1. We identify the geometry/shape of each species using VSEPR.

2. Species A: $\mathrm{H_3O^+}$

   • Central atom: O
   • Oxygen has 6 valence electrons.
   • In $\mathrm{H_3O^+}$, O forms 3 bonds and has 1 lone pair.
   • Electron pair geometry is tetrahedral, but molecular shape is trigonal pyramidal.
   • So, $A \to \text{III (Pyramidal)}$.

3. Species B: Acetylide anion

   • Acetylide anion is $\mathrm{C_2^{2-}}$ type or commonly represented as $\mathrm{HC\equiv C^-}$ with the carbon framework being $sp$ hybridised.
   • Around the carbon atoms involved in the triple bond, geometry is linear.
   • So, $B \to \text{II (Linear)}$.

4. Species C: $\mathrm{NH_4^+}$

   • Central atom: N
   • Nitrogen forms 4 bonds and has no lone pair.
   • Geometry is tetrahedral.
   • So, $C \to \text{I (Tetrahedral)}$.

5. Species D: $\mathrm{ClO_2^-}$

   • Central atom: Cl
   • Total valence electrons: $7 + 2(6) + 1 = 20$
   • Around Cl, there are 2 bond pairs and 2 lone pairs.
   • Electron pair geometry is tetrahedral, but molecular shape is bent.
   • So, $D \to \text{IV (Bent)}$.

6. Final matching: $A \to III, \quad B \to II, \quad C \to I, \quad D \to IV$

7. Checking options:

   • Option A: $B \to I$ incorrect
   • Option B: $A \to III, B \to II, C \to I, D \to IV$ correct
   • Option C: $B \to IV$ incorrect
   • Option D: multiple mismatches

Therefore, the correct option is B.