The correct order of dipole moment of the different methyl halides is:

The correct answer is Option (3) → CH₃Cl > CH₃F > CH₃Br > CH₃I

Note: None of the given options was correct in the NTA question paper. To provide additional question practice to the students, the question has been slightly modified so that the question has one clear answer.

The dipole moment of a molecule is the product of the charge separation and the bond length (μ = q × d). Down the group from F to I, the electronegativity of the halogen falls, so the charge on the atoms falls, but the C–X bond length rises. The two effects work against each other.

• \( \ce{CH3Cl} \) (1.860 D) — first. The much longer C–Cl bond (178 pm against 139 pm) more than makes up for the smaller charge separation compared with C–F.

• \( \ce{CH3F} \) (1.847 D) — second. Fluorine is the most electronegative halogen, but the C–F bond is the shortest.

• \( \ce{CH3Br} \) (1.830 D) — third. From Cl to Br the fall in electronegativity wins over the longer bond.

• \( \ce{CH3I} \) (1.636 D) — last. Iodine has the lowest electronegativity, so the charge separation is the smallest.

NCERT (2026-27 edition):
Nature of C–X Bond – …