Final answer:
A resonance contributor for a molecule is drawn by moving pi bonds or lone pairs within the molecule's Lewis structure, ensuring the octet rule is obeyed, and all positions of the atoms remain unchanged. In the case of benzene, double bonds are shifted to illustrate the concept of resonance, which implies electron delocalization throughout the molecule.
Step-by-step explanation:
To draw a resonance contributor for a molecule, firstly, determine its initial Lewis structure. This will give you a clear starting point. For the resonance structure, pi bonds (π bonds) and lone pairs will be our focus since they can be rearranged while keeping the positions of the atoms and the sigma bonds (σ bonds) intact.
Take the example of benzene. Benzene is a hexagonal ring with alternating single and double bonds. To create a resonance structure, we can shift the position of the double bonds. In the resonance form, one double bond moves over by one position, turning the adjacent single bond into a double bond, and the double bond it replaced becomes a single bond. Remember, all atoms must still obey the octet rule, and you should not move or add any atoms.
Let's take a closer look at the process:
- Identify all bonds and lone pairs that can be moved within the molecule without breaking the octet rule.
- Select a pi bond or a lone pair to move.
- Shift the selected electrons to form a new pi bond or lone pair, ensuring that the overall charge and the number of unpaired electrons, if any, remain unchanged.
While drawing, keep in mind that these structures are hypothetical and help us understand the nature of electron delocalization in the molecule. Every resonance structure should reflect a valid, potential arrangement of electrons.
Note: The real electronic structure of a molecule is a hybrid of all its resonance forms, with the molecule adopting an electron distribution that is intermediate between those forms.