Posted by: Mark Foreman | May 21, 2013

The banana bond

Bananas are fun things, those of you who might have thought that I have changed into Sean Banan can relax and breathe easy again. I am here to talk about the banana bond. This is the informal name for a bent bond, the banana bond I want to discuss now is a very bent bond which is found in diborane.

Now after we have seen how to draw the Lewis structure of a borohydride anion we can try to make a similar diagram for diborane using conventional two atom and two electron bonds. This is the best I can do and it is clear that it is a dismal failure, the borons only have six electrons around them and this falls short of the eight electrons which the atoms need to attain a noble gas like full outer shell.

Bad version of diborane, it has only six valence electrons around each boron.

Bad version of diborane, it has only six valence electrons around each boron.

Now what we have in this molecule are two three centred two electron bonds, these are bonds which involve three atoms and only have two electrons in them. With two sp3 and one s orbital we can draw out the three ways the orbitals can be combined to make three molecular orbitals. Here are the three molecular orbitals associated with the three atoms.

The three orbitals in the banana bond

The three orbitals in the banana bond

So after we back the electrons into the two bonding orbitals for the banana bonds, we can now get eight electrons around each boron atom. Thus even for these electron deficient molecules we can still have eight electrons around each boron.

Electrons in a diborane molecule

Electrons in a diborane molecule

We can move onto bigger and better things, for example if we replace the main group boron with a transition metal (such as copper) then to have a noble gas configuration we need 18 electrons around the transition metal. For example we can have a copper complex. How about this complex.

[Cu(BH4)(PPh3)2]

[Cu(BH4)(PPh3)2]

Now I hold the view that in organometallic chemistry that oxidation state can be a bit of fluid, at times it is not clear. It is possible to assign two different oxidation states to a metal in the same complex. Now if we assume that the copper in our complex has an oxidation state of zero then the zero valent copper has 11 electrons, ten of these electrons will stay around the copper in its atomic orbitals while one will enter one of the 3 centre 2 electron bonds. Two electrons from the bridging hydrogens will be in these banana bonds while one of the boron electrons will also be in a banana bond. The bonds to the phosphines contain two electrons from the phosphorus atoms (dative bonds) while the terminal hydrogen to boron bonds are simple normal two centre two electron bonds.

The copper borohydride complex with a zero valent copper and a zero valent boron

The copper borohydride complex with a zero valent copper and a zero valent boron

If we make the complex a copper(I) complex which has a boron in the (-I) oxidation state then the boron will have four electrons to start with. The complex will still have four electrons in the banana bonds, four in the dative bonds from the phosphines and four in the “normal” bonds to the terminal hydrides. Again we have 18 valence electrons around the copper and eight around the boron.

 

The copper borohydride complex with a copper(I) centre in it.

The copper borohydride complex with a copper(I) centre in it.

We will get onto another fun type of bond soon.

 

 

 

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