Match the LIST-I (Species) with LIST-II (Properties) and choose the correct answer

Match the LIST-I (Species) with LIST-II (Properties) and choose the correct answer | JEE Main Chemistry
QMatching
Match the LIST-I with LIST-II and choose the correct answer:

LIST-I:
A. Carbocation
B. C-Free radical
C. Nucleophile
D. Electrophile

LIST-II:
I. Species that can supply a pair of electrons.
II. Species that can receive a pair of electrons.
III. $sp^2$ hybridized carbon with empty p-orbital.
IV. $sp^2/sp^3$ hybridized carbon with one unpaired electron.
✅ Correct Answer
A-III, B-IV, C-I, D-II
Step-by-Step Solution
1
Analyze A (Carbocation)

A carbocation ($R^+$) has a carbon atom with only 6 valence electrons. It uses three $sp^2$ hybrid orbitals for bonding and has one unhybridized **empty p-orbital**. This perfectly matches description **III**.

2
Analyze B (C-Free radical)

A carbon free radical ($R\cdot$) is a neutral species containing one **unpaired electron**. Depending on the environment, the carbon can be $sp^2$ (planar) or $sp^3$ (pyramidal). This matches description **IV**.

3
Analyze C (Nucleophile)

The term “nucleophile” means nucleus-loving. These are species with high electron density (lone pairs or negative charge) that **supply/donate a pair of electrons** to form a chemical bond. This matches description **I**.

4
Analyze D (Electrophile)

The term “electrophile” means electron-loving. These are electron-deficient species (positive charge or incomplete octet) that **receive/accept a pair of electrons** during a reaction. This matches description **II**.

5
Synthesize the Matches

Combining the analysis: A-III, B-IV, C-I, D-II.

6
Compare with Options

Referring to the provided screenshot, Option C matches the sequence A-III, B-IV, C-I, D-II.

7
Verify the Conclusion

The definitions align with standard IUPAC terminology for reaction intermediates and reagents.

Final Result: A-III, B-IV, C-I, D-II (Option C)
📚
Theory
1. Structure of Carbocations
Carbocations are trivalent carbon species with a positive charge. The central carbon is $sp^2$ hybridized, leading to a trigonal planar geometry. The bond angles are approximately 120°. The key feature is the unhybridized p-orbital, which lies perpendicular to the plane of the three hybrid orbitals and is completely empty. This vacancy makes carbocations highly electrophilic and reactive. They are stabilized by inductive effects (+I) and hyperconjugation from adjacent alkyl groups, leading to the stability order: $3^\circ > 2^\circ > 1^\circ > \text{methyl}$. They are common intermediates in $S_N1$ and $E1$ reactions.
2. Carbon Free Radicals
Carbon free radicals are neutral species formed by homolytic fission of a covalent bond. They possess an odd, unpaired electron in the valence shell (total 7 electrons). While methyl radicals are typically $sp^2$ hybridized and planar, substituted radicals can sometimes adopt a shallow pyramidal shape due to $sp^3$ character. Like carbocations, they are stabilized by resonance and hyperconjugation. They are highly reactive because they seek to complete their octet by pairing their lone electron. Radical reactions typically involve three stages: initiation, propagation, and termination.
3. Nucleophiles and Lewis Bases
A nucleophile is a reagent that brings an electron pair to a reaction site. It is essentially a Lewis base. Nucleophiles can be negatively charged (anions like $OH^-$, $CN^-$, $Cl^-$) or neutral molecules with lone pairs (like $H_2O$, $NH_3$, $ROH$). The “nucleophilicity” depends on factors like basicity, polarizability, and steric hindrance. In organic mechanisms, nucleophiles attack electron-deficient centers (like carbocations or carbonyl carbons) to form new covalent bonds. Anionic nucleophiles are generally more reactive than their conjugate neutral counterparts.
4. Electrophiles and Lewis Acids
An electrophile is a reagent that is attracted to electrons. It is essentially a Lewis acid. Electrophiles can be positively charged ions ($H^+$, $NO_2^+$, $CH_3^+$) or neutral molecules with incomplete octets or vacant orbitals ($BF_3$, $AlCl_3$, $FeCl_3$). Even molecules with polar bonds, where a carbon has a partial positive charge ($\delta^+$), act as electrophilic centers (e.g., in alkyl halides). In a chemical reaction, the electrophile accepts an electron pair from a nucleophile. Electrophilic addition and electrophilic aromatic substitution are fundamental reaction types in organic chemistry.
FAQs
1
Is a carbocation a Lewis acid or base?
A carbocation is a Lewis acid because it can accept an electron pair due to its empty p-orbital.
2
What is the difference between a nucleophile and a base?
While both donate electrons, a base targets a proton ($H^+$), whereas a nucleophile targets a carbon atom or other nuclei.
3
Are free radicals paramagnetic?
Yes, because they contain an unpaired electron, they are attracted to magnetic fields.
4
Why is $BF_3$ an electrophile?
Boron in $BF_3$ has only 6 electrons in its valence shell, making it electron-deficient and capable of accepting a lone pair.
5
Is $NH_3$ a nucleophile?
Yes, Ammonia is a neutral nucleophile because it has a lone pair on the Nitrogen atom.
6
Can a molecule be both an electrophile and a nucleophile?
Yes, species like water or alcohols have lone pairs (nucleophilic) and polar H-bonds (electrophilic at H).
7
What does ‘homolytic fission’ produce?
Homolytic fission produces two free radicals by equally splitting the bonding electron pair.
8
What does ‘heterolytic fission’ produce?
Heterolytic fission produces a cation (electrophile) and an anion (nucleophile).
9
Which is more stable: $CH_3^+$ or $(CH_3)_3C^+$?
$(CH_3)_3C^+$ (tert-butyl carbocation) is much more stable due to 9 hyperconjugative structures and +I effect.
10
Does sp2 hybridization imply planarity?
Yes, $sp^2$ hybrid orbitals are oriented at 120° in a single plane.
📤 Share with friends!
Scroll to Top