A valid Lewis structure of __________ cannot be drawn without violating the octet rule.
Answer: BeH2
Why would BeH2 break the 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.
H:Be:H
To understand why a valid Lewis structure of BeH2 cannot be drawn without violating the octet rule, let's first review the octet rule. The octet rule states that atoms tend to gain, lose, or share electrons in order to achieve a stable arrangement of eight valence electrons (except for hydrogen, which can only achieve two valence electrons).
Beryllium (Be) is an element located in Group 2 of the periodic table, which means it has two valence electrons. Hydrogen (H) is in Group 1, so it has one valence electron.
When we try to draw a Lewis structure for BeH2, we would start by placing the beryllium atom at the center and hydrogen atoms around it. Since each hydrogen atom needs to share one electron to achieve the stable configuration of two electrons (like helium), we would connect each hydrogen atom to the beryllium atom with a single bond.
However, in this arrangement, beryllium would only have four electrons (two from the bond with each hydrogen) around it. This is fewer than the octet rule suggests, as beryllium can only accommodate four electrons due to its small atomic size.
Since BeH2 cannot satisfy the octet rule while maintaining a valid Lewis structure, it is an example of a compound that violates the octet rule. In such cases, compounds often form with less than eight electrons around the central atom (referred to as electron-deficient compounds) due to the unique characteristics of the elements involved.