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 $s{{p}^{3}}{{d}^{2}}$

Answer 1

Answer: (C)

square planar and $s{{p}^{3}}{{d}^{2}}$

Answer 2

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.