Question: The shape of \[Xe{O_2}{F_2}\] molecule is A. trigonal bipyramidal B. Square planer C. tetrahed...

Question

The shape of $Xe{O_2}{F_2}$ molecule is
A. trigonal bipyramidal
B. Square planer
C. tetrahedral
D. See saw

Answer 1

Solution

In order to solve this question, first we must know the molecular geometry and bond angles of Xenon Dioxide Difluoride $\left( {Xe{O_2}{F_2}} \right)$ . Then we will use the molecular geometry of $Xe{O_2}{F_2}$ to find its position and shape so that we get the required answer.

Complete step by step answer:
Firstly, we must know the concept of $Xe{O_2}{F_2}$ which is the molecular formula of Xenon Dioxide Difluoride. In Xenon Dioxide Difluoride, Its components are xenon, oxygen and fluoride.
Xenon will be the central atom which will have 8 valence electrons. The fluoride atom will be the monovalent surrounding atom. The oxygen atom will be the divalent surrounding atom.

In order to do hybridization of $Xe{O_2}{F_2}$ :
We will take the eight valence electrons of Xenon and add 2 monovalent fluorine atoms. The whole sum will be divided by 2 at the end. The hybridization of Xenon (which is the central atom) is $s{p^3}d$ . In Xenon Dioxide Difluoride, there will be $5s{p^3}d$ . There are five electron pairs around the central atom where it will contain 4 bond pairs and 1 lone pair.
Xenon belongs to a noble gas family which means that it does not have to form any bonds to complete its octet.

$Xe{O_2}{F_2}$ molecular geometry and bond angles-
$Xe{O_2}{F_2}$ molecular geometry is originally said to be trigonal bi-pyramidal but due to the presence of lone pair on equatorial position, the actual shape becomes see-saw. The repulsion between the bond pair and lone pair of electrons will be more. Here, fluoride will be axial atoms and oxygen will be equatorial atoms. As for the angles, the $O - Xe - O$ angle will be 105.7 degrees, $O - Xe - F$ will be 91.6 degrees, and $F - Xe - F$ will be 174.7 degrees.

From the above properties of $Xe{O_2}{F_2}$ , we conclude that the shape of $Xe{O_2}{F_2}$ is see-saw. So, the correct answer is “Option D”.

Note: Whenever we get this problem, one can get confused between trigonal bi-pyramidal and see - saw shape as the molecular geometry of $Xe{O_2}{F_2}$ is originally said to be a trigonal bi-pyramid but due to the presence of lone pairs on equatorial positions, the actual shape will be see – saw.