Question

Find out the shape and hybridization of the following molecules ($PC{l}_5$,$N{H}_3$, $H_{2}O$, $S{F}_6$)

Answer 1

Hint: Shape and hybridization of a molecule can be calculated by using some rules i.e. first write the Lewis structure then calculate the number of sigma bonds and lone pair then determine the steric number (steric number is composed of number of lone pairs and sigma bonds) and then finally assign the hybridization and shape of the molecule.

Complete step by step answer: Firstly, we will learn about the concept of hybridization, which can be defined as the mixing of orbitals of the same energy to give new degenerate orbitals.
In $PC{l}_5$ , the central atom phosphorus contains five valence electrons and each electron is shared by chlorine atoms which results in five sigma bonds and no lone pairs. On the basis of sigma bonds and lone pairs, the steric number will be 5. Hence, $PC{l}_5$ molecule has $s{p}^{3}d$ hybridization and trigonal bipyramidal shape.
In $N{H}_3$ , the central atom nitrogen contains five valence electrons and only three electrons are shared by hydrogen atoms which results in three sigma bonds and one lone pairs (from the remaining two electrons). On the basis of sigma bonds and lone pairs, the steric number will be 4. Hence, $N{H}_3$ molecule has $s{p}^3$ hybridization and shape will be trigonal planar.
In $H_{2}O$ , the central atom oxygen contains six valence electrons and only two electrons are shared by hydrogen atoms which results in two sigma bonds and two lone pairs (remaining 4 electrons will be present as lone pair). On the basis of sigma bonds and lone pairs, the steric number will be 4. Hence, $H_{2}O$ molecule has $s{p}^3$ hybridization and shape will be bent.
In $S{F}_6$ , the central atom sulphur contains six valence electrons and each electron is shared by fluorine atoms which results in six sigma bonds and no lone pairs. On the basis of sigma bonds and lone pairs, the steric number will be 6. Hence, $S{F}_6$ molecule has $s{p}^3{d}^2$ hybridization and the shape will be octahedral.

Note:
We need to remember that the shape does not count lone pairs but the shape is an outcome of a lone pair. For example, in $N{H}_3$ molecule the structure will be tetrahedral but due to the presence of lone pair the shape of ammonia molecule is comes out to be trigonal pyramidal.