Match List I with List II :

Choose the correct answer from the options given below :

Species are classified in organic chemistry on the basis of two criteria:
1. Whether the species seeks electrons (electrophile) or donates electrons (nucleophile).
2. Whether the species contains an unpaired electron (free radical) or a positively charged carbon (carbocation).

Case A → (IV)
Electrophiles are electron-deficient species that possess an empty orbital and are ready to accept an electron pair. $$AlCl_3$$ has an incomplete octet on aluminium, so it can accept a lone pair; hence it is an electrophile.

Case B → (III)
Nucleophiles are electron-rich species that donate a lone pair to an electrophile. $$NH_3$$ has a lone pair on nitrogen and therefore behaves as a nucleophile.

Case C → (I)
Free radicals are neutral species containing one unpaired electron. $$\cdot Cl$$ (chlorine atom with one unpaired electron) is the classic example of a free-radical species.

Case D → (II)
Carbocations are positively charged carbon species; $$CH_3^+$$ (methyl cation) is the simplest carbocation.

Thus the correct matching is:
(A) Electrophile → (IV) $$AlCl_3$$
(B) Nucleophile → (III) $$NH_3$$
(C) Free radical → (I) $$\cdot Cl$$
(D) Carbocation → (II) $$CH_3^+$$

The option that shows this combination is Option A: (A)-(IV), (B)-(III), (C)-(I), (D)-(II).