Question

The bond angle of$Xe{F_2}$molecule is:
A.${120^0}$
B.${109^0}25'$
C.${180^0}$
D.${90^0}$

Answer 1

Hint: This question can be answered from the knowledge of the VSEPR theory and the Hybridization of the central atom. According to the VSEPR theory, the arrangement of atoms around the central atom is decided on the basis of the bond pairs and lone pairs present in the central atom and the repulsion between them.

Complete step by step answer:
According to the VSEPR theory, there is electrostatic repulsion between any two electron pairs and thus they tend to push the atomic orbitals as far as possible and the order of this repulsion is:
Lone pair-lone pair $ > $ Lone pair-bond pair $ > $ bond pair- bond pair.
In the trigonal bipyramidal arrangement, as the axial angles are at ${90^0}$while planer angles are at ${120^0}$, so the lone pair occupies the planer position and the fluorine atom occupies the planer position so that there is maximum distance between the lone pair of electrons. Hence the structure of $Xe{F_2}$ is linear and the bond angle of $F - Xe - F$ is ${180^0}$

So, option C is correct.

Note: In the $Xe{F_2}$ molecule, the central atom is $Xe$ whose electronic configuration is: $\left[ {Kr} \right]4{d^{10}}5{s^2}5{p^6}$. So it has 8 electrons in its valence shell.
According to the formula of hybridization:$H = \dfrac{{V + X - C + A}}{2}$, Where, H is the hybridization of the atom, X is the no. of number of monovalent atoms around the central atom, C is the positive charge on the cationic species if the compound is cationic while A is the anionic charge if the compound is anionic. For $Xe{F_2}$, the number of monovalent groups are two and it has no cationic or anionic charge. Hence, $H = \dfrac{{8 + 2 - 0 + 0}}{2}$=$\dfrac{{10}}{2}$ = 5.
So, the hybridization is $s{p^3}{d^2}$using the 5s, three 5p and two of the 5d orbitals. In these 5 hybrid orbitals, two are occupied by the bond pair electrons that form the bond with the fluorine atoms while the other three contain the three lone pairs of electrons. The shape for a $s{p^3}{d^2}$ hybridized orbital is “trigonal bipyramidal”.