Question

Arrange the following in the decreasing dipole moment: $C{H_3}Cl,C{H_2}C{l_2},CHC{l_3},CC{l_4}$ is: $C{H_3}Cl > C{H_2}C{l_2} > CHC{l_3} > CC{l_4}$ .
if true enter 1, else enter 0.

Answer 1

In order to answer the question, to know whether the given arrangement is true or false, we will explain on what basis the dipole moment is decreasing and then we will rearrange the given compounds in the decreasing dipole moment sequence.

Complete answer:
Decreasing dipole moment:
$C{H_3}Cl > C{H_2}C{l_2} > CHC{l_3} > CC{l_4}$
The bond dipoles $C - H$ and $C - Cl$ are in opposite directions.
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 $2C - Cl$ bond dipoles. Hence, the dipole moment of $C{H_2}C{H_2}$ is lower than the dipole moment of $C{H_3}Cl$ .
Similarly, $CHC{l_3}$ has a lower dipole moment than in $C{H_2}C{H_2}$ .
The net dipole moment of $CC{l_4}$ is zero because four $C - Cl$ bond dipoles cancel each-other.
Hence, the given arrangement is true.

Note:
The electronegativity difference between chemically bound atoms or elements is the primary cause of dipole moment growth. The separation of positive and negative charges in a compound is known as polar character. Bond dipole moment is a calculation of the polar character of a chemical bond in a molecule between two atoms.