Diborane has the interesting property of having two 3-centered bonds that are each held together by only 2 electrons (see the diagram below, from Wikipedia). These are known as "banana bonds."

I'm assuming there is some sort of bond hybridization transpiring, but the geometry doesn't seem like it is similar to anything I'm familiar with Carbon doing. What sort of hybridization is it, and why don't we see many (any?) other molecules with this bond structure?

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3 Answers 3


Look carefully, it's (distorted) tetrahedral--four groups at nearly symmetrically positions in 3D space{*}. So the hybridization is $sp^3$.

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As you can see, the shape is distorted, but it's tetrahedral. Technically, the banana bonds can be said to be made up of orbitals similar to $sp^3$ but not exactly (like two $sp^{3.1}$ and two $sp^{2.9}$ orbitals--since hybridization is just addition of wavefunctions, we can always change the coefficients to give proper geometry). I'm not too sure of this though.

$\ce{B}$ has an $2s^22p^1$ valence shell, so three covalent bonds gives it an incomplete octet. $\ce{BH3}$ has an empty $2p$ orbital. This orbital overlaps the existing $\ce{B-H}$ $\sigma$ bond cloud (in a nearby $\ce{BH3}$), and forms a 3c2e bond.

It seems that there are a lot more compounds with 3c2e geometry. I'd completely forgotten that there were entire homologous series' under 'boranes' which all have 3c2e bonds (though not the same structure)

And there are Indium and Gallium compounds as well. Still group IIIA, though these are metals. I guess they, like $\ce{Al}$, still form covalent bonds.

So the basic reason for this happening is due to an incomplete octet wanting to fill itself.

Note that "banana" is not necessarily only for 3c2e bonds. Any bent bond is called a "banana" bond.

Regarding similar structures, $\ce{BeCl2}$ and $\ce{AlCl3}$ come to mind, but both of them have the structure via dative(coordinate) bonds. Additionally, $\ce{BeCl2}$ is planar.

Sneaks off and checks Wikipedia. Wikipedia says $\ce{Al2(CH3)6}$ is similar in structure and bond type.

I guess we have less such compounds because there are comparatively few elements ($\ce{B}$ group pretty much) with $\leq3$ valence electrons which form covalent bonds(criteria for the empty orbital). Additionally, $\ce{Al}$ is an iffy case--it like both covalent and ionic bonds. Also, for this geometry (either by banana bonds or by dative bonds), I suppose the relative sizes matter as well--since $\ce{BCl3}$ is a monomer even though $\ce{Cl}$ has a lone pair and can form a dative bond.

*Maybe you're used to the view of tetrahedral structure with an atom at the top? Mentally tilt the boron atom till a hydrogen is up top. You should realize that this is tetrahedral as well.

  • $\begingroup$ I see how it could have a tetrahedral shape, but it seems like it wouldn't due to the strain. $\endgroup$
    – jonsca
    May 4, 2012 at 7:05
  • $\begingroup$ @jonsca: Distorted tetrahedral. Yes, one can say that the hybridization isn't exactly $sp^3$ $\endgroup$ May 4, 2012 at 7:10
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    $\begingroup$ For that matter: a good number of the compounds formed by the boron group elements display 3c2e bonds... Apart from the multitudes of boron (the boranes already show a rich diversity!) and aluminum compounds, there are gallium and indium compounds that display 3c2e bonds; e.g. in here, here, here, here and here. I'm sure there are more... $\endgroup$
    – user95
    May 4, 2012 at 10:45
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    $\begingroup$ One should be careful talking about hybridisation. It may be used as a concept explaining a certain bonding situation that results from a certain geometrical composition of a molecule. Also most covalent bound molecules have more-centre bonds. $\endgroup$ Apr 14, 2014 at 12:23
  • $\begingroup$ AFAIK, Banana bonds happens in hydrides of metal carbonyls and in metal carbonyls themselves. They are accompanied by direct sigma-bond, though. $\endgroup$
    – permeakra
    Jun 28, 2014 at 15:37

Here is a plot of the Quantum Theory of Atoms in Molecules answer to you question. I have shown the bond paths of $\ce{B2H6}$. Indeed, they are "banana-like" but interestingly they are curved inward, unlike the case of cyclopropane which are curved outward.

(Hybridization does not exist. Also, I'm not sure if there is a point of ascribing "number of electrons"--like they are aliquots--to any bonding interaction.)

Also, notice that I have drawn the bond paths between the B's and the four similar hydrogens as solid (covalent), and the set of bond paths along the "bridge" as dotted (not covalent). This is because the sign of the Laplacians of the electron density at their respective bonc critical points (yellow spheres) are opposite.

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  • $\begingroup$ By bond paths, I assume you mean a curve of maximum electron density between atoms? $\endgroup$ May 4, 2013 at 19:10
  • $\begingroup$ Technically, the steepest ascent path through the electron density connecting the two atoms. $\endgroup$
    – Eric Brown
    May 4, 2013 at 21:23
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    $\begingroup$ Could you add level of theory, please. I am not sure what other kind of bond there could be between boron and hydrogen, certainly not ionic. $\endgroup$ Apr 14, 2014 at 12:14
  • $\begingroup$ @Martin I can't recall what the level of theory is, probably B3LYP/6-31G* $\endgroup$
    – Eric Brown
    Jun 2, 2014 at 0:42
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    $\begingroup$ Hydribization 'does not exist' may be true, but it also does not not exist. The concept is useful for explanations, so this answer could be greatly improved by addressing why the view of hybridization leads to an answer disjoint from the physical chemistry of the situation. $\endgroup$
    – Lighthart
    Feb 19, 2015 at 18:43

Multicentre bonds like these (in this case, a 3-centre-2-electron or '3c2e' bond) are pretty fascinating and there is a lot of theoretical work that goes on to understand these. I am a chemist as well, and had to go through learning Wade's rules and all of that. There will be a time for an answer such as that.

But for now I want to post a very, very, simple answer that aims to be more intuitive towards the amateur chemist; because I don't see anything that fills this role, and I don't think it is that hard to explain.

What makes banana bonds possible in diborane?

A "normal" covalent bond is two atoms sharing two electrons. Think about $\ce{F2}$. Isn't it a bit weird how those electrons can belong to both atoms "at the same time", to fulfil both their octets? But you can believe that, because you were taught it at an early stage in chemistry. That's what we could call a 2-centre-2-electron, or '2c2e' bond: there are two electrons being held, or shared, between two centres (two nuclei).

covalent bond in F2

Now, fluorine has a lot of valence electrons to form covalent bonds with. Consider what happens if there aren't enough electrons to go around: you need to start sharing the electrons with more atoms at once. So, the 3c2e bond in diborane is just the same, except that it's three atoms sharing two electrons, and somehow those two electrons belong to all the atoms at the same time.

3c2e bond in diborane

Just like that, everybody can fulfil their complete octet (or duet):

  • Each boron has 2 "normal" covalent bonds to the outer hydrogens (4 electrons), plus 2 orange electrons, plus 2 green electrons, which makes 8.
  • The hydrogens at the top and the bottom each have one pair of electrons.

What's the catch? Why can't we share the electrons between even more atoms? Well, the more you share the electrons, the weaker the bond gets. This is already evident in the case of diborane: the inner ('bridging') B–H bonds are longer, and weaker, than the outer ('terminal') B–H bonds.

It's like only having two slices of cake to feed three people. Because you don't want anybody to throw a tantrum, you cut the slices up and give everybody 2/3 of a slice, and everybody still gets to have cake, but it's just not as much as they would have liked to have.

But to return to the original question of what makes banana bonds possible: it's not as magic as it seems to be at first glance. The idea behind its formation is exactly the same as a normal covalent bond: all the atoms would prefer to have a complete octet. The only difference is that we've got three bonding partners participating in the bond, instead of just two.

What sort of hybridization is it?

You could figure this out from the bond angles and the geometry. However, IMO, it's not worth finding this out: even if you calculated it and found that it was $\mathrm{sp}^{2.7182818}$ in one bond and $\mathrm{sp}^{1.5}$ in the other, you aren't going to do anything with that information. Note that these are B–H bonds, not C–H bonds, and anyway the two types of bonds (2c2e / 'terminal' and 3c2e / 'bridging') are different in terms of their strengths, so you can't use the hybridisation info to compare it against something like methane.

why don't we see many (any?) other molecules with this bond structure?

Most elements have enough valence electrons to not have to resort to these weaker bonds. For example, carbon already has four valence electrons and is happy to just form four normal 2c2e covalent bonds; but boron only has three valence electrons. Even if it uses these to form three 2c2e bonds, that's only six electrons: not a complete octet.

Sure, it could hope to get a dative bond from somewhere else, but that's contingent on it bumping into a molecule that could give it a dative bond, like $\ce{NH3}$ (see https://en.wikipedia.org/wiki/Ammonia_borane). But if it's trapped a container without any ammonia in sight, then it does the next best thing, which is to try to stretch out the electrons between more bonding partners.

In fact, boron is pretty notorious in this regard: it doesn't have enough electrons to form bonds, but still wants to fill up some kind of 'octet'. The problem of not having enough electrons leads to very interesting bonding patterns in boranes; diborane is merely the tip of the iceberg. For more info, see https://en.wikipedia.org/wiki/Boranes.


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