Aluminum and Boron  [ENDORSED]

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Haorui Li 1A
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Joined: Mon Jun 17, 2019 7:23 am

Aluminum and Boron

Postby Haorui Li 1A » Sun Jul 14, 2019 2:19 pm

I learned this in high school, but can someone explain why when we draw the Lewis structures, Aluminum and Boron have only 6 electrons around them and they satisfy the octet rule?

Ethan McCarthy 1F
Posts: 31
Joined: Wed Feb 27, 2019 12:17 am

Re: Aluminum and Boron  [ENDORSED]

Postby Ethan McCarthy 1F » Mon Jul 15, 2019 12:19 am

From the example in lecture (with BF3), the fluorine is more electronegative than the boron, so when finding the most stable Lewis diagram for BF3 we use electronegativity to determine that fluorine is unlikely to give up electrons to fill the valence shell of boron (the less electronegative atom). The valence electron shell of 6 allows boron to fill its octet by forming a coordinate covalent bond, where the atom/species that forms the bond donates both of the electrons that participate in the covalent bond (F- in the case of the reaction BF3 --> BF4-).

hannabarlow1A
Posts: 37
Joined: Mon Jun 17, 2019 7:23 am

Re: Aluminum and Boron

Postby hannabarlow1A » Tue Jul 16, 2019 8:16 am

Boron and Aluminum need 5 electrons to complete an octet, however, they are exceptions to the octet rule because they are involved in lewis acid-base reactions. They can have a complete octet if another atom provides both electrons for a coordinate covalent bond. In such cases, the electron pair donator is the Lewis base and the electron pair receiver is the Lewis acid.


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