Final answer:The hybridization of the central atom in XeF5+ is sp³d², which accounts for the five bonding pairs and two lone pairs of electrons, forming a pentagonal bipyramidal electron pair geometry.Therefore, the correct answer is: option B. sp^3d^2Explanation:The hybridization of the central atom in XeF5+ can be determined by examining the electron configuration around xenon (Xe). In addition to the five fluorine atoms bonded to Xe, we must consider that Xe has a positive charge, which means one fewer electron.By applying the VSEPR model, we can consider the presence of one more imaginary atom to explain the arrangement of electron pairs. This gives us a total valuation of AX5E2, which indicates five bonding pairs and two lone pairs around the central atom.The XeF5+ molecule is then considered to have seven regions of electron density forming a pentagonal bipyramidal structure.Based on the VSEPR theory and the number of electron pairs, we expect the hybridization of Xe in XeF5+ to involve one s orbital, three p orbitals, and two d orbitals, resulting in sp³d² hybridization.Rather than each fluoride being equidistant from Xe as in a regular octahedron, the electron pair geometry is distorted due to the lone pairs, forming the pentagonal bipyramidal arrangement....

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