To determine the shapes of molecules, we use the VSEPR theory, which depends on the total number of electron pairs (bond pairs + lone pairs) around the central atom.

For IF₅:

• Central atom: Iodine (I)
• Valence electrons of I = 7
• Number of F atoms bonded = 5
• Number of bond pairs = 5
• Number of lone pairs = $\frac{1}{2}$(Valence electrons - Number of bonded atoms) = $\frac{1}{2}$(7 - 5) = 1
• Total electron pairs = 5 (bond pairs) + 1 (lone pair) = 6
• According to VSEPR theory, 6 electron pairs lead to an octahedral electron geometry. With 5 bond pairs and 1 lone pair, the molecular geometry is square pyramidal (or tetragonal pyramidal).

For IF₇:

• Central atom: Iodine (I)
• Valence electrons of I = 7
• Number of F atoms bonded = 7
• Number of bond pairs = 7
• Number of lone pairs = $\frac{1}{2}$(7 - 7) = 0
• Total electron pairs = 7 (bond pairs) + 0 (lone pairs) = 7
• According to VSEPR theory, 7 electron pairs with no lone pairs lead to a pentagonal bipyramidal molecular geometry.

Option Analysis:

• A) Tetragonal pyramidal and pentagonal bipyramidal: This matches our derived shapes for IF₅ and IF₇, respectively.
• B) Octahedral and pyramidal: Octahedral is the electron geometry for IF₅, not the molecular shape. Pyramidal is incorrect for IF₇.
• C) Trigonal bipyramidal and square antiprismatic: Trigonal bipyramidal corresponds to 5 electron pairs, not 6. Square antiprismatic is incorrect for IF₇.
• D) Distorted square planar and distorted octahedral: These shapes are incorrect for both molecules.

Correct Answer: (A)