• # question_answer 27) Explain the structures of diborane and boric acid.

Structure of diborane: Diborane is an example of electron deficient compound. Boron atom has three half-filled orbitals in excited state, i.e., it can link with three hydrogen atoms. Thus, while each boron atom in diborane can link to itself three hydrogen atoms, there are no electrons left to form a bond between two boron atoms as shown below: Dilthey, in 1921, proposed a bridge structure for diborane. Four hydrogen atoms, two on the left and two on the right, known as terminal hydrogens and two boron atoms lie in the same plane. Two hydrogen atoms forming bridges, one above and other below, lie in a plane perpendicular to the rest of the molecule. This structure shows that there are two types of hydrogen atoms-terminal and bridging. Four terminal hydrogen atoms can easily be replaced by methyl groups but when two bridging hydrogen atoms are attacked, the molecule is ruptured. According to molecular orbital theory, each of the two boron atoms is in $s{{p}^{3}}$-hybrid state. Of the four hybrid orbitals, three have one electron each while the fourth is empty. Two of the four orbitals of each of the boron atoms overlap with two terminal hydrogen atoms forming two normal B-H $\sigma$-bonds. One of the remaining hybrid orbital (either filled or empty) of one of the boron atoms, k-orbital of hydrogen atom(bridge atom) and one of the hybrid orbital of the other boron atom, overlap to form a delocalized orbital covering the three nuclei with a pair of electrons. Such a bond is known as three centered two electrons bonds. Similar overlapping occurs in one hydrogen atom(bridging) and fourth hybrid orbital of each boron atom. Thus, the formation of diborane molecule can be depicted as shown in the figures. On account of repulsion between the two hydrogen nuclei, the delocalised orbitals of bridges are drifted away from each other giving the shape of a banana. The three centre two electrons bonds are also known as banana bonds. Structure of boric acid The electronic configuration of boron in the ground state is$1{{s}^{2}},2{{s}^{2}},2{{p}^{1}}$. In the excited state, one of the 2s-electrons is promoted to the vacant $2p$-orbital. The three half-filled orbitals undergo $s{{p}^{2}}$-hybridization to give three hybridized orbitals. Each one of the hybridized orbitals overlaps with 2p-orbital of${{O}^{-}}$forming $B-{{O}^{-}}$bonds. In boric acid, planar $BO_{3}^{3-}$ units are joined by hydrogen bonds to give a layer structure as shown in the figure given ahead.