The structure of IF7 is

The correct option is Pentagonal bipyramid.

To determine the structure of iodine heptafluoride (IF₇), we can apply the Valence Shell Electron Pair Repulsion (VSEPR) theory:
1. Identify the central atom and its valence electrons: The central atom is Iodine (I). Since iodine is a halogen belonging to Group 17, it has 7 valence electrons in its outermost shell.

2. Determine the number of bonding pairs: Iodine is bonded to 7 fluorine (F) atoms. Each fluorine atom shares one electron with iodine to form a single covalent bond. This gives a total of 7 bonding pairs of electrons.

3. Determine the number of lone pairs: The number of lone pairs on the central iodine atom can be calculated as follows:

$\text{Number of lone pairs}=\frac{7-7}{2}=0$

Thus, the central iodine atom has 7 bonding pairs and 0 lone pairs.

4. Calculate the steric number and hybridization: The steric number (SN) is the sum of bonding pairs and lone pairs:
Steric Number = 7 (bond pairs) + 0 (lone pairs) = 7
A steric number of 7 corresponds to sp³d³ hybridization.

5. Determine the molecular geometry: According to VSEPR theory, a molecule with a steric number of 7 and zero lone pairs adopts a pentagonal bipyramidal geometry to minimize electron-pair repulsion. In this arrangement:
• Five fluorine atoms are positioned at the corners of a regular pentagon in the equatorial plane, with F–I–F bond angles of 72°.
• The remaining two fluorine atoms occupy the axial positions, one directly above and one directly below the pentagonal plane, forming a 90° angle with the equatorial fluorine atoms.