Choose the correct answer from the options given below :

For every cyclic, conjugated ring the nature of delocalisation is decided by Hückel’s rule.

Hückel’s rule (for a planar, completely conjugated ring):
  • If the number of π‐electrons is $$4n+2\;(n=0,1,2,\dots)$$ the system is aromatic (very stable).
  • If the number of π‐electrons is $$4n\;(n=1,2,\dots)$$ and the ring is planar, the system is anti-aromatic (highly unstable).
  • If the ring is not planar (or conjugation is broken) the compound is simply non-aromatic, irrespective of electron count.

With that in mind, examine each species.

Case A: Cyclooctatetraene, $$C_8H_8$$
It contains eight π-electrons (formal count  $$8 = 4n\;(n=2)$$). However cyclooctatetraene is not planar: it adopts a shallow “tub” conformation that prevents continuous overlap of the p orbitals. Hence conjugation is interrupted and the molecule is non-aromatic.
⇒ (A) ⟶ (III)

Case B: Cyclobutadiene, $$C_4H_4$$
It possesses four π-electrons, $$4 = 4n\;(n=1)$$, and the tiny four-membered ring is essentially planar, so conjugation is complete. Therefore it satisfies the anti-aromatic condition.
⇒ (B) ⟶ (IV)

Case C: Benzene, $$C_6H_6$$
Benzene has six π-electrons, $$6 = 4n+2\;(n=1)$$, is perfectly planar and fully conjugated. It is the prototypical aromatic molecule.
⇒ (C) ⟶ ( I )

Case D: Tropylium cation, $$C_7H_7^+$$
The ion has seven carbon atoms, each contributing one p electron except the positively charged carbon which lacks one. Total π-electrons = 6, again $$6 = 4n+2\;(n=1)$$. The cation is planar and conjugated, so it is aromatic. (Although aromatic like benzene, it is treated separately here because it is a carbocation.)
⇒ (D) ⟶ (II)

Collecting the matches:
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Hence the correct choice is
Option C which is: (A)-(III), (B)-(IV), (C)-(I), (D)-(II).