Question

Draw the structure of diborane.

Answer 1

Hint: The two atoms of boron left with that of each unpaired electron orbital and empty orbital forms the two bridging (B–H–B) bonds with that of the two 1s hydrogen atoms, is also called the banana bond. Keep this in mind when trying to figure out the resultant structure of diborane.

Step-by-Step Solution:
Let us first analyse the chemical properties of Diborane before moving onto its structural analysis.
Diborane is the chemical compound consisting of boron and hydrogen with the formula \[{{B}_{2}}{{H}_{6}}\]. It is a colourless, pyrophoric gas with a repulsively sweet odour. Synonyms include boro-ethane, boron hydride, and di-boron hexahydride. Diborane is a key boron compound with a variety of applications. It has attracted wide attention for its electronic structure. Its derivatives are useful reagents.
Let us now move onto the structural analysis of Diborane.
The model determined by molecular orbital theory describes the bonds between boron and the terminal hydrogen atoms as conventional 2-center, 2-electron covalent bonds. The bonding between the boron atoms and the bridging hydrogen atoms is, however, different from that in molecules such as hydrocarbons.
Each boron uses two electrons in bonding to the terminal hydrogen atoms, and has one valence electron remaining for additional bonding. The bridging hydrogen atoms provide one electron each. The \[{{B}_{2}}{{H}_{2}}\] ring is held together by four electrons which form two 3-center 2-electron bonds. This type of bond is sometimes called a 'banana bond'.
Therefore mentioned structure is drawn as follows:

Note: The boron atom is known to be \[s{{p}^{3}}\] hybridized and has four hybrid orbitals. From these four hybrid orbitals, three of the orbitals have one electron each, and of which one is an empty orbital.