Question

Determine the geometries using VSEPR theory. State the approximate bond angle(s) in each molecule and determine whether each molecule is polar or nonpolar.
a.\[BC{l_3}\]
b.\[PC{l_5}\]
c.\[NC{l_3}\]
d.\[S{O_3}\]
e.\[Xe{F_4}\]
f.\[HB{F_2}\]

Answer 1

Hint: According to VSEPR theory, the bonding pair of electrons and lone pair of electrons decide the hybridization and geometry of the molecules. When all the bonds in the molecule are equally distributed, all the dipole moments of the bond cancel out and become non-polar.

Complete answer:
\[BC{l_3}\] :
The Hybridization is \[s{p^2}\] , which has three boron-chlorine bonds and the charge is equally distributed which leads to the cancellation of dipole moment. The geometry is trigonal planar and the bond angle is \[{120^0}\] .
\[PC{l_5}\]
The hybridization is \[s{p^3}d\] , which has five phosphorus-chlorine bonds and the charge is equally distributed which leads to the cancellation of the dipole moment. The geometry is trigonal bipyramidal and the bond angle is \[{90^0}\] and \[{120^0}\]
\[NC{l_3}\]
The hybridization is \[s{p^3}\] , which has three phosphorus-chlorine bonds and has the electronegativity difference between nitrogen and chlorine, the molecule is polar. The geometry is trigonal pyramidal and the bond angle is less than \[{109.5^0}\]
\[S{O_3}\]
The hybridization is \[s{p^2}\] , which has three sulphur-oxygen bonds and the charge is equally distributed which leads to the cancellation of the dipole moment and the molecule becomes non-polar. The geometry is trigonal planar and the bond angle is \[{120^0}\] .
\[Xe{F_4}\]
The hybridization is \[s{p^3}d\] , which has four xenon-fluorine bonds as the four bonds are in exactly opposite to each other, the molecule becomes non-polar with the geometry of square planar with bond angle of \[{90^0}\]
\[HB{F_2}\]
The Hybridization is \[s{p^2}\] , which has two boron-fluorine bonds and one boron-hydrogen bond and the charge is not equally distributed, it is a polar molecule. The geometry is trigonal planar and the bond angle is \[{120^0}\]

Note:
When the hybridization is \[s{p^3}d\] , the shape is trigonal bipyramidal, but due to the presence of one lone pair of electrons on Xenon, the geometry will be square planar. The electron geometry is trigonal bipyramidal and the molecular geometry is square planar.