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(B) The dipole moment $(\mu)$ depends on the polarity of bonds and the geometry of the molecule.$1$. $p$-Dichlorobenzene $(IV)$: The two $C-Cl$ bond d

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$IV < I < II < III$

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(B) The dipole moment $(\mu)$ depends on the polarity of bonds and the geometry of the molecule.$1$. $p$-Dichlorobenzene $(IV)$: The two $C-Cl$ bond dipoles are equal and opposite, canceling each other out. Thus, $\mu = 0 \ D$.$2$. Toluene $(I)$: The methyl group is electron-donating, creating a small dipole moment $(\mu \approx 0.4 \ D)$.$3$. $m$-Dichlorobenzene $(II)$: The angle between the two $C-Cl$ bonds is $120^{\circ}$. The resultant dipole is $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos(120^{\circ})} = \mu_1$ (where $\mu_1$ is the dipole of chlorobenzene, $\approx 1.5 \ D$).$4$. $o$-Dichlorobenzene $(III)$: The angle between the two $C-Cl$ bonds is $60^{\circ}$. The resultant dipole is $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos(60^{\circ})} \approx 2.5 \ D$.Comparing these values: $0 (IV) Therefore, the increasing order is $IV < I < II < III$.

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