Explain: The shape of $H_2O$ is $V$ (angular).

(N/A) The structure of the $H_2O$ molecule can be explained using $sp^3$ hybridization of oxygen and the $VSEPR$ theory.
In $H_2O$,the valence shell electronic configuration of oxygen in the ground state is $2s^2 2p_x^2 2p_y^1 2p_z^1$,having two paired electrons and two half-filled orbitals.
These four orbitals undergo $sp^3$ hybridization,resulting in two half-filled orbitals and two completely filled orbitals.
The two half-filled $sp^3$ orbitals overlap with the $1s$ orbital of $H$ to form two $O-H$ sigma bonds. The two fully filled $sp^3$ orbitals contain lone pairs.
According to the $VSEPR$ theory,the repulsion between lone pairs is greater than the repulsion between bond pairs. Due to this increased repulsion,the two $O-H$ bonds are pushed closer together,resulting in a bond angle of $104.5^{\circ}$. Thus,$H_2O$ adopts a $V$-shape (angular geometry).