See the answer below.
Work Step by Step
Since no boron atom has lone electron pairs, electron-pair and molecular geometries are the same. None of them makes $\pi$ bonds. For the ones with three bonds: trigonal planar geometry, $sp^2$ hybridization, formal charge of 0. For the ones with four bonds: tetrahedral geometry, $sp^3$ hybridization, formal charge of -1. ($F.\ Charge= NVE-LPE-BE/2$)