Why is diborane cleaved symmetrically by trimethylamine but unsymmetrically by ammonia?
Excerpts from my answer here.

The mechanism proposed for the cleavage reaction involves an initial attack by the donor on one boron atom in diborane, leading to cleavage of one $\ce{B-H-B}$ bridge.

This is followed by the attack of a second donor molecule, cleaving the remaining $\ce{B-H-B}$ bridge. If the boron atom in the pendant $\ce{BH3}$ group is attacked, two moles of adduct result, whereas attack on the atom already carrying a donor produces cation and borohydride:

Support for this view comes from the fact that the intermediate ($\text{I}$) is detectable by tensimetric titration of amine-borane with diborane and by $\ce{^{11}B}$ nmr.

It can then be interpreted that hard bases like ammonia don't experience the steric effects which softer bases like trimethylamine do. Sterics inhibit the attachment of the second donor at the same boron atom.
Source: Boron Hydride Chemistry, Earl Muetterties