F has the highest electronegativity among the three halogens, Cl, Br, and I. As F and C are closer together in CH₃F, the dipole moment decreases. CH₃Cl follows CH₃F, CH₃Br, and CH₃I. As a result, CH₃Cl will possess the greatest dipole moment.

The differential in electronegativity between carbon and the halogens determines the length of the C-halogen bond in full. The bond length decreases as the difference increases.$\mathrm{I}<\mathrm{Br}<\mathrm{Cl}<\mathrm{F}$ are the order of halogens' electronegativity. As a result, CH is the proper order in which C-halogen bond length increases.
${\mathrm{CH}}_3\mathrm{F}<{\mathrm{CH}}_3\mathrm{Cl}<{\mathrm{CH}}_3\mathrm{Br}<{\mathrm{CH}}_3\mathrm{I}$.

There are no lone pairs in ${\mathrm{CH}}_3\mathrm{F}={\mathrm{CH}}_3\mathrm{Cl}={\mathrm{CH}}_3\mathrm{Br}$.

Thus, the option A is correct.