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(B) The dipole moment $(mu)$ depends on the vector sum of individual bond dipoles.$1$. For $p$-dichlorobenzene $(IV)$,the two $C-Cl$ bond dipoles are

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$IV < I < II < III$

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(B) The dipole moment $(mu)$ depends on the vector sum of individual bond dipoles.$1$. For $p$-dichlorobenzene $(IV)$,the two $C-Cl$ bond dipoles are equal and opposite,so $\mu = 0 \ D$.$2$. For toluene $(I)$,the methyl group is electron-donating,resulting in a small dipole moment $(\mu \approx 0.4 \ D)$.$3$. For $m$-dichlorobenzene $(II)$,the angle between the two $C-Cl$ bonds is $120^{\circ}$,so $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos(120^{\circ})} = \mu_{C-Cl} \approx 1.48 \ D$.$4$. For $o$-dichlorobenzene $(III)$,the angle between the two $C-Cl$ bonds is $60^{\circ}$,so $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos(60^{\circ})} = \sqrt{3} \mu_{C-Cl} \approx 2.54 \ D$.Thus,the increasing order is $IV < I < II < III$.

Arrange the following compounds in the increasing order of their dipole moment:
$(I)$ Toluene
$(II)$ $m$-Dichlorobenzene
$(III)$ $o$-Dichlorobenzene
$(IV)$ $p$-Dichlorobenzene

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Arrange the following compounds in the increasing order of their dipole moment:$(I)$ Toluene$(II)$ $m$-Dichlorobenzene$(III)$ $o$-Dichlorobenzene$(IV)$ $p$-Dichlorobenzene