Question

Boric acid ${H_3}B{O_3}$ and $B{F_3}$ have the same number of electrons. The former is a solid and the latter is gas. The reason is:
A. $B{F_2}$ is a Lewis acid while $B{(OH)_3}$ is not.
B. They have different geometries.
C. In $B{F_3},{F^ - }$ is smaller in size than $O{H^ - }$ in $B(OH)$ .
D. No molecular association is possible in $B{F_3}$ while it is possible in $B{(OH)_3}$ due to hydrogen bonding.

Answer 1

Boric acid, also called hydrogen borate, boracic acid, and orthoboric acid is a weak, monobasic Lewis acid of boron. Boron trifluoride is the inorganic compound with the formula $B{F_3}$ . This pungent colorless toxic gas forms white fumes in moist air. It is a useful Lewis acid and a versatile building block for other boron compounds.

Complete step by step answer:
The three oxygen atoms form a trigonal planar geometry around the boron. The $B - O$ bond length is $136{\text{ }}pm$ and the $O - H$ is $97{\text{ }}pm$ . The molecular point group is ${C_{3h}}$ .Crystalline boric acid consists of layers of $B{\left( {OH} \right)_3}$ molecules held together by hydrogen bonds of length $272{\text{ }}pm$ . The distance between two adjacent layers is $318{\text{ }}pm$ . Due to these hydrogen bonds present in the crystalline structure of boric acid, it remains intact in the solid state.
In the case of boron trifluoride, there is no interaction among the molecules of the $B{F_3}$ with the help of any kind of cohesive forces. Thus, it exists in the gaseous state.
Thus, the correct option is D. No molecular association is possible in $B{F_3}$ while it is possible in $B{(OH)_3}$ due to hydrogen bonding.

Note:
The geometry of a molecule of $B{F_3}$ is trigonal planar. Its ${D_{3h}}$ symmetry confirms the prediction of VSEPR theory. The molecule has no dipole moment by virtue of its high symmetry. $B{F_3}$ is commonly referred to as "electron deficient," a description that is reinforced by its exothermic reactivity toward Lewis bases.