Structure of diborane
In view of the trivalency of boron, we would expect it to from a simple hydride, BH3. However, H3 is not known. This is due to the fact that hydrogen atom in BH3 has no electron to from p?- p? back bonding. Thus back boron possesses incomplete octet and HN3 is dimerised to form B2H6 molecule with covalent and three centre bond. The simplest boron hydride is diborane, B2H6.
Electrons are required for the formation of conventional covalent bond structure whereas in diborane, three are only 12 valence electrons, three from each boron atoms. Thus B2H6 is an electron drficient compound. The geometry of B2H6 can therefore be explained as follows.
The four terminal hydrogen atoms and two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms. Each boron atom forms four bonds even though it has only three electrons. The terminal B-H bonds are regular bonds but the bridge B-H bonds are different. Each bridge hydrogen is bonded to the two boron atoms only by sharing of two electrons. Such covalent bond is called three centre electron pair bond or a multi centre bond or banana bond.
The positive charge on the two H nuclei suffers strong electrostatic repulsion due to close proximity of the bridged H atoms. In order to minimize this repulsion, the orbitals of bridged H atoms are bent away from each other in the middle.