1. Determine the shapes using VSEPR theory

We match each species in List I with its molecular geometry in List II.

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2. Species A: $\mathrm{SF_4}$

• Central atom: S
• Valence electrons on S = 6
• Four S–F bonds are formed, leaving one lone pair on S.
• So total electron pairs around S = 5
• Electron pair geometry = trigonal bipyramidal
• With one lone pair, molecular shape = see-saw

Hence, $A \to \text{III}$

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3. Species B: $\mathrm{BrF_3}$

• Central atom: Br
• Valence electrons on Br = 7
• Three Br–F bonds and two lone pairs on Br
• Total electron pairs = 5
• Electron pair geometry = trigonal bipyramidal
• With two lone pairs, molecular shape = T-shaped

In the list, this is written as Bent T-Shape.

Hence, $B \to \text{IV}$

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4. Species C: $\mathrm{BrO_3^-}$

• Central atom: Br
• Total valence electrons: $7 + 3(6) + 1 = 26$
• Around Br, there are three bonded oxygen atoms and one lone pair.
• Thus, steric number = 4
• Electron pair geometry = tetrahedral
• With one lone pair, molecular shape = pyramidal

Hence, $C \to \text{II}$

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5. Species D: $\mathrm{NH_4^+}$

• Central atom: N
• Nitrogen forms four N–H bonds and has no lone pair.
• Steric number = 4
• Molecular shape = tetrahedral

Hence, $D \to \text{I}$

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6. Final matching

$A-\text{III},\; B-\text{IV},\; C-\text{II},\; D-\text{I}$

This corresponds to Option B.

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7. Comparison with stored correct answer

Stored correct answer = B

Our derived answer = B

So, the answer agrees with the stored correct answer.