Answer : The correct option is, (C) [tex]H_2O[/tex]
Explanation :
Formula used:
[tex]\text{Number of electrons}=\frac{1}{2}[V+N-C+A][/tex]
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
First we have to determine the hybridization of the following molecules.
(a) [tex]O_2[/tex]
This is diatomic molecule and diatomic molecules always form linear shape.
(b) [tex]CO_2[/tex]
[tex]\text{Number of electrons}=\frac{1}{2}\times [4]=2[/tex]
The number of electrons is 2 that means the hybridization will be [tex]sp[/tex] and the electronic geometry of the molecule will be linear.
(c) [tex]H_2O[/tex]
[tex]\text{Number of electrons}=\frac{1}{2}\times [6+2]=4[/tex]
The number of electrons is 4 that means the hybridization will be [tex]sp^3[/tex] and the electronic geometry of the molecule will be tetrahedral.
But as there are two atoms around the central oxygen atom, the third and fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be bent.
(d) [tex]CH_4[/tex]
[tex]\text{Number of electrons}=\frac{1}{2}\times [4+4]=4[/tex]
The number of electrons is 4 that means the hybridization will be [tex]sp^3[/tex] and the electronic geometry of the molecule will be tetrahedral.
Hence, the molecule has bent structure will be, [tex]H_2O[/tex]
