Identify the species in which the central atom undergoes sp^3d^2 hybridisation and possesses exactly one lone pair of electrons.

Step-by-step solution

For XeF₄ : Central atom Xe has 8 valence electrons. Number of bond pairs = 4 Number of lone pairs = 8 - 4 2 = 2 Steric number = 4 + 2 = 6 Hybridisation is sp^3d^2 with 2 lone pairs. For BrF₅ : Central atom Br has 7 valence electrons. Number of bond pairs = 5 Number of lone pairs = 7 - 5 2 = 1 Steric number = 5 + 1 = 6 Hybridisation is sp^3d^2 with exactly 1 lone pair. For PCl₅ : Central atom P has 5 valence electrons. Number of bond pairs = 5 Number of lone pairs = 5 - 5 2 = 0 Steric number = 5 + 0 = 5 Hybridisatio