Question

What is the hybridization and shape for an XeF6 2+ molecule?

Answer 1

Answer:

Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²

Explanation:

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²

Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48

Ca. #Substrate e⁻ = 6F = 6(8) = 48

#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs

#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs

BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron

Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²    

XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)