To solve this problem, we need to determine the hybridization and the number of electrons around the central atom in $$BF$_3. \\$Step 1: Determine the central atom and its valence electrons.$\\$• Boron (B) is the central atom in $$BF$_3. \\$• Boron belongs to group 13 of the periodic table and has 3 valence electrons.$\\$Step 2: Determine the hybridization of the central atom.$\\$• In $$BF$_3,$ boron forms three sigma bonds with three fluorine atoms.$\\$• The geometry of $$BF$_3$ is trigonal planar, which corresponds to $sp^2$ hybridization.$\\$Step 3: Calculate the number of electrons around the central atom.$\\$• Each B-F bond involves a pair of electrons, one from boron and one from fluorine.$\\$• Boron contributes 3 electrons, one for each bond.$\\$• Therefore, there are 3 pairs of electrons (6 electrons in total) around boron.$\\$Conclusion:$\\$• The hybridization of the central atom (B) in $$BF$_3$ is $sp^2. \\$• The number of electrons around the central atom is 6.$\\$Therefore, the correct option is Option 3: $sp^2$ and 6.