Match List-I with List-II.

List-I	List-II
A	I. Spiro compound
B	II. Aromatic compound
C	III. Non-planar Heterocyclic compound
D	IV. Bicyclo compound

Structure $$\mathrm{A}$$ represents tetrahydrofuran, a saturated five-membered ring containing an oxygen atom.

Since the ring contains an atom other than carbon, it is a heterocyclic compound.

All carbon atoms are $$\mathrm{sp^3}$$ hybridised, so the ring adopts a puckered non-planar geometry to minimise ring strain.

Therefore:

$$\mathrm{A \rightarrow III}$$

$$\mathrm{(Non\text{-}planar\ Heterocyclic\ Compound)}$$

Structure $$\mathrm{B}$$ contains two triangular rings sharing a common bond.

Thus, the two rings share two adjacent carbon atoms.

Compounds in which rings share two or more atoms are called bicyclo compounds.

Therefore:

$$\mathrm{B \rightarrow IV}$$

$$\mathrm{(Bicyclo\ Compound)}$$

Structure $$\mathrm{C}$$ contains two rings joined through exactly one common carbon atom.

Such compounds are called spiro compounds.

Therefore:

$$\mathrm{C \rightarrow I}$$

$$\mathrm{(Spiro\ Compound)}$$

Structure $$\mathrm{D}$$ represents furan.

It is cyclic, planar and fully conjugated.

One lone pair of oxygen participates in delocalisation, giving:

$$\mathrm{6\ \pi\ electrons}$$

Thus, it satisfies Hückel’s rule:

$$\mathrm{(4n+2)\ \pi\ electrons}$$

Hence, furan is aromatic.

Therefore:

$$\mathrm{D \rightarrow II}$$

$$\mathrm{(Aromatic\ Compound)}$$

Correct matching:

$$\boxed{\mathrm{A-III,\ B-IV,\ C-I,\ D-II}}$$