Answer:
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.
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