Why is diborane cleaved symmetrically by trimethylamine but unsymmetrically by ammonia?

I know that soft bases cleave diborane symmetrically but hard ones don't. However, what makes ammonia harder than trimethylamine? And why does hardness/softness affect this? 




Excerpts from my answer [here](https://chemistry.stackexchange.com/a/73547/43942).



> 
> 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.
> 
> 
> [![enter image description here](https://i.stack.imgur.com/k1tfL.jpg)](https://i.stack.imgur.com/k1tfL.jpg)
> 
> 
> 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:
> 
> 
> [![enter image description here](https://i.stack.imgur.com/md8Z9.jpg)](https://i.stack.imgur.com/md8Z9.jpg)
> 
> 
> 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](https://books.google.co.in/books?id=hLUy6or0In8C&dq=Boron+Hydride+Chemistry&source=gbs_navlinks_s), Earl Muetterties


 5