See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: A molecule has zero dipole moment when all individual bond dipoles (or group dipoles) cancel each other due to molecular symmetry. Step 1 – Analyze option (b): The structure shown is 3,4-dichlorocyclohex-1-ene (or equivalently a cyclohexene with Cl on the two sp2 carbons of the double bond). The two C–Cl bonds are on adjacent carbons of the double bond. Because both Cl atoms are on a planar alkene system and are arranged on opposite sides (trans) of the double bond, the two C–Cl dipoles point in opposite directions within the plane and cancel vectorially. The symmetrical placement about the axis of the double bond means the net dipole moment is zero. Step 2 – Analyze option (a): 1,3-dichlorocyclopentane with both Cl groups on wedge bonds (cis) — the two C–Cl dipoles both point in the same direction (same face of the ring), so they add rather than cancel. Net dipole ≠ 0. Step 3 – Analyze option (c): 1,2,3-trichlorocyclopropane with all three Cl on the same face (all wedge bonds) — the dipoles do not fully cancel because the geometry is not symmetric enough to give zero net moment with three identical substituents all on the same side. Net dipole ≠ 0. Step 4 – Analyze option (d): 1,3-dichlorocyclobutane with one Cl on wedge (up) and one Cl on dash (down) — trans configuration. In the cyclobutane ring, the 1,3-trans isomer does not have a plane of symmetry that makes the two C–Cl vectors cancel completely; the molecule still has a net dipole moment ≠ 0. Step 5 – Conclusion: Only option (b), where the two C–Cl bonds on the alkene carbons are arranged symmetrically (trans on a planar double bond within a ring), results in complete cancellation of bond dipoles, giving a net dipole moment of zero. Therefore, the correct answer is B.