Question

Draw the structure of $\text{I}{\text{F}}_{\text{7}}$. Write its geometry and the type of hybridization.

Answer 1

Hint For drawing the structure of $\text{I}{\text{F}}_{\text{7}}$, we have to know about the bonds present in the structure and, no. of bonds present in the structure depends on the hybridization of central atom of $\text{I}{\text{F}}_{\text{7}}$ i.e. Iodine.

Complete step by step solution:
Chemically $\text{I}{\text{F}}_{\text{7}}$ is known as Iodine heptafluoride. For determining the structure & geometry of $\text{I}{\text{F}}_{\text{7}}$, firstly we have to know about the hybridization of the central atom of $\text{I}{\text{F}}_{\text{7}}$.
-For determining the hybridization first, we write the electronic configuration of iodine. As we know that atomic number of iodine is 53, so its electronic configuration is written as $\left[\text{Kr}\right]\text{4}{\text{d}}^{\text{10}}\text{5}{\text{s}}^{\text{2}}\text{5}{\text{p}}^{\text{5}}$.
-In the outermost shell of iodine 7 electrons are present & in ground state it is displayed as follow:

-In the $\text{I}{\text{F}}_{\text{7}}$ molecule, seven fluorine atoms are present & for the formation of $\text{I}{\text{F}}_{\text{7}}$ we have to make sure that iodine atom will present in the excited state so that we get seven unpaired electrons and these electrons will pair with seven fluorine atoms. So, electronic configuration of iodine in the excited state is displayed as follow:

-Now seven fluorine atoms will do pairing with seven unpaired electrons of iodine and form $\text{I}{\text{F}}_{\text{7}}$, whose hybridization is $\text{s}{\text{p}}^{\text{3}}{\text{d}}^{\text{3}}$.

-Hence, the geometry of $\text{I}{\text{F}}_{\text{7}}$ is pentagonal pyramidal and its structure is shown as below:

Note: In this question, some of you may predict wrong hybridization of iodine if you pair the fluorine atoms in the ground state of iodine. So, always keep in mind that incoming atoms always pair with the unpaired electron not with the paired electron.