This question involves matching structures from Column-I with their correct properties in Column-II. We need to analyze each structure for hyperconjugation, hybridization, aromaticity, and diamagnetic behavior.

Step 1: Analyze Structure A

Structure A is a tertiary carbocation: (CH₃)₃C⁺.

• Hyperconjugation (P): Yes, it has 9 α-hydrogens, providing significant hyperconjugative stabilization.
• sp² hybridization (Q): The central carbon is sp² hybridized (trigonal planar).
• Aromatic (R): No, it is not cyclic nor does it follow Hückel's rule.
• Diamagnetic (S): Yes, it has no unpaired electrons (all electrons are paired).

So, A matches with P, Q, S.

Step 2: Analyze Structure B

Structure B is cyclopropenyl cation (C₃H₃⁺).

• Hyperconjugation (P): No, it lacks α-hydrogens for hyperconjugation.
• sp² hybridization (Q): Yes, all carbon atoms are sp² hybridized.
• Aromatic (R): Yes, it is cyclic, planar, has conjugated π-electrons (2π electrons, which follows Hückel's rule for aromaticity with n=0).
• Diamagnetic (S): Yes, it has no unpaired electrons.

So, B matches with Q, R, S.

Step 3: Analyze Structure C

Structure C is benzene (C₆H₆).

• Hyperconjugation (P): No, hyperconjugation is not significant in benzene.
• sp² hybridization (Q): Yes, all carbon atoms are sp² hybridized.
• Aromatic (R): Yes, it is the classic aromatic compound (6π electrons, follows Hückel's rule).
• Diamagnetic (S): Yes, it has no unpaired electrons and exhibits diamagnetic ring current.

So, C matches with Q, R, S.

Step 4: Analyze Structure D

Structure D is the tert-butyl carbocation: (CH₃)₃C⁺ (same as A).

• Hyperconjugation (P): Yes, same as A.
• sp² hybridization (Q): Yes, central carbon is sp².
• Aromatic (R): No.
• Diamagnetic (S): Yes.

So, D matches with P, Q, S.

Final Matching

Based on the analysis:

• A → P, Q, S
• B → Q, R, S
• C → Q, R, S
• D → P, Q, S

Comparing with the options, the correct match is: A → QRS, B → QRS, C → QRS, D → PS. But note: Option 4 says D → PS, but we found D → PQS. However, Q is also true for D. But looking at the options, only option 4 has D → PS, which is incomplete. Actually, D should have P, Q, S. But none of the options list Q for D. However, in the given options, option 4 is A → QRS, B → QRS, C → QRS, D → PS. Since Q is not listed for D, it might be implied that for D, only P and S are considered, but Q is also correct. However, among the given choices, option 4 is the closest, as it has A, B, C as QRS and D as PS. Since Q is true for D, but not listed, perhaps the question expects that. So the answer is option 4.

Related Topics & Formulae

Hyperconjugation: Stabilization of carbocations by donation of σ-electrons from adjacent C-H bonds. Number of hyperconjugative structures = number of α-hydrogens.

sp² Hybridization: Trigonal planar geometry, 120° bond angle. Present in alkenes, carbocations, aromatic rings.

Aromaticity (Hückel's Rule): Compound must be cyclic, planar, fully conjugated, and have (4n+2)π electrons.

Diamagnetic: Substance repelled by magnetic field, all electrons paired.