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Hint: Boric acid (${{H}_{3}}B{{O}_{3}}$) in solid crystalline state has two dimensional layered structure. Hydrogen bond is a weak bond between an electronegative atom like N, O, F and a hydrogen atom bonded to another electronegative atom.
Complete answer:
To understand the structure of boric acid (${{H}_{3}}B{{O}_{3}}$), consider the ground state electronic configuration of boron.
B (in ground state): $1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}$
One electron from 2s moves to 2p-orbital in the excited state and the electronic configuration becomes: $1{{s}^{2}}2{{s}^{1}}2{{p}^{2}}$
One s and two p-orbitals are now available to bond with three oxygen atoms. Therefore, the hybridization of the central atom B is$s{{p}^{2}}$.
Now, since only one electron of each oxygen atom is used in the bond formation. The borate ion formed is a trivalent ion,$BO_{3}^{-3}$. Due to $s{{p}^{2}}$ the geometry of $BO_{3}^{-3}$ is trigonal planar. In the triangular $BO_{3}^{-3}$, three oxygen atoms are present at the corners of an equilateral triangle.
Boron is attached to three oxygen atoms and each oxygen is bonded to one hydrogen. Each such unit is ${{H}_{3}}B{{O}_{3}}$. These units are bonded together through hydrogen bonds to give a two-dimensional layered structure of boric acid.
So, the correct answer is “Option D”.
Additional Information: Boric acid or orthoboric acid is a weak monobasic acid. It does not donate protons but accepts a pair of electrons ${{H}_{3}}B{{O}_{3}}$ from and thus behaves as a Lewis acid. It has a soapy touch and is moderately soluble in water.
Note: Bonds between boron and oxygen atoms are covalent. One ${{H}_{3}}B{{O}_{3}}$ unit is joined to another through hydrogen bonding between oxygen of one ${{H}_{3}}B{{O}_{3}}$ to the hydrogen of another ${{H}_{3}}B{{O}_{3}}$.
Complete answer:
To understand the structure of boric acid (${{H}_{3}}B{{O}_{3}}$), consider the ground state electronic configuration of boron.
B (in ground state): $1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}$
One electron from 2s moves to 2p-orbital in the excited state and the electronic configuration becomes: $1{{s}^{2}}2{{s}^{1}}2{{p}^{2}}$
One s and two p-orbitals are now available to bond with three oxygen atoms. Therefore, the hybridization of the central atom B is$s{{p}^{2}}$.
Now, since only one electron of each oxygen atom is used in the bond formation. The borate ion formed is a trivalent ion,$BO_{3}^{-3}$. Due to $s{{p}^{2}}$ the geometry of $BO_{3}^{-3}$ is trigonal planar. In the triangular $BO_{3}^{-3}$, three oxygen atoms are present at the corners of an equilateral triangle.
Boron is attached to three oxygen atoms and each oxygen is bonded to one hydrogen. Each such unit is ${{H}_{3}}B{{O}_{3}}$. These units are bonded together through hydrogen bonds to give a two-dimensional layered structure of boric acid.
So, the correct answer is “Option D”.
Additional Information: Boric acid or orthoboric acid is a weak monobasic acid. It does not donate protons but accepts a pair of electrons ${{H}_{3}}B{{O}_{3}}$ from and thus behaves as a Lewis acid. It has a soapy touch and is moderately soluble in water.
Note: Bonds between boron and oxygen atoms are covalent. One ${{H}_{3}}B{{O}_{3}}$ unit is joined to another through hydrogen bonding between oxygen of one ${{H}_{3}}B{{O}_{3}}$ to the hydrogen of another ${{H}_{3}}B{{O}_{3}}$.
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