Question

Shape of $XeO{F_4}$ is:
A. Octahedral
B. Square pyramidal
C. Pyramidal
D. T-shaped

Answer 1

At first, the hybridisation state of the $XeO{F_4}$ molecule needs to be determined. This state of hybridisation will determine the shape of the given molecule and hence the shape will be according to the given configuration of the electrons in the orbitals.

Complete answer:
The shape of the molecule $XeO{F_4}$ has the electronic configuration in such a manner so that there is a chemistry of square pyramidal. The internal molecule which is the central atom is $Xe$ and there are four $F$ residues which are associated with the central atom. This forms the basal structure of the $XeO{F_4}$ molecule. The associated $O$ residue forms a double-bonded structure with the $Xe$ residue. Therefore, the internal $Xe$ residue forms six bonds with different molecules. There are $8$ valence electrons in the electronic configuration of $Xe$ residue. Therefore as electrons are already forming bonds, there is a lone pair of electrons which is left without any possible bond formation. This lone pair is present as the trans structure associated with $Xe$ residues. The hybridisation is based on the electron filling of the orbitals. The hybridisation of this chemical molecule $XeO{F_4}$ is $s{p^3}{d^2}$, which means according to the nature of hybridisation the structure is square pyramidal. The pyramidal structure is because of the $O$ residue which is present on a different plane. The resultant structure of the $XeO{F_4}$ is given in the diagram.

Hence, option B is correct.

Note:
The $XeO{F_4}$ molecule is a structure which has the super-octet character. This is because as a result of the covalent interactions of all the atoms with the central $Xe$ atom the number of electrons is more than the natural octet filled level. Hence this molecule does not follow the octet rule.