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Q.
The correct decreasing order of dipole moment in $ CH_{3}Cl,\,CH_{3} Br$ and $ CH_{3}F $ is
Delhi UMET/DPMTDelhi UMET/DPMT 2011Chemical Bonding and Molecular Structure
Solution:
As we move down the group in the Periodic Table from $F$ to $I$, the electronegativity of halogen decreases, therefore the polarity of the $C$ ? $X$ bond and hence the dipole moment of the haloalkane should also decrease accordingly. But the dipole moment of $CH _{3} F$ is slightly lower than that of $CH _{3} Cl$. The reason being that although the magnitude of -ve charge on the $F$ atom is much higher than that on the $C l$ atom but due to small size of $F$ as compared to $Cl$, the $C?F$ bond distance is so small that the product of charge and distance $(\mu=q \times d)$, ie, dipole moment of $CH _{3} F$ turns out to be slightly lower than that of $CH _{3} Cl$. Therefore, the order of dipole moment is
$\underset{1.860\,D}{CH_3Cl} > \underset{1.847\,D}{CH_3F} > \underset{1.830\,D}{CH_3Br}$