beryllium violates octet?

[blinktonheim]

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I just read that in the Exam Krackers Ochem book, has anyone heard of this before??

 

[UndergradGuy7]

I just read that in the Exam Krackers Ochem book, has anyone heard of this before??

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.

 

[blinktonheim]

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??

 

[casper22]

Agree with Undergradguy...

I heard it yesterday in Chad's video too...

 

[blinktonheim]

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?

 

[blinktonheim]

CaO2 would exist just not both O's bonded to Ca, it would have the polyatomic peroxide ion..

 

[UndergradGuy7]

CaO2 would exist just not both O's bonded to Ca, it would have the polyatomic peroxide ion..

nvm

 

[Nat426]

It helps if at some point you memorize the neutral state of one row of molecules. Meaning, now many bonds does C have to be neutral 4, N? 3 O? 2...

 

[CANgnome]

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.