To determine the hybridisation of the central atom (iodine) in iodine heptafluoride (IF₇), we can use the steric number method.

Concept: The hybridisation of a central atom can be determined by calculating its steric number, which is the sum of the number of sigma bonds formed by the central atom and the number of lone pairs on the central atom.

Step 1: Determine the number of valence electrons of the central atom.
Iodine (I) is in Group 17, so it has 7 valence electrons.

Step 2: Determine the number of sigma bonds.
In IF₇, iodine is bonded to 7 fluorine atoms. Each fluorine atom forms one single bond (sigma bond) with iodine. So, the number of sigma bonds = 7.

Step 3: Determine the number of lone pairs on the central atom.
Number of electrons used in bonding = 7 (for 7 F atoms) × 1 (electron per bond) = 7 electrons.
Number of lone pairs = (Total valence electrons - Electrons used in bonding) / 2
Number of lone pairs = (7 - 7) / 2 = 0 / 2 = 0.

Step 4: Calculate the steric number.
Steric number = Number of sigma bonds + Number of lone pairs
Steric number = 7 + 0 = 7.

Step 5: Determine the hybridisation based on the steric number.
Steric number 7 corresponds to $s{p}^3{d}^3$ hybridisation.

Option Analysis:

• A) $s{p}^3{d}^3$: This corresponds to a steric number of 7, which is correct for IF₇.
• B) $s{p}^{3}d$: This corresponds to a steric number of 5 (e.g., PCl₅).
• C) $s{p}^3{d}^2$: This corresponds to a steric number of 6 (e.g., SF₆).
• D) $ds{p}^2$: This corresponds to a steric number of 4, typically found in square planar complexes (e.g., [Ni(CN)₄]²⁻).

Correct Answer: (A)