Import Question JSON

HINT: Dipole moment $ \propto $ Electronegativity difference $ \propto \frac{1}{\text{Bond angle}} $ Explanation: The dipole moment of a molecule depends on the electronegativity difference between the bonded atoms and the geometry of the molecule. Groups attached to a benzene ring can either withdraw electrons from the ring (electron-withdrawing groups, -I or -M effect) or donate electrons to the ring (electron-donating groups, +I or +M effect), thereby affecting the overall polarity and dipole moment of the molecule. Let's analyze the given molecules: **Molecule I:** o-nitrophenol ($ \text{C}_6\text{H}_4(\text{OH})(\text{NO}_2) $) * The $ \text{NO}_2 $ group is a strong electron-withdrawing group (both -I and -M effect). It pulls electron density away from the benzene ring. * The $ \text{OH} $ group is an electron-donating group by resonance (+M effect) but electron-withdrawing by induction (-I effect). However, the -I effect of $ \text{OH} $ is less significant than the electron-withdrawing power of $ \text{NO}_2 $. * The $ \text{NO}_2 $ and $ \text{OH} $ groups are ortho to each other. The dipole moments due to these groups will add up significantly because they are both polarizing the molecule in a way that contributes to a large net dipole moment, with the $ \text{NO}_2 $ group having a very strong pulling effect. **Molecule II:** o-dichlorobenzene ($ \text{C}_6\text{H}_4\text{Cl}_2 $) * Chlorine ($ \text{Cl} $) is an electron-withdrawing group due to its high electronegativity (-I effect), although it has a weak +M effect. * In o-dichlorobenzene, two $ \text{Cl} $ atoms are ortho to each other. Their individual bond dipoles will contribute to the overall molecular dipole moment. The vector sum of these two bond dipoles will result in a significant net dipole moment. **Molecule III:** o-xylene ($ \text{C}_6\text{H}_4(\text{CH}_3)_2 $) * The $ \text{CH}_3 $ group is an electron-donating group (+I effect and hyperconjugation). * In o-xylene, two $ \text{CH}_3 $ groups are ortho to each other. Since $ \text{CH}_3 $ groups are electron-donating, they push electron density into the ring. The overall polarity developed by these groups will be much smaller compared to electron-withdrawing groups. * The net dipole moment for molecules with electron-donating groups is generally lower than those with electron-withdrawing groups, especially strong ones. **Comparison of Dipole Moments:** * **Molecule I (o-nitrophenol):** The $ \text{NO}_2 $ group is a very strong electron-withdrawing group, causing a large separation of charge and thus a high dipole moment. The $ \text{OH} $ group also contributes to the overall polarity. * **Molecule II (o-dichlorobenzene):** Chlorine atoms are electronegative, leading to significant bond dipoles. While less strong than $ \text{NO}_2 $, the two $ \text{Cl} $ atoms in the ortho position still create a substantial net dipole moment. * **Molecule III (o-xylene):** Methyl groups are electron-donating groups. The polarity induced by these groups is much weaker, resulting in a much smaller dipole moment compared to molecules with strong electron-withdrawing groups. Therefore, the order of dipole moment will be: $ \text{I} > \text{II} > \text{III} $ Polarity develops when an electron is pulled from the benzene ring, $ \text{NO}_2 $ is having the highest pulling ability in given molecules, and $ \text{CH}_3 $ pushes the electron. The correct answer is option 1.