Hybridization of XeOF4

Explanation:
- The hybridization of XeOF4 can be understood by looking at the electron configuration of Xenon (Xe) and the structure of the molecule.
- Xenon has the atomic number 54, and its electron configuration is [Kr]5s24d105p6. In its ground state, Xenon has three unpaired electrons in its 5p orbital.
- The oxygen atom, on the other hand, has the atomic number 8 and the electronic configuration is 1s22s22p4. It requires two more electrons to attain a stable configuration.
- In XeOF4, the oxygen atom forms a double bond with Xenon, requiring two electrons. This leaves four unpaired electrons on Xenon.
- To accommodate these four electrons, Xenon undergoes hybridization, where the valence orbitals mix to form new hybrid orbitals.
- In the case of XeOF4, Xenon undergoes sp3d2 hybridization.

Steps to determine the hybridization:
1. Count the number of sigma bonds and lone pairs around the central atom (Xenon). In this case, there are 4 sigma bonds and 2 lone pairs.
2. Add these numbers together: 4 + 2 = 6.
3. Determine the hybridization based on the total number obtained:
- 2: sp hybridization
- 3: sp2 hybridization
- 4: sp3 hybridization
- 5: sp3d hybridization
- 6: sp3d2 hybridization

Explanation of sp3d2 hybridization:
- In sp3d2 hybridization, the 5s, 4p, and 2d orbitals of Xenon undergo hybridization to form six sp3d2 hybrid orbitals.
- These hybrid orbitals are arranged in an octahedral geometry around the Xenon atom, with four orbitals oriented in the equatorial plane and two orbitals oriented in the axial positions.
- The oxygen atom forms a double bond with one of the equatorial hybrid orbitals of Xenon, resulting in a linear arrangement.

Summary:
- The hybridization of XeOF4 is sp3d2.
- The Xenon atom undergoes sp3d2 hybridization to accommodate the four unpaired electrons and form six sp3d2 hybrid orbitals.
- The oxygen atom forms a double bond with one of the equatorial hybrid orbitals of Xenon.