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From ChemWiki by UC Davis:

As a side note, it is important to note that $\ce{BF3}$ frequently bonds with a $\ce{F-}$ ion in order to form $\ce{BF4-}$ rather than staying as $\ce{BF3}$. This structure completes boron's octet and it is more common in nature.

Wait, what? Boron has three electrons. Bonding with four fluoride ions gives it four electrons — these are covalent bonds (shared electrons). That makes for a grand total of 7 valence electrons. Did the author mean $\ce{BF_5^2-}$?

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As you point out, the $\ce{B}$ in $\ce{BF_3}$ is sharing a total of 6 electrons, 3 from the boron and 3 from the three fluorines. Now add a fluoride anion that brings 2 more electrons to share. This now makes for a total of 8 electrons around the boron.

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