1. Determine the shape of each molecule using VSEPR theory

We match each molecule in List I with its molecular shape in List II.

2. Molecule (A): $\mathrm{BrF_5}$

• Central atom: Br
• Valence electrons on Br = 7
• Bonded to 5 F atoms, so 5 bond pairs
• Remaining electrons on Br = 1 lone pair

Thus, steric number = 6, corresponding to octahedral electron-pair geometry. With one lone pair, the molecular shape is:

$\boxed{\text{Square pyramidal}}$

So,

$(A) \to (IV)$

3. Molecule (B): $\mathrm{H_2O}$

• Central atom: O
• Oxygen has 6 valence electrons
• Forms 2 bonds with H, leaving 2 lone pairs

So total electron pairs = 4, tetrahedral electron-pair geometry. With 2 bond pairs and 2 lone pairs, the molecular shape is:

$\boxed{\text{Bent}}$

So,

$(B) \to (III)$

4. Molecule (C): $\mathrm{ClF_3}$

• Central atom: Cl
• Chlorine has 7 valence electrons
• Forms 3 bonds with F, leaving 2 lone pairs

So total electron pairs = 5, trigonal bipyramidal electron-pair geometry. With 2 lone pairs occupying equatorial positions, the molecular shape is:

$\boxed{\text{T-shape}}$

So,

$(C) \to (I)$

5. Molecule (D): $\mathrm{SF_4}$

• Central atom: S
• Sulfur has 6 valence electrons
• Forms 4 bonds with F, leaving 1 lone pair

So total electron pairs = 5, trigonal bipyramidal electron-pair geometry. With 1 lone pair, the molecular shape is:

$\boxed{\text{See saw}}$

So,

$(D) \to (II)$

6. Final matching

This corresponds to Option B.

7. Comparison with stored correct answer

Stored correct answer = B

Our derived answer = B

So the answers agree.