Question

Hybridisation state of $Xe$ in $Xe{{F}_{2}}$, $Xe{{F}_{4}}$, and $Xe{{F}_{6}}$ respectively are:
(A) $s{{p}^{2}}$$s{{p}^{3}}d$$s{{p}^{3}}{{d}^{2}}$
(B) $s{{p}^{3}}d,s{{p}^{3}}{{d}^{2}},s{{p}^{3}}{{d}^{3}}$
(C) $s{{p}^{3}}{{d}^{2}},s{{p}^{3}}d,s{{p}^{3}}{{d}^{3}}$
(D) $s{{p}^{2}},s{{p}^{3}},s{{p}^{3}}d$

Answer 1

Hint:The term hybridisation, as the name suggests, involves mixing up of the individual orbitals in a way that all the orbitals would acquire equal energy at the end of it. The hybridisation of a certain molecule depends on the number of lone pairs as well as bond pairs that molecule has.

Complete step-by-step answer:Hybridization can be defined as the concept where the mixing of two atomic orbitals takes place, along with the same levels of energy in order to give a degenerated type of orbitals which are now. This intermixing of the orbitals is based on quantum mechanics. The atomic orbitals which have the same energy level could only take an active part in the process of hybridization and both fully filled as well as half-filled orbitals could also take a part in the same process, provided each of them have an equal amount of energy.Throughout the process of hybridization or the intermixing of the orbitals, the atomic orbitals having similar energy are mixed with each other such as the mixing of two of the ‘$s$’ orbitals or two ‘$p$’ orbitals together or mixing of an ‘$s$’ orbital along with a ‘$p$’ orbital or ‘$s$’ orbital along with a ‘$d$’ orbital.Now, we will discuss some of the hybridisations, which are being mentioned in the question.$s{{p}^{3}}d$ Hybridization, is a type which involves the mixing of $1-s$ orbital, $3-p$ orbitals along with $1-d$ orbital in order to form $5$ $s{{p}^{3}}d$ hybridized orbitals of equal energy. They have trigonal bi pyramidal geometry. The mixture of $s$, $p$and $d$ orbital forms trigonal bi pyramidal symmetry.For instance, Hybridization in Phosphorus pentachloride ($PC{{l}_{5}}$), shows $s{{p}^{3}}d$ hybridisation.\[s{{p}^{3}}{{d}^{2~}}\] hybridization is a type which occurs because of the mixing of $1-s$ orbital, $3-p$ orbitals along with $2-d$ orbital, each of which undergo intermixing with each other in order to form six identical hybrid orbitals having same energy. These six orbitals are all pointed towards the corners of an imaginary octahedron.\[s{{p}^{3}}{{d}^{3}}\] hybridisation is a type of hybridisation where one $s$, three $p$ and three $d$ atomic orbitals in order to form seven individual, but equivalent hybrid orbitals having equal energy. Now, in order to find out the hybridisation of a molecule, we need to know the number of lone pairs and the number of bond pairs, present in the molecule. In the case of $Xe{{F}_{2}}$, the number of lone pairs are three and the number of bond pairs are two. So, if we add those, we will get the total number of the bonds, which is five. So the hybridisation corresponding to five bonds is $s{{p}^{3}}d$.Second molecule is $Xe{{F}_{4}}$, number of bond pair is four and lone pair is two, which becomes six, on addition and we know that this corresponds to the \[s{{p}^{3}}{{d}^{2~}}\] hybridization.
Now, for $Xe{{F}_{6}}$, number of bond pairs is six, and lone pairs is one, which becomes seven on addition, which corresponds to \[s{{p}^{3}}{{d}^{3}}\] hybridisation.

So, the most appropriate answer would be option (B).

Note:The hybridisation of $Xe{{F}_{2}}$ is $s{{p}^{3}}d$ as it has four as total numbers of bond pair and lone pairs. In case of $Xe{{F}_{4}}$ the hybridisation is \[s{{p}^{3}}{{d}^{2~}}\] as it has six as the sum of total number of bond pairs and lone pairs. And in case of $Xe{{F}_{6}}$ it is \[s{{p}^{3}}{{d}^{3}}\] because of the same reason. After hybridisation, each of the hybrid orbitals have equal energy, and are degenerate in nature.