Question

Question: What are the types of Hybridization of Iodine in interhalogen compounds $I{{F}_{3}}$,$I{{F}_{5}}$and $I{{F}_{7}}$, respectively?

Answer 1

Hint: question is based on VSEPR theory. To find out the Hybridization in any compound we will use the following relation: Degree of hybridization= number of sigma bonds number of lone pairs

Complete answer: electronic configuration of Iodine is \[[Kr]4{{d}^{10}}5{{S}^{2}}5{{P}^{5}}\]
So, it is having 7 electrons in its valence shell and 5d orbitals are vacant, so it can change its valency by transferring electrons into vacant 5d orbitals.
$I{{F}_{3}}$: as Fluorine always shows 1 valency, therefore Iodine is forming three sigma bonds with Fluorine. To form 3 sigma bonds it needs, 3 half filled orbitals. For that it will transfer one of its 5P electrons into vacant d-orbital.
Electronic configuration of valence shell of iodine after this electron transferring will be
\[5{{S}^{2}}5{{P}^{4}}5{{d}^{1}}\]
As you can see, 5d orbital is and out of 3 orbitals of P, two are half filled and one is having lone pair. 5S is also having lone pairs.
So total number of lone pairs= $2$
Total number of sigma bonds= $3$
Degree of hybridization=$3+2$
Degree of hybridization=$5$
Also it is clear that in these 5 orbitals, 1 orbital is S, 3 orbitals are P and one orbital is d, therefore $I{{F}_{3}}$ is having $S{{P}^{3}}d$ hybridization.
$I{{F}_{5}}$: As Iodine over here is attached to 5 fluorine atoms, so it is forming 5 sigma bonds with them. To form 5 sigma bonds, it needs 5 half filled orbitals, therefore two p electrons will be transferred to vacant d orbitals.
Electronic configuration of valence shell of Iodine
\[5{{S}^{2}}5{{P}^{3}}5{{d}^{2}}\]
All three P orbitals are half filled along with two d orbitals. S orbital is having a lone pair.
Number of lone pairs= $1$
Number of sigma bonds=$5$
Degree of hybridization=$5+1$
Degree of hybridization=$6$
Also you can see that one S orbital, three P orbitals and two d orbitals are involved in hybridization, therefore $I{{F}_{5}}$ is having $S{{P}^{3}}{{d}^{2}}$ hybridization.
\[I{{F}_{7}}\]: As Iodine over here is attached to 7 fluorine atoms, so it is forming 7 sigma bonds with them. To form 7 sigma bonds, it needs 7 half filled orbitals, therefore two P electrons and one S electron will be transferred to vacant d orbitals.
Electronic configuration of valence shell of Iodine
\[5{{S}^{1}}5{{P}^{3}}5{{d}^{3}}\]
All three P orbitals are half filled along with three d orbitals and one S orbital.
Number of lone pairs=$0$
Number of sigma bonds=$7$
Degree of hybridization=$7+0$
Degree of hybridization=$7$

Also you can see that one S orbital, three P orbitals and three d orbitals are involved in hybridization, therefore $I{{F}_{7}}$ is having $S{{P}^{3}}{{d}^{3}}$ hybridization.

Additional Information: Always remember that the formula we used is only applicable for covalent compounds, not for coordination compounds because in coordination compounds we use crystal field theory to explain the structure of them and VSEPR theory is not applicable over there.

Note: VSEPR theory accounts for the presence of lone pairs and its impact on the structure of molecules. It considers that the presence of lone pair distorts the structure because lone pair repels bond pairs more than a bond pair repels a bond pair and that readjustment within the molecule takes place to minimize the repulsion and that alters the bond angle.