A) octahedral
B) trigonal bipyramidal
C) tetrahedral
D) pentagonal bipyramidal
Correct Answer: D
Solution :
Number of hybrid orbitals, \[H=\frac{1}{2}[V+M+A-C]\](where, V = number of valence electrons of central atom) M = number of monovalent atoms A = negative charge C = positive charge] \[\therefore \] \[H=\frac{1}{2}[7+7+0-0]=7\] Thus, the hybridisation of\[\text{I}{{\text{F}}_{\text{7}}}\]is \[s{{p}^{3}}{{d}^{3}}\]and its geometry is pentagonal bipyramidal.
You need to login to perform this action.
You will be redirected in
3 sec
