1. Determine the shapes one by one using VSEPR theory.

2. For $\mathrm{ICl}$

• It is a diatomic molecule.
• Any diatomic molecule is linear.

So, $\mathrm{ICl} \rightarrow \text{Linear}$ This matches IV.

3. For $\mathrm{ICl}_3$

• Central atom: Iodine
• Valence electrons on iodine = 7
• It forms 3 bonds with 3 chlorine atoms, leaving 2 lone pairs.
• Thus total electron pairs around iodine = 5

So the electron pair geometry is trigonal bipyramidal. With 2 lone pairs occupying equatorial positions, the molecular shape becomes: $\text{T-shape}$

Thus, $\mathrm{ICl}_3 \rightarrow \text{T-shape}$ This matches I.

4. For $\mathrm{ClF}_5$

• Central atom: Chlorine
• Valence electrons on chlorine = 7
• It forms 5 bonds with fluorine atoms, leaving 1 lone pair.
• Total electron pairs = 6

So the electron pair geometry is octahedral. With one lone pair, the molecular shape becomes: $\text{Square pyramidal}$

Thus, $\mathrm{ClF}_5 \rightarrow \text{Square pyramidal}$ This matches II.

5. For $\mathrm{IF}_7$

• Central atom: Iodine
• Valence electrons on iodine = 7
• It forms 7 bonds with fluorine atoms and has no lone pair.
• Total electron pairs = 7

Hence the geometry is: $\text{Pentagonal bipyramidal}$

Thus, $\mathrm{IF}_7 \rightarrow \text{Pentagonal bipyramidal}$ This matches III.

6. Final matching

So we get:

• $A \rightarrow IV$
• $B \rightarrow I$
• $C \rightarrow II$
• $D \rightarrow III$

This corresponds to Option B.

7. Comparison with stored answer

Stored correct answer = B

Our derived answer = B

So, the answer agrees with the stored answer.