Question

Arrange the following in the decreasing dipole moment : ${CH}_{3}Cl,{CH}_2{Cl}_2,CH{Cl}_3,C{Cl}_4$ is: ${CH}_{3}Cl>{CH}_2{Cl}_2>CH{Cl}_3>C{Cl}_4$.
if true enter 1, else enter 0.

• A. 1

Answer 1

The correct option is A 1
Decreasing dipole moment : ${CH}_{3}Cl>{CH}_2{Cl}_2>CH{Cl}_3>C{Cl}_4$.

The bond dipoles$C-H$ and $C-Cl$are in opposite direction.

In $C{H}_{3}Cl$, the resultant of $3C-H$ bond dipoles is reduced by one C-Cl bond dipole.

In $C{H}_{2}Cl$, the resultant $2C-H$ bond dipoles is reduced by the resultant 2 C-Cl bond dipoles. Hence, the dipole moment of $C{H}_{2}C{l}_2$ is lower than the dipole moment of $C{H}_{3}Cl$.

Similarly, $CHC{l}_3$ has lower dipole moment than In $C{H}_{2}C{l}_2$,

The net dipole moment of $CC{l}_4$ is zero because four $C-Cl$ bond dipoles cancel each other.