BF₃ · boron trifluoride
Boron is stable with only six electrons here. Forcing a B=F double bond would put a positive formal charge on fluorine, the most electronegative element, so the electron-deficient structure wins.
Begin by counting , the outer-shell electrons that participate in bonding: boron is a Group 13 element, so it has 13 - 10 = 3 valence electrons; each fluorine is a Group 17 element, so it has 17 - 10 = 7 valence electrons, and with 3 of them that is 7 × 3 = 21, which totals 24 valence electrons, the only electrons available to build the structure, so every bond and placed later must be drawn from this count.
The is the least electronegative , since it must share its electrons among the most neighbors, and here that atom is boron. A more electronegative atom holds its electrons too tightly to serve as the hub, so it stays terminal.
Begin the structure with a from B to each of the 3 terminal . Each single bond is a shared pair, so these bonds together use 6 electrons and 18 of the 24 are still available.
Place on the terminal before the , since a completes its with lone pairs once its is drawn. The F atoms take 9 lone pairs in total (18 electrons), and 0 electrons remain.
With the terminal octets complete, no electrons remain, so B receives no at this stage.
Now verify the electron count on every , terminal atoms included, not only the . B: 6 shared + 0 nonbonding = 6 (6 electrons); each F: 2 shared + 6 nonbonding = 8 (a full ).
Check the on each . Formal charge equals an atom's , minus its nonbonding (lone-pair) electrons, minus half of its ; the bonding electrons are halved because the two atoms in a bond share that pair equally, so each is assigned one electron of it. Here all 4 atoms have a formal charge of zero, which is the ideal outcome and confirms this is the preferred structure.
B is surrounded by only 6 electrons, an , and that is correct here. Forming an additional bond to reach eight would place a positive on a more electronegative , which is less stable than leaving B electron-deficient.
Watch out
It's tempting to force a so B reaches eight, but that would push a positive onto a more electronegative , which costs more than the does. Resist the urge.
Determine the using : B is surrounded by 3 bonding domains and 0 , for 3 in total. Because electron domains repel one another, 3 domains arrange themselves as far apart as possible, giving trigonal planar and sp² .
Every domain carries an here, so and agree: trigonal planar, 120° angles throughout.
: each B-F bond is genuinely polar on its own, but the arrangement points those pulls symmetrically against each other. They cancel to nothing. Nonpolar overall.
Worth knowing
Boron sits at six electrons and is comfortable there, which leaves it hungry for an electron pair. That hunger makes BF3 a classic Lewis acid used to drive organic reactions.