beryllium violates octet?
Started by blinktonheim
Yes Be only has 2 electrons in the s subshell. So if can only make two bonds. So after it makes 2 bonds it only has 4 electrons.
Same thing with boron. It has 3 electrons before bonding so after it has 6 electrons.
so is that the same as like CaO2? would Ca be violating the "octet" rule there? or is it because Be can form covalent bonds but Ca cant??
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now that i think about it, CaO2 wouldnt exist, but CaO, when that bonds it loses 2 electrons thus having a full octet outer shell? but when Be bonds it doesnt have a full shell?
CaO2 would exist just not both O's bonded to Ca, it would have the polyatomic peroxide ion..
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they mention this in KBB. These elements can violate the octet:
H: 2e-
Li: 2e-
Be: 4e-
B: 6e-
Al: 6e-
S & P: >8 is possible.
i might be missing some uncommon ones. I know there are questions with noble gases that end up with >8 electrons too.
H: 2e-
Li: 2e-
Be: 4e-
B: 6e-
Al: 6e-
S & P: >8 is possible.
i might be missing some uncommon ones. I know there are questions with noble gases that end up with >8 electrons too.
