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Aluminum and Boron

Posted: Sun Jul 14, 2019 2:19 pm
by Haorui Li 1A
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?

Re: Aluminum and Boron  [ENDORSED]

Posted: Mon Jul 15, 2019 12:19 am
by Ethan McCarthy 1F
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-).

Re: Aluminum and Boron

Posted: Tue Jul 16, 2019 8:16 am
by hannabarlow1A
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.