Question

Read the following statements:

(A) Q has more $\delta^-$ on chlorine than P.
(B) Q has more dipole moment than P.
(C) In Q, C–Cl bond has double bond character.
(D) In Q, Cl is attached to sp$^2$ hybridised carbon but in P, Cl is attached to sp$^3$.
(E) In Q, C–Cl bond length is more due to repulsion between lone pair on chlorine and $\pi$ electron in aromatic ring.

• A. A, B, D, E
• B. C, D
• C. B, C, D
• D. B, C, D, E

Hint:

The electron density on chlorine and the hybridisation of carbon in aromatic compounds significantly affect the dipole moment, bond character, and bond length.

Answer 1

The Correct Option is B

Step 1: Analyzing Statement (A).
In Q, the chlorine is attached to a sp$^2$ hybridised carbon, which is more electronegative than sp$^3$ hybridised carbon in P. As a result, chlorine in Q has more electron density ($\delta^-$) than chlorine in P. Therefore, Statement (A) is correct.

Step 2: Analyzing Statement (B).
Since Q has more electron density on chlorine due to the sp$^2$ hybridisation of the carbon, Q will have a higher dipole moment compared to P. This makes Statement (B) correct.

Step 3: Analyzing Statement (C).
In Q, the C–Cl bond exhibits partial double bond character due to resonance with the aromatic ring. This leads to an increase in bond strength and decreases the bond length. Hence, Statement (C) is correct.

Step 4: Analyzing Statement (D).
In Q, chlorine is attached to a sp$^2$ hybridised carbon (from the benzene ring), which is more electronegative and forms a stronger bond with chlorine. In P, chlorine is attached to sp$^3$ hybridised carbon, making it less electronegative. Thus, Statement (D) is correct.

Step 5: Analyzing Statement (E).
In Q, due to the resonance and the proximity of chlorine's lone pair and the $\pi$ electron density of the aromatic ring, there is repulsion that increases the C–Cl bond length. This makes Statement (E) incorrect.

Step 6: Conclusion.
The correct statements are C and D. Thus, the correct answer is (2).