what is the hybridization in XeOF4 is it sp3d2 or sp3d3
The correct answer and explanation is :
The hybridization in XeOF₄ is sp³d².
Explanation:
To determine the hybridization, we need to consider the structure of XeOF₄ (xenon oxytetrafluoride). The central atom is xenon (Xe), which belongs to Group 18 of the periodic table. Xenon is a noble gas and typically has a complete octet. However, it can form compounds with highly electronegative elements like oxygen and fluorine.
- Electron Configuration of Xenon:
Xenon has the electron configuration of [Kr] 4d¹⁰ 5s² 5p⁶. In XeOF₄, xenon utilizes some of its d-orbitals to form bonds. - Bonding in XeOF₄:
In XeOF₄, xenon is bonded to one oxygen atom (O) and four fluorine atoms (F). The oxygen atom has a double bond with xenon, while each fluorine atom forms a single bond with xenon. - Electron Domains:
The number of electron domains around the xenon atom can be determined from the bonds and lone pairs:
- One double bond with oxygen.
- Four single bonds with fluorine atoms.
- Xenon also has two lone pairs of electrons. This gives a total of 6 electron domains (1 from the O double bond, 4 from F single bonds, and 2 lone pairs).
- Geometry and Hybridization:
With 6 electron domains, the molecular geometry around xenon is octahedral, which suggests sp³d² hybridization. In sp³d² hybridization, one s orbital, three p orbitals, and two d orbitals from the xenon atom mix to form six equivalent hybrid orbitals. These orbitals form the six bonds and accommodate the lone pairs.
Thus, the hybridization in XeOF₄ is sp³d², not sp³d³. The sp³d³ hybridization occurs in cases where there are more than six electron domains (like in compounds with seven bonds), which is not the case for XeOF₄.
In summary, XeOF₄ has sp³d² hybridization, consistent with an octahedral geometry around xenon.