Boron trifluoride, BF3, has measured F–B–F bond angles of approximately 120∘. Which of the following best explains this observation?
a) Boron forms sp hybrid orbitals that orient themselves 120∘ apart to minimize repulsion.
b) Boron is surrounded by three electron domains that give rise to sp2 hybrid orbitals arranged 120∘ apart.
c) Boron has four electron domains (three bonds and one lone pair) which adopt sp3 hybridization; strong lone-pair repulsion pushes the B–F bonds to 120∘.
d) The central boron uses dsp3 hybridization, and the three equatorial B–F bonds in a trigonal-bipyramidal geometry are 120∘ apart.