A) Octahedral, \[s{{p}^{3}}{{d}^{2}}\]
B) Trigonal bipyramidal, \[s{{p}^{3}}d\]
C) Planar triangle, \[s{{p}^{3}}{{d}^{3}}\]
D) Square planar, \[s{{p}^{3}}{{d}^{2}}\]
Correct Answer: A
Solution :
\[Xe{{F}_{4}}\], has octahedral geometry where hybridization of Xe is \[s{{p}^{3}}{{d}^{2}}\].
You need to login to perform this action.
You will be redirected in
3 sec
