Match the species in List I with their geometry in List II List I List II A. PCl_5 I. Tetrahedral B. BrF_5 II. Square Planar C. BF_4^- III. Trigonal bipyramidal D. [Ni(CN)_4]^ 2- IV. Square pyramidal Choose the correct answer from the options given below:
C. A-III, B-IV, C-I, D-II
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 atoms and has 0 lone pairs. The hybridization is sp^3, and the geometry is tetrahedral. (C I) For [ Ni ( CN )_4]^ 2- , the central metal ion Ni ^ 2+ has a 3d^8 electronic configuration. Since CN ^- is a strong field ligand, it causes the pairing of electrons in the 3d orbitals. The hybridization is dsp^2, and the geometry is square planar. (D II) The correct matching is A-III, B-IV, C-I, D-II. Answer: A-III, B-IV, C-I, D-II
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