Arrange the following in the decreasing order of their Dipole moments. a. Chlorobenzene b. 1,2-Dichlorobenzene c. 1,3-Dichlorobenzene d. 1,4-Dichlorobenzene.

Let $\mu$ be the dipole moment of a single C-Cl bond. The net dipole moment is $\mu_{net} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta}$, where $\theta$ is the angle between the two C-Cl bonds.

(b) 1,2-Dichlorobenzene (ortho): $\theta = 60°$
$\mu_{net} = \sqrt{\mu^2 + \mu^2 + 2\mu^2\cos60°} = \sqrt{3}\mu$ (Highest)

(d) 1,4-Dichlorobenzene (para): $\theta = 180°$
$\mu_{net} = \sqrt{\mu^2 + \mu^2 + 2\mu^2\cos180°} = 0$ (Lowest)

(c) 1,3-Dichlorobenzene (meta): $\theta = 120°$
$\mu_{net} = \sqrt{\mu^2 + \mu^2 + 2\mu^2\cos120°} = \mu$

(a) Chlorobenzene: theoretically $\mu_{net} = \mu$, but due to the $-I$ effect of the second Cl in 1,3-dichlorobenzene slightly reducing individual bond dipoles, chlorobenzene has a slightly higher dipole moment than 1,3-dichlorobenzene.

Decreasing order: $b > a > c > d$