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

\(CH_3Cl > CH_3F >CH_3Br > CH_3I \)

The correct answer is Option (3) → \(CH_3Cl > CH_3F >CH_3Br > CH_3I \)

The dipole moment (μ) of a bond is defined by the following formula:

$μ = q× d$

Where:

q is the magnitude of the partial charge (determined by electronegativity).

d is the bond length (the distance between the charge centers).

The Tug-of-war

1. Electronegativity (q): Fluorine is the most electronegative element ($F> Cl > Br> I$). Based on charge alone, $CH_3F$ should have the highest dipole moment.

2. Bond Length (d): As you go down the halogen group, the atomic size increases significantly. The $C-Cl$ bond is much longer than the $C - F$ bond.

In the case of methyl chloride ($CH_3Cl$), the increase in bond length (d) is large enough to overcompensate for the slightly lower electronegativity compared to Fluorine. This results in a higher product ($q × d$).

For Bromine and lodine, the electronegativity drops so significantly that even the longer bond lengths cannot make up the difference, which is why they fall to the end of the list.

Hence, the correct order is \(CH_3Cl > CH_3F >CH_3Br > CH_3I \)