Question

Mention the type of hybridisation in the following compounds:
(a) BrF₅ (b) BrF₃

Hint:

Use the formula: H = (1)/(2)[V + M - C + A] - V: Valence electrons of central atom - M: Number of monovalent atoms (like F) - C, A: Charge of cation/anion For BrF₅: (1)/(2)[7+5] = 6 (sp³d²). For BrF₃: (1)/(2)[7+3] = 5 (sp³d).

Answer 1

Concept: Hybridisation is determined by the number of steric units (sigma bonds + lone pairs) around the central atom. Bromine (Br) is the central atom in both interhalogen compounds and belongs to Group 17, having 7 valence electrons.
(a) BrF₅ (Bromine Pentafluoride)
Counting electrons and steric number Bromine uses 5 of its 7 valence electrons to form 5 sigma bonds with Fluorine atoms. This leaves 2 electrons, which form 1 lone pair. Steric Number = 5 (bonds) + 1 (lone pair) = 6 Hybridisation: A steric number of 6 corresponds to sp³d² hybridisation.
Geometry/Shape: The geometry is octahedral, but due to the lone pair, the molecular shape is square pyramidal.
(b) BrF₃ (Bromine Trifluoride)
Counting electrons and steric number Bromine uses 3 of its 7 valence electrons to form 3 sigma bonds with Fluorine atoms. This leaves 4 electrons, which form 2 lone pairs. Steric Number = 3 (bonds) + 2 (lone pairs) = 5 Hybridisation: A steric number of 5 corresponds to sp³d hybridisation.
Geometry/Shape: The geometry is trigonal bipyramidal, but due to the two lone pairs occupying equatorial positions, the molecular shape is Bent T-shape.