On the basis of VSEPR theory, match the molecules listed in Column I with their shapes given in Column II. No. Column I No. Column II A ClF₃ P See-saw B BrF₅ Q Pentagonal bipyramidal C SF₄ R Square pyramidal D IF₇ S T-shaped

Step-by-step solution

For ClF₃ , the central atom Cl has 7 valence electrons. It forms 3 bonds with F atoms and has 2 lone pairs. The steric number is 3 + 2 = 5 , which corresponds to trigonal bipyramidal geometry. With 2 lone pairs in equatorial positions, the shape is T-shaped. Thus, A = S . For BrF₅ , the central atom Br has 7 valence electrons. It forms 5 bonds with F atoms and has 1 lone pair. The steric number is 5 + 1 = 6 , which corresponds to octahedral geometry. With 1 lone pair, the shape is square pyramidal. Thus, B = R . Fo