Xeof2 Hybridization

The hybridization of XeOF2 can be determined by examining the electronic and molecular geometry of the molecule. Xe has a total of 8 valence electrons, while O has 6 and F has 7, making the total number of valence electrons in XeOF2 equal to 28.

Using VSEPR theory, we can predict that the Xe atom in XeOF2 adopts a linear electron pair geometry due to the presence of two lone pairs and two bonding pairs of electrons around it. This results in sp hybridization of the Xe atom, with the two sp hybrid orbitals oriented in opposite directions along the molecular axis.

Meanwhile, the oxygen atom in XeOF2 is sp3 hybridized, forming four hybrid orbitals, one of which forms a sigma bond with the Xe atom and the remaining three form sigma bonds with the fluorine atoms. The fluorine atoms, being electronegative, contribute their unhybridized p orbitals to form three pi bonds with the oxygen atom.

Therefore, the hybridization of Xe in XeOF2 is sp, while the hybridization of O is sp3.