Question

The hybrid orbitals used by central atom in \[{\rm{BeC}}{{\rm{l}}_{\rm{2}}}\] , \[{\rm{BC}}{{\rm{l}}_{\rm{3}}}\] and \[{\rm{CC}}{{\rm{l}}_{\rm{4}}}\]molecules are respectively
A) \[s{p^2},s{p^3}\] and \[sp\]
B) \[sp,s{p^2}\] and \[s{p^3}\]
C) \[s{p^2},s{p^3}\] and \[s{p^2}\]
D) \[s{p^2},sp\] and \[s{p^3}\]

Answer 1

Hint: According to the VSEPR theory, the hybridization of a central atom is to be identified by counting the groups around the atom. The surrounding groups include the atoms bonded to the atom and the lone pairs of the atom.

Complete step by step solution:If the count of surrounding groups around an atom is 4, then the central atom’s hybridization is \[s{p^3}\] .
If the count is 3, then the central atom’s hybridization is \[s{p^2}\] . The count of total groups is 2, then the hybridization is \[sp\] .
Let’s find out hybridization in \[{\rm{BeC}}{{\rm{l}}_{\rm{2}}}\]. Two chlorine atoms form bonds with the beryllium atom. And there is no lone pair. So, hybridization in \[{\rm{BeC}}{{\rm{l}}_{\rm{2}}}\]is \[sp\] . And the molecule has a linear shape.
Now, we have to find out hybridization \[{\rm{BC}}{{\rm{l}}_{\rm{3}}}\]. Here, three chlorine atoms and no lone pair are present. So, the count of groups is three, that means, the hybridization of the molecule is \[s{p^2}\] . And the molecule has the trigonal planar shape.
 In\[{\rm{CC}}{{\rm{l}}_{\rm{4}}}\], four chlorine atoms form bonds with the carbon atom. And there is no lone pair present. Therefore, the count of groups is four and the molecule has the hybridization of \[s{p^3}\] . And the shape of the molecule is tetrahedral.

Therefore, option B is right.

Note: The VSEPR theory states that repulsion of lone pair-lone pair is highest. And the bond pair-bond pair is lowest. And the repulsion of bond pair-lone pair is in the intermediate between the two. Therefore, the presence of lone pairs in an atom gives different shapes.