How will you account for the $104.5^{\circ}$ bond angle in water?

(N/A) In water $(H_2O)$,the oxygen atom undergoes $sp^3$ hybridization,which theoretically predicts a tetrahedral geometry with a bond angle of $109^{\circ} 28^{\prime}$.
However,the oxygen atom in water is bonded to two hydrogen atoms and also possesses two lone pairs of electrons.
According to the $VSEPR$ theory,the repulsion between lone pair-lone pair $(lp-lp)$ is stronger than the repulsion between lone pair-bond pair $(lp-bp)$,which in turn is stronger than bond pair-bond pair $(bp-bp)$ repulsion.
Due to the strong repulsion exerted by the two lone pairs on the two $O-H$ bond pairs,the bond angle is compressed from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $104.5^{\circ}$.