Consider the experimentally determined dipole moments of the various positional isomers of dichlorob — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

A) True. In $m$ -dichlorobenzene, the angle $\theta$ between the two $C-Cl$ bonds is $120^\circ$. The resultant vector is $\mu_R = \sqrt{\mu^2 + \mu^2 + 2\mu^2\cos(120^\circ)} = \sqrt{2\mu^2 + 2\mu^2(-0.5)} = \sqrt{\mu^2} = \mu$. B) True. In $p$ -dichlorobenzene, $\theta = 180^\circ$. The vectors are perfectly opposed, so $\mu_R = 0$. C) True. Since $\mu_R = 2\mu\cos(\theta/2)$, the resultant dipole moment decreases as the angle $\theta$ increases. $o$ -isomer ($60^\circ$) > $m$ -isomer ($120^\circ$) > $p$ -isomer ($180^\circ$). D) False. For $o$ -dichlorobenzene ($\theta = 60^\circ$), $\mu_R = \sqrt{\mu^2 + \mu^2 + 2\mu^2\cos(60^\circ)} = \sqrt{2\mu^2 + \mu^2} = \sqrt{3}\mu$, not $\sqrt{2}\mu$.