Final answer:
The hybridization of bromine (Br) in bromine pentafluoride (BrF5) is sp3d2, as it needs to accommodate five bonding pairs and one lone pair for its square pyramidal geometry.
Step-by-step explanation:
Understanding the Hybridization of Br in BrF5
The question involves determining the hybridization of bromine (Br) in the molecule bromine pentafluoride (BrF5). In BrF5, the bromine atom is surrounded by five fluorine atoms and has one lone pair, making it an AX5E system according to the VSEPR model, with a total of six electron pairs. The molecule exhibits a square pyramidal geometry. To accommodate both bonding pairs and the lone pair, the valence shell electron pair repulsion (VSEPR) theory predicts that the central atom would have a hybridization that allows for five bonding domains and one lone pair.
When we take into account these electron domains, we see that the bromine atom must undergo sp3d2 hybridization. This involves the mixing of one s orbital, three p orbitals, and two d orbitals to create six sp3d2 hybrid orbitals. Out of these six hybrid orbitals, five are used for forming bonds with fluorine atoms, and one houses the lone pair of electrons. The arrangement of these orbitals leads to the unique molecular geometry observed in BrF5. Therefore, the correct answer to the hybridization of Br in BrF5 is E, sp3d2.