1,2-dihydroxybenzene is obtained when two of the adjacent hydrogen atoms in benzene are replaced with an OH group. A skeleton of the molecule is shown here. (a) Complete a Lewis structure for the molecule using bonds and electron pairs as needed. (b) Are there any resonance structures for the molecule? If so, sketch them. (c) Are the resonance structures in (a) and (b) equivalent to one another as they are in benzene?

Since the carbon atoms in benzene are sp2 hybridized, they are capable of forming three sigma bonds. By replacing two adjacent hydrogen atoms with hydroxyl (-OH) groups, we get 1,2-dihydroxybenzene. We can start by drawing the benzene ring with the carbon and hydrogen atoms and then place the OH groups on the adjacent carbon atoms. First, we need to draw a hexagonal ring (benzene ring), and each corner of the hexagon represents a carbon atom connected to each other using either single or double bonds. Next, connect each carbon atom with a hydrogen atom except for those adjacent carbons that have the hydroxyl groups attached. Now, replace two adjacent hydrogen atoms with hydroxyl groups. To complete the Lewis structure, each carbon atom should form 3 sigma bonds, and the OH group should be drawn with the oxygen atom bonded to a hydrogen atom and a carbon. Finally, complete the pi bonds within the benzene ring according to aromaticity rules.

Now that we have the Lewis structure for 1,2-dihydroxybenzene, we can proceed to look for possible resonance structures. An alternative resonance form can be drawn by shifting the pi bond, and if there is still scope for rearrangement of the pi electrons, we can draw another form. Considering the given molecule, we can see that the pi bonds on the benzene ring can be rearranged by shifting the double bond on the left side of the hydroxyl groups. This results in another resonance form for 1,2-dihydroxybenzene.

Finally, we have to compare the resonance structures in terms of their equivalence. Two resonance structures are said to be equivalent if the arrangement of atoms, sigma bonds, and electrons are identical in both of them, and only the positions of the pi bonds are different. In the case of benzene, the resonance structures look identical in terms of atom arrangement and bonds, except for the location of pi bonds. However, in the case of 1,2-dihydroxybenzene, the presence of the two OH groups prevents the pi bonds from being distributed symmetrically. As a result, the two resonance structures are not equivalent.