Beryllium Octet Rule
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Beryllium Octet Rule
Can someone please explain why beryllium does not follow the octet rule in some cases like BeCl2???
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Re: Beryllium Octet Rule
The first four elements of the periodic table up to boron never obtain an octet of electrons
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Re: Beryllium Octet Rule
The Boron in BF3 does not have a full octet because BF3 is a Lewis Acid, and the empty p-orbital on B can accept an electron lone pair from a Lewis Base.
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Re: Beryllium Octet Rule
To add to this, you can tell that beryllium will now have a full octet because of formal charge. In BeCl2, Be has 2 shared bonds, making its formal charge 0, and Cl has 6 unshared electrons and 1 shared bond, making the formal charge of both Cl atoms in BeCl2 0. This proves to be the most stable structure for BeCl2.
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Re: Beryllium Octet Rule
Is the reason Be doesn't need an octet because of the amount of valence electrons it has?
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Re: Beryllium Octet Rule
Beryllium doesn't fulfill the octet rules because of its valence electrons and the orbitals they occupy. Just like H, He, Li, and Be, they are all exceptions to the octet rules mainly because they don't need 8 to feel complete. Their s-orbitals do not need 8 electrons to complete the shell to feel full, so they don't follow the octet rule.
Boron is a weird exception. Boron and Aluminum (as well as some other group 13 compounds) do not require an octet as well in their compounds. The textbook was not clear on why this occurred in group 13 compounds, but they said that they may have an incomplete octet, or require a halogen atom to act as bridge to give them more.
Boron is a weird exception. Boron and Aluminum (as well as some other group 13 compounds) do not require an octet as well in their compounds. The textbook was not clear on why this occurred in group 13 compounds, but they said that they may have an incomplete octet, or require a halogen atom to act as bridge to give them more.
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