To determine the shape and hybridization of IF₅, we first need to find the steric number (SN) of the central atom, Iodine (I).

Step 1: Calculate the steric number (SN)

The steric number is the sum of the number of bond pairs and lone pairs around the central atom. For IF₅:

• Central atom: Iodine (I)
• Valence electrons of I = 7
• Number of surrounding atoms (F) = 5
• Number of electrons used in bonding = 5 (one for each F)
• Number of remaining electrons = 7 - 5 = 2
• Number of lone pairs = 2 / 2 = 1
• Steric Number (SN) = Number of bond pairs + Number of lone pairs = 5 + 1 = 6

Step 2: Determine hybridization

A steric number of 6 corresponds to sp³d² hybridization.

Step 3: Determine the molecular geometry (shape)

With 5 bond pairs and 1 lone pair (SN = 6), the electron geometry is octahedral. However, the molecular geometry (shape) is determined by the arrangement of atoms, considering the lone pair's repulsion. The lone pair occupies one position in the octahedral arrangement, distorting the shape to a square pyramidal geometry.

Option Analysis:

• A) Sea-saw, sp³d: Sea-saw shape corresponds to SN=5 with 4 bond pairs and 1 lone pair (e.g., SF₄), and sp³d hybridization. This is incorrect for IF₅.
• B) Trigonal bipyramidal, sp³d: Trigonal bipyramidal shape corresponds to SN=5 with 5 bond pairs and 0 lone pairs (e.g., PCl₅), and sp³d hybridization. This is incorrect for IF₅.
• C) Square pyramidal, sp³d²: This matches our calculated shape and hybridization for IF₅.
• D) Pentagonal pyramidal, sp³d³: Pentagonal pyramidal shape corresponds to SN=7 with 6 bond pairs and 1 lone pair (e.g., XeOF₅⁻), and sp³d³ hybridization. This is incorrect for IF₅.

Correct Answer: (C)