Describe the hybridization state of phosphorus in phosphorus pentabromide.
The correct answer and explanation is:
In phosphorus pentabromide (PBr₅), the hybridization state of phosphorus is sp³d. This hybridization occurs because phosphorus is in the central position and forms five bonds with five bromine atoms.
To explain further, phosphorus has an atomic number of 15, and its electron configuration is [Ne] 3s² 3p³. In the case of PBr₅, phosphorus needs to form five bonds, which requires five unpaired electrons. To achieve this, one of its 3s electrons is promoted to the empty 3d orbital, making five valence electrons available for bonding.
The five orbitals involved in bonding are the one 3s orbital, three 3p orbitals, and one 3d orbital. These five atomic orbitals hybridize to form five sp³d hybrid orbitals. The five sp³d hybrid orbitals arrange themselves in a trigonal bipyramidal geometry to minimize electron repulsion. Three of these orbitals are arranged in a plane, forming 120-degree angles, and the remaining two orbitals are aligned axially, with 90-degree angles between them. This arrangement allows for maximum spatial separation between the bonding pairs of electrons.
Each of the sp³d hybrid orbitals forms a sigma bond with the bromine atoms. The bromine atoms each bring one electron to form a bond with phosphorus, completing its valence shell and giving phosphorus a formal oxidation state of +5 in PBr₅.
The trigonal bipyramidal geometry of PBr₅ is characteristic of molecules where the central atom is sp³d hybridized, and the five bonds around phosphorus are arranged to minimize electron pair repulsion, following the VSEPR theory.