posted by Anon on .
A valid Lewis structure of __________ cannot be drawn without violating the octet rule.
Why would BeH2 break the octet?
Is it because it only has four electrons to begin with with only two valence electrons? There's no way it can possible even make an octet?
Yes but BeH2 is not that common. I think BeCl2 is a better example. Boron is another that breaks the octet rule; e.g., BF3.
Be and B are the huge ones on breaking the octet rule. But, for that question they gave BeH2 so, I guess reasoning is that they are the exceptions? It's multiple choice. but I still want to understand it.