Final answer:
Resonance involves drawing multiple contributing structures to represent electron delocalization. Stabilization is affected by p-orbital sizes, electronegativity, and charge distribution. Resonance is exemplified in molecules like sulfur dioxide and the carbonate ion.
Step-by-step explanation:
The concept of resonance in chemistry involves drawing multiple resonance contributors or contributing structures to represent the delocalization of electrons within a molecule. In the process of drawing a resonance contributor produced by the indicated flow of electrons, it's important to follow the resonance rules. One must move π-electrons or lone pairs, without moving σ-electrons, to create a new π-bond or a lone pair.
Resonance stabilization is affected by various factors, such as the size of the p-orbitals involved in delocalization, the electronegativity of the atoms, and the equal or unequal distribution of charge in the resulting structures. For instance, a negative charge is more stable on a more electronegative atom, like oxygen, and this will be a major contributing structure. Conversely, the contribution is minor if the negative charge is on a less electronegative atom or if there is a positive charge on an electronegative atom.
In the provided examples, resonance is less significant when there is size mismatch between the p-orbitals of the atoms involved (like 3p-orbital of Cl and 2p-orbital of C). Additionally, an atom's willingness to donate a lone pair for resonance is also a determining factor, making resonance more significant in cases where similar sized p-orbitals and less electronegative atoms (like N in comparison to O) are involved.
For compounds like sulfur dioxide (SO2) and the carbonate ion (CO32-), drawing resonance structures is essential to describe the distribution of electrons and to understand the stabilization provided by resonance effects.