Consider the experimentally determined dipole moments of the various positional isomers of dichlorob — Haloalkanes and Haloarenes Chemistry Question
Question
Consider the experimentally determined dipole moments of the various positional isomers of dichlorobenzene. Which of the following statements correctly interpret their dipole characteristics using basic vector principles?
💡 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 $\sqrt{2}\mu$.