Consider the following species: XeF ₂, ClF ₃, SF ₄, PCl ₅, BrF ₅, XeF ₄, I ₃^- Find the sum of the number of lone pairs present on the central atoms of ONLY those species which are sp ^3 d hybridized.

Step-by-step solution

First, determine the hybridization and number of lone pairs on the central atom for each species. The steric number (SN) is given by 1 2 (V + M - C + A) , where V is the number of valence electrons of the central atom, M is the number of monovalent atoms, C is the cationic charge, and A is the anionic charge. The number of lone pairs is SN - Bond Pairs . XeF ₂ : SN = 1 2 (8 + 2) = 5 ( sp ^3 d ). Lone pairs = 5 - 2 = 3 . ClF ₃ : SN = 1 2 (7 + 3) = 5 ( sp ^3 d ). Lone pairs = 5 - 3 = 2 . SF ₄ : SN = 1 2 (6 + 4) = 5 (