Match List I with List II

Choose the correct answer from the options given below:

First write down the two lists exactly as they appear in the question.

List I (Types of reactive intermediates / reagents)
A. Electrophile
B. Nucleophile
C. Free-radical
D. Carbenium (carbocation) ion

List II (Typical examples)
I. $$CH_3^+$$  (methyl carbocation)
II. $$Cl\cdot$$  (chlorine radical)
III. $$NH_3$$  (ammonia)
IV. $$BF_3$$  (boron trifluoride)

Now decide which example fits each description.

Case A (Electrophile)
An electrophile is an electron-deficient Lewis acid that can accept an electron pair. $$BF_3$$ has an incomplete octet on boron and readily accepts electron density, so it is a classic electrophile.
Hence A → IV.

Case B (Nucleophile)
A nucleophile is electron-rich and donates an electron pair. $$NH_3$$ possesses a lone pair on nitrogen and behaves as a Lewis base, i.e. a nucleophile.
Hence B → III.

Case C (Free-radical)
A free-radical carries one unpaired electron. $$Cl\cdot$$ clearly has a single unpaired electron.
Hence C → II.

Case D (Carbenium / Carbocation)
A carbenium ion is a positively charged carbon species such as $$CH_3^+$$.
Hence D → I.

Collecting all matches: A-IV, B-III, C-II, D-I.

The only option that contains this combination is:
Option C which is: A-IV, B-III, C-II, D-I.