Question

The shape of $Xe{F}_4$ molecule and hybridisation of xenon in it are

• A. tetrahedral and $s{p}^3$
• B. square planar and $ds{p}^2$
• C. square planar and $s{p}^3{d}^2$
• D. octahedral and $s{p}^3{d}^2$
• E. octahedral and $d^{2}s{p}^3$

Answer 1

Answer: (C)

In $Xe{F}_4,$ the central atom, $Xe,$ has eight electrons in its outermost shell. Out of these four are used for forming four a bonds with F and four remain as lone pairs. $\therefore Xe{F}_4\Rightarrow 4\sigma$ bonds + 2 lone pairs $\Rightarrow$ 6 hybridized orbitals, ie, $s{p}^3{d}^2$ hybridisation Since, two lone pairs of electrons are present, the geometry of $Xe{F}_4$ becomes square planar from octahedral.