Question

The molecule $B{F_3}$ has a similar geometrical structure with which of the following molecules?
A) $BeC{l_2}$
B) $N{H_3}$
C) $PC{l_5}$
D) None of the above.

Answer 1

Find out the number of valence electrons of Boron and check if there are any lone pairs or unpaired electrons are present around it after bonding with three Fluorine atoms. The hybridization and shape of the molecule can be found in this manner.

Complete answer: In the given molecule $B{F_3}$ the central atom Boron has three valance electrons. These three electrons bond with three Fluorine atoms so there are no lone pairs but Boron has one vacant orbital.
Therefore they hybridization of $B{F_3}$ is $s{p^2}$
We know that the geometrical shape associated with $s{p^2}$ hybridization is Trigonal planar.
Hence the geometrical shape of $B{F_3}$ is Trigonal planar.
Similarly, we can find the geometrical shape for $BeC{l_2}$ . The hybridization of $BeC{l_2}$ is $sp$ and hence the shape associated with it is Linear.
Since Nitrogen in $N{H_3}$ contains a lone pair its hybridization becomes $s{p^3}$ and therefore its geometrical shape is trigonal pyramidal.
In $PC{l_5}$ we can see that there are five bonds around Phosphorous and therefore its hybridization is $s{p^3}d$ and hence its geometrical shape would be Trigonal bipyramidal.
Since none of the given molecules have the same structure as $B{F_3}$ the correct answer would be option D i.e none of the above.

Hence the correct option is(D).

Note: To find the geometrical shape of the given molecule, we have to find its hybridization. The number of bonds and any presence of lone pairs defines the hybridization of the molecule and hence its shape. This accounts for the trigonal planar shape of $B{F_3}$ .