Question

The correct geometry and hybridisation for $Xe{F}_4$ are:

• A. octachedral, $s{p}^3{d}^2$
• B. trignoal bipyramidal, $s{p}^{3}d$
• C. planer triangle. $s{p}^3{d}^3$
• D. square planer, $s{p}^3{d}^2$

Answer 1

The correct option is A octachedral, $s{p}^3{d}^2$

Xenon being a noble gas has 8 valance electrons. There are 4 Flourine atoms to be bonded with Xe, each of single bond. After subtracting the bonded number of electrons from the total number of atoms in the valence shell we will get the non bonded or lone pair of atoms surrounding the central atom Xe. so the lone pair of electrons surrounding Xe is

$8–4=4$ ( ie, 2 pairs)

Thus, molecule has 4 bond pairs and 2 lone pairs in the molecule. By adding the electron pairs surrounding the atom,

$4+2=6$ , where 6 represents octahedral geometry

so the geometry of the molecule is octahedral $s{p}^3{d}^2$