Final answer:
The hybridization of the central atom in the phosphorus tetrabromate (PBr4-) anion is D. sp3d, because phosphorus has four bonds to bromine and one lone pair, indicating a trigonal bipyramidal geometry.
Step-by-step explanation:
The hybridization of the central atom in the phosphorus tetrabromate (PBr4-) anion is d. sp3d.
In PBr4-, phosphorus (P) is the central atom. The electron configuration of phosphorus is 1s2 2s2 2p6 3s2 3p3. To determine the hybridization, we count the number of regions of electron density around the central atom, which in this case is 5. This suggests that phosphorus is sp3d hybridized, meaning it will have 5 hybrid orbitals formed by the combination of one s, three p, and one d orbital.
These hybrid orbitals will result in a trigonal bipyramidal geometry, with the 3p orbitals pointing towards the equatorial positions and the 3d orbital pointing towards one of the axial positions. The four bromine atoms will be bonded to the phosphorus atom through these hybrid orbitals, resulting in a negatively charged tetrabromate ion (PBr4-).
The hybridization of the central atom in the phosphorus tetrabromate (PBr4-) anion can be determined by examining the number of electron pairs around the central phosphorus atom. Phosphorus in this ion has four bonds to bromine atoms and one additional lone pair due to the negative charge, making a total of five electron pairs. This arrangement suggests a trigonal bipyramidal geometry, which is consistent with sp3d hybridization. Therefore, the correct answer is D. sp3d. Similar hybridization occurs in other molecules such as phosphorus pentachloride (PCl5), which also exhibits sp3d hybridization to accommodate five pairs of valence electrons around the central phosphorus atom.