Chemistry - How to determine the worst resonance structures out of a given set?
Solution 1:
(C) is technically a valid resonance structure. You can derive it from (B) by moving the electron pair from the double bond between the carbons 2 and 3 (numbering from the left: this may not be correct from a nomenclature standpoint, but I'm using it for this answer) onto carbon 2 and moving the electron pair from the double bond between the oxygen and the carbon onto the oxygen.
(C) is least stable because it is, simply put, absolutely disgusting. Charge separation (carbons two and three) is gross, you have two cations in close proximity (carbons three and four), a terminal carbocation (carbon 1), and three atoms with empty valence shells (carbons 1, 3, and 4). So while the resonance structure is valid enough for (C) to be considered a resonance structure, that resonance structure happens to be the spawn of the devil.
That said, the compound in the answers isn't the same as the compound named in the question, like ron said.
As to the first part of your question, the filled valence shells are more strictly necessary for the second row elements because they don't have d orbitals to confound the picture.
Solution 2:
As ron stated in the question’s comments, none of the resonance structures display 1-methoxybuta-1,3-diene. In fact, they show an extremely unstable (if generateable) dication. Choose a random $+$, replace it with a $-$ (don’t choose the one on the oxygen) to get resonances for the structure the question asks for. That should also semi-answer your question about the two positive charges.
As for the question on octets: It is the more traditional viewpoint that elements of the third and higher periods such as sulphur could expand their octets (e.g. in $\ce{SO2}$). This was generally explained with contribution by d-orbitals (and Pauling’s dislike for charge separation).
Nowadays, however, these structures are more generally considered as abiding by the octet rule invoking charge separation where necessary and multicentred bonds where applicable. The reason for this is that d-orbitals are too far removed energetically from s and p of the same shells to actively take part in bonding.
Finally, there are sub-octet structures to consider (heptets in radicals or sextets in carbocations, carbenes and the like). These exist and are undebated. This violation of the octet rule is ‘less bad’ (by a mile) than a super-octet. Why? Well, you’re not filling electrons into high-energy contributed orbitals but rather removing them from stable ones.
That said, when considering the first of these octet paragraphs, it is a lot harder for me to think of compounds that have sub-octets for elements of higher periods — most examples I know are carbon or nitrogen centred.
On the topic of the ugliness of (c), Breaking Bioinformatics has answered that nicely. But you could have arrived at the same conclusion by checking your book’s rules:
Does not apply here. No resonance has an octet on every carbon.
(c) has four formal charges, all the others have two. Stop here, mark (c) and grab a piece of cake.
Solution 3:
Following rules are 100%-accurate and can be used to determine stability:
more the no. of π-bonds more stable the contributing structure is.
if the no. of π-bonds are equal then … the molecule with no charge or with no charge separation is more stable.
if the molecules are not containing any charges i.e neutral then … the molecule in which all atoms have a complete octet is more stable.
if the atoms in a molecule have an incomplete octet or have charge separation then … the negative charge is more stable on a more electronegative atom (like N,O) and positive charge on a less electronegative atom.
if above the condition is satisfied, too, in a molecule then … if in a molecule unlike charges are closer then it is more stable than the molecule which has unlike charges far away.
And if like charges are closer to each other then it is more unstable than in which like charges are far away.