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

Hint : Dipole moment can be defined as the polarity between the two atoms in the molecule which are bonded by a chemical bond. It is a vector quantity that means it has magnitude as well as direction.

Complete Step By Step Answer:
First of all, we start with the definition of dipole moment:
Dipole moment: Dipole moment is the polarity between the two atoms in the molecule which are bonded by a chemical bond. It is a vector quantity as it consists of magnitude as well as direction.
Now, how we will analyse the total dipole moment:
As the dipole moment is a vector quantity, if the direction of the dipole moment is opposite, they are cancelled, and if same, they are going to be added.
Now, analysing the dipole moments of all the given elements.
As in $CC{l_4}$ , there are four dipole moments of $C - Cl$ bond, each two cancel each other. Hence, the total dipole moment will be zero.
In $CHC{l_3}$ , there are three dipole moments of $C - Cl$ bond, two will cancel each other and the one $C - Cl$ bond dipole moment is reduced by $C - H$ . Hence, it will be greater than $CC{l_4}$ .
In $C{H_2}C{l_2}$ , there are two dipole moments of $C - Cl$ bond and two of $C - H$ bond dipole moments, which are reduced by each other. Hence, it will be greater than $CHC{l_3}$ .
And in the last $C{H_3}Cl$ , the one dipole moment of $C - Cl$ is reduced by three $C - H$ dipole bonds. Hence, has the highest dipole moment.
Hence, the correct order of the decreasing order of the dipole moment as follows:
$C{H_3}Cl > C{H_2}C{l_2} > CHC{l_3} > CC{l_4}$
Hence, the statement is true.
Hence, the answer will be $1$ .

Note :
Dipole moment is numerically calculated as the charge multiplied by the distance between the bonds between the two given atoms. The unit of the dipole moment is coulomb meter $\left( {Cm} \right)$ also known as Debye $(D)$ .