Which of the following methyl halide will have the highest value of dipole moment?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(CH_3-Cl\)
The dipole moment of a bond is the product of the charge separated and the distance between the charges, \( \mu = q \times d \). Going from F to I, two effects pull in opposite directions: the electronegativity of the halogen falls, so the partial charges become smaller, while the C–X bond becomes longer.
From \( \ce{CH3-F} \) to \( \ce{CH3-Cl} \), the bond length rises sharply, which more than makes up for the smaller charge, so \( \ce{CH3Cl} \) has a slightly larger dipole moment than \( \ce{CH3F} \). After chlorine the falling charge wins, and the dipole moment decreases:
\( \displaystyle \ce{CH3Cl} \gt \ce{CH3F} \gt \ce{CH3Br} \gt \ce{CH3I} \)
So the methyl halide with the highest dipole moment is \( \ce{CH3-Cl} \).
Why the other options are incorrect
Option (1) \(CH_3-I\) — Incorrect. Iodine is the least electronegative halogen, so \( \ce{CH3I} \) has the smallest dipole moment of the four.
Option (2) \(CH_3-Br\) — Incorrect. Its dipole moment is lower than that of \( \ce{CH3Cl} \), because the charge separation falls more than the bond length rises.
Option (3) \(CH_3-F\) — Incorrect. Fluorine is the most electronegative, but the C–F bond is so short that the dipole moment stays just below that of \( \ce{CH3Cl} \).
NCERT (2026-27 edition):
Nature of C–X Bond – …
