JEE Chemistry 2025 Question with Solution

Given: LIST-I contains Carbocation, C-Free radical, Nucleophile and Electrophile. LIST-II contains their definitions or characteristics.

Find: The correct matching between LIST-I and LIST-II.

A carbocation is a positively charged carbon species. It is electron deficient and has an empty $p$-orbital. Therefore, A (Carbocation) matches III.

A C-Free radical has one unpaired electron. Such a carbon species can be described as $\text{sp}^2/\text{sp}^3$ hybridized with one unpaired electron. Therefore, B (C-Free radical) matches IV.

A nucleophile donates a pair of electrons. Therefore, C (Nucleophile) matches I.

An electrophile accepts a pair of electrons because it is electron deficient. Therefore, D (Electrophile) matches II.

So the correct matching is:

• A - III
• B - IV
• C - I
• D - II

Therefore, the correct option is D.

Given:

• Carbocation
• C-Free radical
• Nucleophile
• Electrophile

Find: Which description in LIST-II corresponds to each species.

Key ideas:

• Carbocation: $\text{sp}^2$, planar, electron deficient, empty $p$-orbital
• Free radical: one unpaired electron
• Nucleophile: electron-pair donor
• Electrophile: electron-pair acceptor

Now match one by one:

1. Carbocation corresponds to $\text{sp}^2$ hybridized carbon with empty $p$-orbital, so A - III.
2. C-Free radical corresponds to $\text{sp}^2/\text{sp}^3$ hybridized carbon with one unpaired electron, so B - IV.
3. Nucleophile corresponds to species that can supply a pair of electrons, so C - I.
4. Electrophile corresponds to species that can receive a pair of electrons, so D - II.

Hence the final matching is A-III, B-IV, C-I, D-II, so the correct option is D.