Question

In Water molecule, Oxygen is:
A.sp-hybridized
B. $s{p^3}$ hybridized
C. $s{p^2}$ hybridized
D.None of these

Answer 1

Hint : In valence bond theory, orbital hybridisation (or hybridization) is the concept of combining atomic orbitals into new hybrid orbitals (with different energies, shapes, and other properties than the component atomic orbitals) suitable for electron pairing to form chemical bonds.

Complete Step By Step Answer:
The central oxygen atom in a water molecule is $s{p^3}$ hybridized. Tetrahedral geometry is expected as a result. It does, however, have two bond pairs and two lone pairs. As a result, it is a V-shaped molecule.
The central atom in ${H_2}O$ is oxygen, which has $6$ Valence electrons. Two hydrogen atoms form two bonded pairs, while the remaining two electrons form a lone pair. Thus, oxygen is surrounded by four pairs of electrons and adopts a tetrahedral structure, but it is bent due to the presence of a lone pair.
The state of hybridization of oxygen in ${H_2}O$ is $s{p^3}$ . Hydrogen, on the other hand, is not hybridized. Hydrogen's $1s$ atomic orbitals play a role in the overall geometry of the water molecule. Actually we can say that via $s{p^3}$ -s sigma bonds, two $s{p^3}$ hybridized orbitals of oxygen combine with two - $1s$ orbital(s) of two hydrogen atoms (which will be two in number, of course).
The remaining hybridized oxygen orbital will repel the other two $s{p^3}$ -s bonds, causing the bond angle to “shrink” from ${109^0}\,to\,{104^0}$ . (approx.).
So the correct option is found to be Option B- $s{p^3}$ hybridized.

Note :
The tetrahedral arrangement of two lone pairs and two bonding pairs of electrons in water's molecular structure is consistent. Thus, the oxygen atom is $s{p^3}$ hybridised, with lone pairs occupying two of the hybrid orbitals and bonding pairs occupying the other two.