Question

What is the structure of $ I{F_7} $ ?

Answer 1

Hint :Iodine hypo fluoride or iodine (VII) chloride are other names for $ I{F_7} $ . It's a strong-bonding interhalogen molecule. According to VSEPR theory, it has a pentagonal bi-pyramidal structure with $ s{p^3}{d^3} $ hybridization.

Complete Step By Step Answer:
The core atom I(iodine) is connected to 7 fluorine atoms via 7 sigma bonds in $ I{F_7} $ . So, in this case, the steric number is 7. As a result, I(iodine) hybridization in $ I{F_7} $ is $ s{p^3}{d^3} $ . As a result, both the electron pair and molecular geometry are pentagonal bi-pyramidal.
Let’s look how the structure comes:
Iodine(atomic no.=53)
  $ \left[ {E.C} \right] = 4{d^{10}}5{s^2}5{p^5} $ where E.C=electronic configuration
Look at the following image:
In first excited state its electronic configuration changes:
  $ \left[ {E.C} \right] = 5{s^1}5{p^3}5{d^3} $
Hence we can see that there are 7 fluorine molecules. One enters $ 5s $ , 3 can enter $ 5p $ and 3 will enter $ 5d $ .
Hence we can make out that the hybridization is $ s{p^3}{d^3} $ . And hence geometry is pentagonal bi-pyramidal
Let's look at the structure:

This is pentagonal bi-pyramidal. Also not every bonding angle is the same. Five electron pairs are on the same plane at a 72-degree angle, while the other two are perpendicular to the plane and make a 90-degree angle with it. Even bond lengths differ. The axial bonds are 186pm long, whereas the equatorial bonds are 179pm long.

Note :
There can be many ionization states of an element. But whenever we draw structures we should only consider the first excited state and calculate the hybridization.
Also The primary distinction between axial and equatorial positions is that axial bonds are vertical, whereas equatorial bonds are horizontal. The phrases axial and equatorial are critical when illustrating the real 3D orientation of chemical bonds.
It is not necessary to mention bond angles and bond lengths while drawing the structure.