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Boron is an exception to the octet rule. This occurs because a coordinate covalent bond is formed. In coordinate covalent bonds, one of the two atoms involved in the bond provides both of the shared electrons as opposed to each atom involved in the bond providing one electron. Additionally, if you were to calculate Boron's formal charge in a molecule like BF3, it would equal zero. Therefore, you know that even without an octet, Boron is stable in this molecule.
What's already been said is true, but I also wanted to add that it's possible for B (boron) to complete its octet if another atom provides two electrons in what's called a coordinate covalent bond. To reiterate the example from today's lecture, BF3 + F^- --> BF4^-. In the product, B forms an octet due to the two electrons given by F^- (remember that F^- has 8 valence electrons).
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