Introduction:
In chemistry, hybridization is the process of combining atomic orbitals to form hybrid orbitals that are suitable for the bonding of atoms in molecules. The hybridization of Xef2 is determined by combining the atomic orbitals of Xenon and Fluorine.

Steps to Find Hybridization of Xef2:

1. Determine the number of valence electrons in Xe atom:
- Xenon (Xe) belongs to the noble gas family and has 8 valence electrons.

2. Determine the number of valence electrons in F atoms:
- Fluorine (F) has 7 valence electrons each.

3. Calculate the total number of valence electrons in the molecule:
- Xe has 8 valence electrons, and there are two F atoms, each with 7 valence electrons. Therefore, the total number of valence electrons is 8 + 2(7) = 22.

4. Determine the molecular geometry of Xef2:
- The molecular geometry of Xef2 is linear, with the two F atoms arranged in a straight line on opposite sides of the Xe atom.

5. Determine the hybridization of Xef2:
- To determine the hybridization of Xe in Xef2, we need to count the number of electron pairs around the central atom. In Xef2, there are two electron pairs around the central Xe atom. Therefore, the hybridization of Xe in Xef2 is sp.

Conclusion:
In conclusion, the hybridization of Xef2 is sp. This is because there are two electron pairs around the central Xe atom, making it necessary to combine one s and one p orbital to form two sp hybrid orbitals.

Attached ions + no. of lone pairs
so that, would be 2+6e that is 2+3lp , hence it is sp3d2