On the basis of VBT explain the structure of Diborane .

Hint: Valence bond theory explains the chemical bonding of a molecule along with the molecular orbital theory. To answer the question one must be aware of the Valence Bond Theory and how the atomic orbitals combine together to give individual chemical bonds during molecule formation.

Complete answer:
Before we understand the structure of Diborane based on the valence bond theory let us first understand this theory.
Valence bond theory (VBT) assumes that the electrons occupy atomic orbitals of individual atoms within a molecule, and that electrons of one atom are attracted to the nucleus of another atom. As they move closer to each other due to attractive force, they start to repulse once they reach the minimum distance where electron density begins to cause repulsion between the two atoms. In addition to that all bonds are localised bonds formed between two atoms by the donation of an electron from each atom.
Based on the above VBT we will explain the structure of Diborane . The chemical formula of Diborane is $ {B_2}{H_6} $. Borane has an atomic number $ 5 $ and the electronic configuration is $ 1{s^2}2{s^2}2{p^1} $. During bond formation the one electron gets excited to the $ 2p $ orbital and will have four $ s{p^3} $ hybrid orbital, out of which three have one electron each and one is an empty orbital. Each Borane forms two covalent bonds with Hydrogen and is now left with two orbitals, one having an empty orbital and other having one electron. This is known as the banana bond formed due to unpaired electrons orbital and empty orbitals of two $ Borane $ atoms. These orbitals form the two bridging $ B - H - B $ bonds.

Note:
  Diborane is formed by water and Boric acid. It is colourless gas at room temperature with a repulsive, sweet order. A pure Diborane does not react with air or oxygen but in an impure state on mixture with air it forms a very explosive mixture.