Chapter 9 Bonding and Molecular Structure: Orbital Hybridization and Molecular Orbitals - Study Questions - Page 369d: 51

See the answer below.

a,b) Since boron makes respectively 3 and 4 bonds and has no lone pairs, it displays respectively trigonal planar with $sp^2$ hybridization and tetrahedral geometry with $sp^3$ hybridization. c) Nitrogen has a partially negative charge because it is more electronegative than hydrogen, boron has a partially positive charge because it is less electronegative than fluorine. The two are attracted by electrostatic forces. d) It donates an oxygen lone pair to the boron, for the same reason as in item C.