Question

\[{\rm{Xe}}{{\rm{F}}_{\rm{2}}}\] molecule is
A) Square planar
B) Trigonal bipyramidal
C) Trigonal planar
D) Linear

Answer 1

Hint: We know that xenon forms compounds with smaller electronegative atoms like fluorine and oxygen. Some of the compounds of xenon are \[{\rm{Xe}}{{\rm{F}}_{\rm{2}}}\], \[{\rm{Xe}}{{\rm{F}}_4}\], \[{\rm{Xe}}{{\rm{O}}_{\rm{3}}}\] etc. Here, we have to identify the shape of \[{\rm{Xe}}{{\rm{F}}_{\rm{2}}}\].

Complete Step by Step Solution:
Let's understand the hybridization of \[{\rm{Xe}}{{\rm{F}}_{\rm{2}}}\]. The configuration of xenon at the ground state is \[5{s^2}5{p^6}\] . The configuration of xenon changes to \[5{s^2}5{p^5}5{d^1}\]at the excited state. Thus, excited state configuration of xenon, that is, \[5s,5{p^y},5{p^x},5{p^z},5{d^z}\] undergoes hybridization to form 5 \[s{p^3}d\] orbitals. Therefore, xenon difluoride has a hybridization of \[s{p^3}d\].

Let's discuss the structure of xenon difluoride. The formation of two sigma bonds F-Xe-F occurs due to the overlap of two half-filled orbitals of fluorine atoms. The three atomic orbitals which contain lone pairs have no participation in bonding. So, the molecule has two bond pairs and three numbers of lone pairs surrounding it. So, the geometry of the molecule is linear in nature. The shape is such that the positions of lone pairs are in equatorial positions. And the measure of the bond angle is \[180^\circ \].

Fig: Structure of \[{\rm{Xe}}{{\rm{F}}_{\rm{2}}}\]

Therefore, the xenon difluoride molecule has a linear shape.
Hence, option D is right.

Note: It is to be noted that only xenon amongst the noble gases forms many compounds. Mostly they are non-reactive in nature because of the fully filled valence shell. The exception in the case of xenon is due to its bigger size which causes weak attraction of the electrons to the nucleus and helps in compound formation.