Dihydroxybenzene, \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}\), exists in three forms (isomers) called ortho, meta, and para: Which of these has a nonzero dipole moment?

First, we need to draw the Lewis structures for each of the isomers. Here are the Lewis structures for ortho, meta, and para dihydroxybenzene: Ortho: ``` O || H₁O -- C₁ C₂ -- O₂H // \\ \ HC₄──C₃──C₆──C₅ \\ / -------- H₆ ``` Meta: ``` O₁H \ C₁ // \ HC₄ C₂ -- O₂H \ // C₃──C₆ // \ HC₅ C₁ \ O₃H ``` Para: ``` O₁H \ C₁ // \ HC₄ C₂ \\ C₃ \ // C₆ ---O₂H // HC₅ ```

We need to look at the molecular geometry of the structures to understand their shape and symmetry to help determine their polarity. Ortho: The molecular geometry of ortho dihydroxybenzene is not symmetrical. The hydroxyl groups are on adjacent carbon atoms. Meta: The molecular geometry of meta dihydroxybenzene is more symmetrical than ortho, but not completely symmetrical. The hydroxyl groups are separated by a single carbon atom. Para: The molecular geometry of para dihydroxybenzene is symmetrical. The hydroxyl groups are on opposite sides of the benzene ring.

To determine the polarity of each isomer, we need to consider the electronegativity of the atoms involved and the molecular geometry we determined in Step 2. Ortho: Due to the asymmetrical molecular geometry, there will be an uneven distribution of electron density, resulting in a nonzero dipole moment. Meta: The molecular geometry is not completely symmetrical, but the hydroxyl groups are separated by a single carbon atom. Due to this separation, there is a stronger combined dipole moment, resulting in a nonzero dipole moment. Para: The molecular geometry is symmetrical, meaning the dipole moments for the hydroxyl groups are equal and opposite, ultimately canceling each other out. This results in a zero dipole moment. So, ortho and meta dihydroxybenzene have nonzero dipole moments, while para dihydroxybenzene has a zero dipole moment.