Match the species in List I with their geometry in List II List I List II A. PCl₅ I. Tetrahedral B. BrF₅ II. Square Planar C. BF₄^- III. Trigonal bipyramidal D. [Ni(CN)₄ ] ²⁻ IV. Square pyramidal Choose the correct answer from the options given below:

Step-by-step solution

For PCl ₅ , the central atom P has 5 valence electrons. It forms 5 bond pairs with Cl atoms and has 0 lone pairs. The hybridization is sp^3d , and the geometry is trigonal bipyramidal. (A III) For BrF ₅ , the central atom Br has 7 valence electrons. It forms 5 bond pairs with F atoms and has 1 lone pair. The hybridization is sp^3d^2 , and the geometry is square pyramidal. (B IV) For BF ₄^- , the central atom B has 3 valence electrons plus 1 from the negative charge, giving 4 electrons. It forms 4 bond pairs with F