### Related Formula mutextnet = sqrtmu_1^2 + mu_2^2 + 2mu_1mu_2costheta $\mu{\text{net}} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta} $ textBoiling point propto textDipole-dipole interactions + textVan der Waals forces$\text{Boiling point} \propto \text{Dipole-dipole interactions} + \text{Van der Waals forces}$ ### Core Logic Let's analyze the visual structures alongside their scientific orientations: - Statement (I) compares 1,2-dichlorobenzene and 1,2-dibromobenzene. Chlorine has a higher electronegativity than bromine, creating a larger bond dipole. The vacant d-orbital interactions do not invert this baseline dipole trend. Thus, 1,2-dichlorobenzene is more polar, making Statement I correct. - Statement (II) evaluates dihalobenzene isomers. For the para-isomer, individual bond dipoles are oriented at 180^circ$180^{\circ}$, cancelling out completely: mutextpara = 0 $\mu{\text{para}} = 0 $ Since mu_textmeta > 0$\mu_{\text{meta}} > 0$, the para-isomer is *less* polar than the meta-isomer. This directly falsifies Statement II. ### Step 1: Spatial Alignments The geometric configurations map out as follows:

Hence, Statement I is correct, but Statement II is incorrect. ### Pattern Recognition Dipole tracking rule: Para-substituted benzenes with identical groups possess a structural center of inversion, guaranteeing a net dipole moment of exactly zero (mu = 0$\mu = 0$). They can never be more polar than any asymmetric ortho or meta structural isomer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes