# Bonds in diborane

1. Dec 6, 2011

### Acut

I was reading about the chemistry of diborane, and there's something puzzling me. John B. Russell, in his "General Chemistry" book states that the boron atom in the diborane "uses all its orbitals in the valence shell forming four hybrid sp3 orbitals, two of which will be used in the bonding with the terminal H atoms, in conventional two centered bonds. The four remaining hybrid orbitals (two from each boron) will superpose to the 1s orbitals of two H atoms to form three-centered molecular orbitals" (my translation)

I was wondering how it is possible for those bonds to be made. In the ground state, each boron atom should have the following configuration:

$\uparrow\downarrow$$\uparrow$
2s2px

When excited, the configuration will be
$\uparrow\uparrow\uparrow$
2s2px2py2pz

(it is supposed to be one up arrow for each 2s, 2px and 2py orbital and an empty 2pz)
And it will hybridize:
$\uparrow\uparrow\uparrow$
sp3

(sorry, I don't know how to format this properly with tex)

I can't see how this sp3 hybridization explains the bonds. The configurations I got suggest that boron would make 3 simple bonds and still accept a lone pair from another atom (if it is supposed to complete the octet) or it would form three simple bonds only (if an incomplete octet is enough to make it stable). Where am I getting wrong? How can hybridization be used in this case?

2. Dec 7, 2011

### DrDu

Yes, you can look at it that way. Only that here it does not accept a lone pair but a bonding pair from a hydrogen from the other BH3 unit.