question_answer

The type of hybridization in \[Xe{{F}_{4}}\] is:

A) \[ds{{p}^{2}}\]                                 

B) \[s{{p}^{3}}d\]

C) \[s{{p}^{3}}{{d}^{2}}\]                  

D) \[s{{p}^{3}}{{d}^{3}}\]

Correct Answer: C

Solution :

Electronic configuration of\[Xe\]in ground state: Electronic configuration of\[Xe\]in excited state: Electronic configuration of\[Xe\]in\[Xe{{F}_{4}}\]: Note: The expected geometry of \[Xe{{F}_{4}}\] is octahedral. On account of the fact that \[lp-lp\] repulsion \[>lp-bp\] repulsion, there is some distortion in the shape of the molecule. Thus \[Xe{{F}_{4}}\] has distorted octahedral geometry with two lone pair of electrons. In other words, it has a square planar geometry.