See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1: Identify each molecule and its geometry. (I) cis-1,2-difluoroethylene (Z): Both F atoms are on the same side of the double bond. The C-F bond dipoles point in similar directions and add up significantly, giving a moderate-to-large net dipole moment. (II) trans-1,2-difluoroethylene (E): The two C-F bond dipoles point in opposite directions and largely cancel each other, giving a very small (near zero) dipole moment. (III) 1,1,2-trifluoro-2-chloroethylene: C1 bears two F atoms (like CF2), and C2 bears one F and one Cl. The vector sum of the C-F and C-Cl dipoles does not cancel completely. However, the three F atoms and one Cl create a net dipole, but partial cancellations occur. The net dipole is moderate. (IV) 1,1-difluoroethylene (CH2=CF2): Both F atoms are on the same carbon. The two C-F dipoles on C1 point in the same general direction (away from C1 toward F), and the CH2 end has small opposing contribution. This gives a significant net dipole moment. Step 2: Rank the dipole moments. - Molecule I (cis-1,2-difluoroethylene): The two C-F dipoles are on opposite carbons but on the same side; they add constructively. Dipole moment ≈ 2.42 D, which is large. - Molecule IV (1,1-difluoroethylene): Both F on same carbon; net dipole ≈ 1.37 D. - Molecule III (1,1,2-trifluoro-2-chloroethylene): Mixed substitution with partial cancellations; net dipole is intermediate but less than I and IV, approximately 1.0–1.2 D. - Molecule II (trans-1,2-difluoroethylene): Near-complete cancellation of the two C-F dipoles; dipole moment ≈ 0 D (actually very small, ~0 D). Thus the order is: I > IV > III > II. Step 3: Why other options fail. - (a) IV > I > III > II is wrong because cis (I) has a larger dipole than 1,1-disubstituted (IV). - (c) III > I > II > IV incorrectly places III highest and IV lowest. - (d) II > III > IV > I is wrong because trans (II) has the smallest dipole due to cancellation. Therefore, the correct answer is B.