Question

The correct decreasing order of dipole moment in $CH_{3}Cl,\,CH_{3} Br$ and $CH_{3}F$ is

• A. $CH_{3}F > CH_{3} Cl > CH_{3}Br$
• B. $CH_{3}F > CH_{3}Br > CH_{3}Cl$
• C. $CH_{3}Cl > CH_{3} F > CH_{3}Br$
• D. $CH_{3}Cl > CH_{3} Br > CH_{3}F$

Answer 1

Answer: (C)

$CH_{3}Cl > CH_{3} F > CH_{3}Br$

Answer 2

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
$\underset{1.860\,D}{CH_3Cl} > \underset{1.847\,D}{CH_3F} > \underset{1.830\,D}{CH_3Br}$