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

The Correct Option is 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.