Formation of water (H2O) molecule on the basis of sp3 hybridization :
i. Water molecule (H2O) has one oxygen atom and two hydrogen atoms.
ii. The ground state electronic configuration of oxygen (Z = 8) is 1s2 2s2 \(2p_x^2\) \(2p_y^1\) \(2p_z^1\).
Electronic configuration of oxygen :
iii. The ground state electronic configuration explains the observed valency of oxygen in H2O molecule which is 2.
iv. The 2s, 2px, 2py and 2pz orbitals of oxygen atom mix and recast to form four sp3 hybrid orbitals of equivalent energy.
These orbitals are tetrahedrally oriented in space.
Two of the sp hybrid orbitals contain lone pair of electrons.
v. Two half-filled sp3 hybrid orbitals of O atom overlap axially with half-filled 1s orbitals of two different hydrogen atoms to form two O-H (sp3 -s) sigma covalent bonds.
vi. Since, there are two lone pairs of electrons in two of the sp hybrid orbitals of oxygen, there is repulsion between lone pair and bonding pair of electrons. As a result, the HO-H bond angle is reduced from regular tetrahedral angle 109°28′ to 104°35′. The geometry of H2O molecule is angular or V shaped.
Diagram :
