Octet question

[LexyDenaburg_3A]

When drawing Lewis structures, how can we tell if it's okay that a molecule doesn't have an octet? For example, Boron on BF3 (lecture example)? How do we know that it's correct if Boron doesn't form an octet?

[Crystal Yu 1D]

I believe that some of the exceptions will simply have to be memorized, but a good rule of thumb when determining the structure is to check formal charges. When the formal charges are 0, the atom is in a stable state. For BF3, the formal charges are 0, so despite not having an octet for B, the structure makes sense. Hope this helps! :)

[Jordan Tatang 3L]

Adding on to that, some other atoms that don't have an octet are hydrogen, helium, lithium, and beryllium.

[Trevor_Ramsey_3H]

Also going off the previous response, he mentioned today in lecture that P, S, and Cl can accommodate more than 8 valence electrons so this goes against the octet rule as well, and like was stated above, it relies on the formal charge and the more stable structure to be able to determine these types of bonding structures.

[Helena Hu 3E]

I would just memorize the exceptions --> hydrogen is pretty easy to get since it is one of the most common elements, but take time to know the ones that do not follow the rule. It'll come naturally with a lot of practice. For the most part, elements need to satisfy the octet rule.
Hope this helps!

[Katie Phan 1K]

I don't think it's too difficult to memorize the exceptions! Also sometimes when you look at a structure it just makes more sense when you think about its stability and formal change.

[Evonne Hsu 1J]

Yes! There are a lot of exceptions for atoms that do not have an octet, as discussed in Lavelle's lectures. For example, N, S, Be, etc.

[Charlotte Adams 1A]

I think we just have to memorize the exceptions. For example, Boron doesn't usually have an octet but has 6 valence electrons. Both Boron and Aluminum are commonly involved in Lewis acids and base reactions.