#### Answer

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.
($F.\ Charge= NVE-LPE-BE/2$)