Q.Explain why $\text{CH}_3\text{F}$ has the lowest boiling point and $\text{CH}_3\text{I}$ the highest among the methyl halides $\text{CH}_3\text{F}$, $\text{CH}_3\text{Cl}$, $\text{CH}_3\text{Br}$ and $\text{CH}_3\text{I}$, even though the $\text{C--F}$ bond is the most polar of the four.

Boiling point depends on the total strength of intermolecular attraction between molecules, which is the sum of a dipole-dipole contribution (from the polar C--X bond) and a London dispersion contribution (which grows with how easily the electron cloud around each atom can be distorted, i.e. polarisability, and generally with atomic/molecular mass). Fluorine has the highest electronegativity of the halogens, so $\text{CH}_3\text{F}$ has the most polar C--X bond of the four methyl halides -- but fluorine is also the smallest and least polarisable halogen, so its dispersion contribution is the weakest. Iodine is the opposite: the least electronegative and least polar bond, but by far the largest, most polarisable, most massive halogen, giving the strongest dispersion contribution. Since the dispersion contribution changes far more dramatically across this series than the dipole contribution does, it dominates the trend, and boiling point rises steadily from $\text{CH}_3\text{F}$ to $\text{CH}_3\text{I}$.

Separate the two contributions to intermolecular attraction (dipole from bond polarity; dispersion from polarisability/mass) and identify which one actually dominates for this series.

Assuming higher bond polarity must always mean a higher boiling point -- true for isolated comparisons of otherwise similar molecules, but not when polarisability changes dramatically across the series, as it does here.