Consider the following chemical species: XeF ₂ , ClF ₃ , SF ₄ , XeF ₄ , and BrF ₅ . Let A be the sum of the number of lone pairs of electrons on the central atoms of the species that have sp^3d hybridization, and let B be the sum of the number of lone pairs of electrons on the central atoms of the species that have sp^3d^2 hybridization. The value of A - B is

Step-by-step solution

First, we determine the steric number (SN = number of bond pairs + number of lone pairs) and the hybridization for each molecule. 1. XeF ₂ : Xe has 8 valence electrons. It forms 2 single bonds with F. Lone pairs = (8 - 2) / 2 = 3 . SN = 2 (bond pairs) + 3 (lone pairs) = 5 sp^3d hybridization. 2. ClF ₃ : Cl has 7 valence electrons. It forms 3 single bonds with F. Lone pairs = (7 - 3) / 2 = 2 . SN = 3 (bond pairs) + 2 (lone pairs) = 5 sp^3d hybridization. 3. SF ₄ : S has 6 valence electrons. It forms 4 single bonds w