To determine the hybridization of XeF6 (Xenon hexafluoride), we need to consider its molecular geometry and the number of electron pairs around the central atom.
- Determine the total number of valence electrons: Xenon (Xe) belongs to Group 18 (or 8A) of the periodic table and has eight valence electrons. Fluorine (F) has seven valence electrons. Since there are six fluorine atoms, the total number of valence electrons is:
8 (Xe) + 6 (F) = 8 + 6 × 7 = 50
Determine the molecular geometry: XeF6 has an octahedral geometry, where the central xenon atom is surrounded by six fluorine atoms arranged in a symmetrical manner around it.
Determine the electron pairs: In an octahedral arrangement, there are six bonding pairs (Xe-F) and no lone pairs on the central atom.
Determine the hybridization: The number of bonding pairs determines the hybridization. In this case, XeF6 has six bonding pairs, which corresponds to the sp³d² hybridization. The central xenon atom hybridizes its s, p, and d orbitals to form six sp³d² hybrid orbitals, allowing it to accommodate the six bonding pairs from the fluorine atoms.
Therefore, the hybridization of XeF6 is sp³d².