Final answer:
The oxidation state of carbon atoms in a molecule like C3O2 is typically derived by rules assigning -2 to oxygen and balancing the rest among the carbons; for C3O2, terminal carbons are often +2 and the central one 0. Non-planar structures do not affect oxidation states, which depend on bond type and count, not molecular geometry.
Step-by-step explanation:
The question concerns the oxidation states of carbon atoms in C3O2, specifically how the molecular structure, including possible non-planar configurations, might affect these oxidation states. In general, oxygen is assigned an oxidation number of -2 and hydrogen +1, unless with a metal.
For carbon, there is no specific rule and its oxidation number is determined by considering the atoms it's attached to and the overall charge of the molecule or ion. In our molecule C3O2, with terminal oxygen atoms at -2 oxidation state and adjacent carbon atoms at +2 oxidation state, it is assumed that the terminal carbons' oxidation numbers are balanced out by the oxygens, while the central carbon, traditionally, would end up with an oxidation state of 0.
However, if the central carbon forms bonds that polarize its electron distribution, the oxidation state could theoretically differ from zero. Yet, typically when considering bonds between the same element, in this case, carbon-carbon bonds, they are not factored into the oxidation state calculations.
So, the structure being bent or non-planar does not directly change the oxidation state, as it is only the number and type of bonds that directly affect it, not the geometrical arrangement.