VSEPR theory states that electron pairs around a central atom arrange themselves to minimise repulsion. The shape of a molecule is determined by the number of bonding pairs and lone pairs on the central atom. In H_2O, the oxygen atom has two bonding pairs and two lone pairs, giving a total of four electron pairs arranged tetrahedrally. However, the molecular geometry (considering only bond positions) is bent. The two lone pairs on oxygen exert stronger repulsion than bonding pairs, compressing the H-O-H bond angle from the ideal 109.5 degrees to approximately 104.5 degrees. BeCl_2 is a linear molecule (sp hybridisation, 180 degrees) because beryllium has only two bonding pairs and no lone pairs. BF_3 is trigonal planar (sp^2, 120 degrees) as boron has three bonding pairs and no lone pairs. CH_4 is tetrahedral (sp^3, 109.5 degrees) with four equivalent bonding pairs and no lone pairs. This question directly tests NCERT Class 11 VSEPR concepts on lone pair versus bond pair repulsion ordering: lp-lp > lp-bp > bp-bp. Final check: bent geometry implies fewer than 180 degrees, confirmed at 104.5 degrees for water.