Take diborane as an example, 3c-2e bonds are required (3 centered, 2 electron) to account for the B-H-B bridges, the terminal B-H bods can be thought of as conventional (2c-2e) bonds. This means each boron atom uses 2 electrons and 2 roughly sp3 orbitals to from the (2c-2e) bonds to the 2 hydrogen atoms. The boron atom in each BH2 group then still has one electron and two hybrid orbitals, for more bonding.
The plane of the two remaining orbitals are perpendicular to the BH2 plane, and when two of these BH2 groups approach each other with hydrogen atoms lying in the plane of the four empty orbitals, two B-H-B (3c-2e) bonds are formed. The electrons are provided by the one electron carried by each H atom and by each BH2 group.
Whereas none of this is possible with halides.
That is probably confusing, but I tried to keep it as simple as possible, it is really hard though without diagrams. For higher boranes there are B-B-B (3c-2e) bonds, in two different forms (open and closed), as well as B-B (2c-2e) bonds.
In short summary not using an example, since discrete BH3 has only six valence electrons which are involved in the B-H bonds that utilize 3 of the 4 atomic orbitals. It is this empty atomic orbital (perpendicular to the plane defined by BH3) that is highly reactive; this is a completely different mechanism then how carbon forms organic compounds.
Boron is a very diverse and interesting element, it was actually boranes that led to the development of the early concepts of multicenter bonding.