$H_{2}O$ is dipolar, whereas $BeF_{2}$ is not. It is because:

The polarity of a molecule depends on both its bond dipoles and its spatial arrangement (molecular geometry). According to the sources:

1. $H_{2}O$ (Water): The central oxygen atom has two bonding pairs and two lone pairs of electrons ($AB_{2}E_{2}$ type). According to VSEPR theory, these four electron pairs are arranged tetrahedrally, but the presence of lone pairs results in a bent or angular geometry with a bond angle of 104.5° . Because of this angular shape, the individual O–H bond dipoles do not cancel each other out, resulting in a net dipole moment of 1.85 D .
2. $BeF_{2}$ (Beryllium Fluoride): The central beryllium atom has only two bonding pairs and no lone pairs ($AB_{2}$ type). This results in a linear geometry where the bond angle is 180° . In this linear arrangement, the two equal Be–F bond dipoles point in exactly opposite directions and completely cancel each other out, leading to a net dipole moment of zero .

Thus, the difference in their dipolar nature is due to their different molecular geometries .