Give the structure of `XeOF_2` and state of hybridization of Xe in it.

Explanation

On the basis of VSEPR Theory

Total number of electron pairs =5
Shape of `XeOF_2` On the basis of hybridisation
Valence shell electronic configuration of Xe (Z =54) is ground and excited states are as follows:

Hybridisation involved in `XeOF_2` is `sp^3 d`. Out of the five hybridised orbitals two orbitals are fully filled while three are half filled. Out of these three half filled orbitals two orbitals form two sigma bonds with two F-atoms, while the third half filled orbitals forms d-orbital is involved in dp-p`pi` bonding with O-atom.
Structure of `XeO_2 F_2`
Valence shell electronic configuration of Xe (Z = 54) is ground and excited states are as follows:

Hybridisation involved in `XeO_2 F_2` is `sp^3 d`. Out of the five hybridised orbital, one is fully filled and four are half filled. Of these five half filled orbitals, two orbitals form two `sigma`-bonds with Fluorine and two orbitals form two `sigma`-bonds, with oxygen atoms. The two unhybridized orbitals containing one electron each form two `pi`-bonds with the oxygen atom as shown above. So the geometry of `XeO_2 F_2` is trigonal bipyramidal.