Which of the following has the highest dipole moment?

The correct answer is Option (4) → \(CH_2Cl_2\)

The dipole moment of a molecule is the vector sum of its bond dipoles. In the four molecules carbon is tetrahedral, and each C–Cl bond has a much larger dipole than a C–H bond, so the answer depends on how the C–Cl dipoles add up.

• \( \ce{CH4} \) and \( \ce{CCl4} \): four identical bonds pointing to the corners of a regular tetrahedron cancel completely, so μ = 0 for both.

• \( \ce{CHCl3} \): the three C–Cl dipoles add to a resultant equal to one C–Cl dipole (the three tetrahedral vectors together equal the fourth bond direction reversed).

• \( \ce{CH2Cl2} \): two C–Cl dipoles at about 109.5° give a resultant of 2 μ_C–Cl cos 54.75° ≈ 1.15 μ_C–Cl.

So \( \ce{CH2Cl2} \) has the largest dipole moment; the measured values are about 1.60 D for \( \ce{CH2Cl2} \) and 1.04 D for \( \ce{CHCl3} \).

Why the other options are incorrect

• Option (1) CH₄ — Incorrect. Symmetric tetrahedron of C–H bonds; dipole moment zero.

• Option (2) CCl₄ — Incorrect. Symmetric tetrahedron of C–Cl bonds; the four large bond dipoles cancel, so μ = 0.

• Option (3) CHCl₃ — Incorrect. A third chlorine partly cancels the other two, so its dipole moment (about 1.04 D) is lower than that of \( \ce{CH2Cl2} \).