? $\endgroup$ Dec 26, 2014 at 12:42
  • $\begingroup$ @StackExchangeLover I am currently on my tablet and cannot answer properly. What do you know about symmetry and point groups? The circles on the bottom represent s orbitals. Black and white represent the different phases of the orbitals. U is fot ungerade and g is for gerade. Best you read a little more about how orbitals combine. Since i am out of the office, i will not be able to update this post for at least a week, sorry. $\endgroup$ Dec 26, 2014 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.