Question

The structure of $I{F_7}$ is
a.) Trigonal bipyramidal
b.) Octahedral
c.) Pentagonal bipyramidal
d.) Tetrahedral

Answer 1

Hint: For finding out the hybridization of the given molecule, we can use the hybridization formula, and by referring to the arrangement of the $spdf$ orbitals we can conclude the structure of the given molecular formula.


Complete step by step answer:
First consider finding out that which is the central atom, in this question our central atom is Iodine. Now, since the central atom has been found we will consider the hybridization formula.
The general formula of hybridization is is given as
$H = (V + X + C + A)/2$
Where,
$V = $ The number of valence electrons of the central atom.
$X = $ The number of monovalent atoms attached to the central atom.
$C = $ Total positive charge on the molecule.
$A = $ Total negative charge on the molecule.
Now, for $I{F_7}$ since we know that central atom is iodine and there are $7$ fluorine atom surrounding it, and the number of valence electron of iodine is $7$, then writing all the values;
Total number of valence electrons of iodine; \[V = 7\]
The number of fluorine attached to iodine; $X = 7$
Total positive charge on the molecule; $C = + 0$
Total negative charge on the molecule; $A = - 0$
By putting all the above given values in the hybridization formula we get,
$H = (7 + 7 + 0 - 0)/2$
$H = 14/2$,
$H = 7$,
- Now referring to the table given below,

VALUES OF H	HYBRIDIZATION	STRUCTURE
2	$sp$	Linear
3	$s{p^2}$	Trigonal planar
4	$s{p^3}$	Tetrahedral
5	$s{p^3}d$	Trigonal bipyramidal
6	$s{p^3}{d^2}$	Octahedral
7	$s{p^3}{d^3}$	Pentagonal bipyramidal

By considering the above table we get to know that since our $H$ is $7$ the hybridization of $I{F_7}$ is pentagonal bipyramidal.

 The correct answer is option “C” .

Note: In questions related to structures always remember to apply hybridization formula, and refer to the table given above, it is applicable to all compounds which have hybridization seven or less than it.