Select the correct statement regarding ${\mathrm{B}}_2{\mathrm{H}}_6$. 

Boron contains three valence electrons, thus when it undergoes hybridization, it should form three bonds in a molecule (${\mathrm{sp}}^2$) since only the s and p orbitals are utilised, and the final p orbital is left empty.

However, the diborane,${\mathrm{B}}_2{\mathrm{H}}_6$ , has three centred bonds and two electrons per bond. Four hydrogen atoms form a connection with each boron atom. Tetrahedral geometry results from this.

Thus, each boron atom has undergone ${\mathrm{sp}}^3$ hybridization.