In a typical carbon double bond $sp^2$ hybidrization is used to explain the resulting bond. When two carbon atoms A and B form a double bond one $sp^2$ orbital from A bonds to another $sp^2$ orbital from B and one $p$ orbital from A bonds to a $p$ orbital from B. This leaves two $sp$ orbitals sticking out the end of the ethene-like molecule. I think one could stick two ethene-like molecules end to end to stick together with two $sp^2$ orbitals from each end but then I think one would have an interfering $p$ orbital that would subtract and leave a bond order of 1. I think one might then have further hybridization and end up with three 1.67 bonds (instead of one single and two double bonds) but not three double bonds. The second possibility is sticking together two $sp^2$ orbitals and one $p$ orbital from A to two $sp^2$ orbitals and one $p$ orbital from B leaving behind one $sp^2$ bond sticking out on each end. I think one could stick together such pair with another pair rotated at an angle. After hybridization I think one ends up with three net 2 bonds together but this is confusing to me. A third possibility is sticking together two ethyne-like molecules together but I think that would be a hybrid of two 3-bonds and a -1 bond which would be 1.67 again.

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    $\begingroup$ en.wikipedia.org/wiki/Cumulene $\endgroup$ Commented Mar 13, 2017 at 19:23
  • $\begingroup$ To make the post easier to read, do not refer to single- , double- and triple bonds as "-1 bond", "2 bond" and "3-bond". $\endgroup$
    – Paul Kolk
    Commented 5 hours ago

2 Answers 2


There's two different ways to stick double bonds next to each other.

The first is to put the two double bonds right next to each other, the "cumulenes":

Cumulenes Image from Wikipedia Commons

In these compounds, the pi orbital of one double bond does NOT hybridize/delocalize with the pi orbital of the adjacent double bond. For example, in the structure on the far left, the three carbon propadiene, the two end carbons are hybridized $\ce{sp^2}$, and the center carbon is hybridized $\ce{sp}$. The remaining two $\ce{p}$ orbitals of the center carbon participate in the double bonds, one to each. But because the two $\ce{p}$ orbitals are perpendicular to each other, there's no communication between the two pi bonds - they're perpendicular to each other, too. One of them is in the plane of the screen, and the other is in the plane going in/out of the screen. (Or, it is in the simplest model of organic bonding that's taught. Once you get into things like molecular orbital theory things get more complicated, but the fact of two non-interacting orbitals remains.)

You can see this by looking at where the hydrogens are located in propadiene.


There's another way double bonds can be combined, and that with an interveneing single bond, the "polyenes":

Example polyene

Image from Wikipedia Commons

In a polyene, all the carbons are $\ce{sp^2}$ hybridized. Unlike a cumulene, the pi bonds of a polyene are all on the same "side" of the polyene. For example, in the polyene shown here, all the pi bonds are sticking in/out of the screen.

This means that you can draw resonance structures where the double bonds "travel" along the carbon chain, effectively swapping the double and single bonds. Like benzene, the distinction between double and single bond starts to break down, and it's more correct to think of a continous pi bond throughout the whole of the polyene chain.

But unlike benzene, the end effects for the bond-swap resonance form means it's less stable than the "normal" bond locations. This means it contributes less to the hybridized form. So while the polyene chain doesn't behave entirely like an alternating single/double bond form, it's not like it's really a "1.67" or "1.5" bond order, either.


Pentacarbon dioxide, $\ce{C5O2}$, actually has four carbon-carbon double bonds cumulated in a chain, with additional carbon-oxygen double bonds tacked on at the ends. The molecule may be kept in solution at ambient temperature and in neat form up to -90°C.

Pentacarbon dioxide, with a $4n$ count of cumulated $\pi$ electrons (specifically, sixteen), may be stabilized by a linear-geometry analogue to aromatic coupling in rings. See this answer.


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