The hybridization of B in the BF3 molecule is
BF3 has a boron atom with three outer-shell electrons in its ground state and three fluorine atoms containing seven outer electrons. Further, if we observe closely, one boron electron is unpaired in the ground state. During the formation of this compound, the 2s orbital and two 2p orbitals hybridize. Only one of the empty p-orbital is left behind as the lone pair. In short, Boron needs 3 hybridized orbitals to make bonds with 3 atoms of F where the 2pz orbitals get overlapped with these hybridized sp2 orbitals and bonds are formed.
RAO, K. V. (1964). Chemical Bonding.