While reading about diborane, I came to know that the two bridging hydrogen atoms in diborane form 3 center 2 electron bonds. On account of repulsion between the two hydrogen nuclei, the delocalised orbitals of bridges are drifted away from each other giving the shape of a banana. Such bonds are called banana bonds. In diborane 3 center 2 electron bonds and banana bonds are same.

What if a compound has only one bridging atom which forms 3 center 2 electron bond. As there is only one bridging atom, it would not face repulsion so it won't form a banana bond.

  • $\begingroup$ In that case, it won't dimerise/polymerise to form 3C-2e- bonds(such type of delocalizations of electrons won't be possible). $\endgroup$
    – user600016
    Nov 3, 2018 at 13:59
  • 1
    $\begingroup$ @thewitness not really, en.wikipedia.org/wiki/Trihydrogen_cation $\endgroup$
    – Mithoron
    Nov 3, 2018 at 16:12

1 Answer 1


The phrase "banana bond" is usually reserved for situations where steric conditions prevent the orbitals of two atoms from aligning properly so that the bond appears "bent" and is weaker than an optimally aligned bond would be. Cyclopropane is the poster child for such bonds, diborane is not because this description of banana involves only two atomic centers.

It's a different concept from three-center two-electron bonds, where typically the bonds are well aligned but the three atoms bonded together are not always collinear. This is what diborane does. The atoms in the diborane bonds are not collinear because we need two such bonds, not because either bond suffers from imperfect overlap.


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