(i)

Dichlormethane (CH${}_2$Cl${}_2$)
μ = 1.60D

(ii) 

Chloroform (CHCl${}_3$)

μ = 1.08D

(iii) 

Carbon tetrachloride (CCl${}_4$)
μ = 0D

CCl${}_4$ is a symmetrical molecule. Therefore, the dipole moments of all four C−Cl bonds
cancel each other. Hence, its resultant dipole moment is zero.
As shown in the above figure, in CHCl${}_3$, the resultant of dipole moments of two C−Cl bonds
is opposed by the resultant of dipole moments of one C−H bond and one C−Cl bond. Since
the resultant of one C−H bond and one C−Cl bond dipole moments is smaller than two
C−Cl bonds, the opposition is to a small extent. As a result, CHCl${}_3$ has a small dipole
moment of 1.08 D.
On the other hand, in case of CH${}_2$Cl${}_2$, the resultant of the dipole moments of two C−Cl
bonds is strengthened by the resultant of the dipole moments of two C−H bonds. As a

result, CH${}_2$Cl${}_2$ has a higher dipole moment of 1.60 D than CHCl3 i.e., CH${}_2$Cl${}_2$ has the highest
dipole moment.
Hence, the given compounds can be arranged in the increasing order of their dipole
moments as:
CCl${}_4$ < CHCl${}_3$ < CH${}_2$Cl${}_2$