In one of his great answers, Martin gave us the calculated molecular orbitals for carbon monoxide I am copying below. Note that while cyanide has a nitrogen atom in place of the oxygen and features a single negative charge, we can still qualitatively consider the MO schemes identical; thus I will base my discussion on what you see in the image.
In this scheme, the HOMO is referred to as MO number 7; since 5 and 6 are π-type MOs, this corresponds to the 5σ you mentioned.
One thing we immediately note is that the carbon-centred lobe of the HOMO is very diffuse and extends far into space; much farther than any other orbital of the molecule. It is also the orbital that an acid (either proton or Lewis acid) would bond to. In Pearson’s HSAB theory, large and diffuse orbitals — often also called polarisable — are associated with soft acids or bases, as the charge per volume value is considered comparatively low. Therefore, just by looking at the size of this orbital, we can identify it as a ‘soft’ one.
It should also be explained why soft bases tend to create more stable complexes with soft acids. This is a direct consequence of the more diffuse nature of the orbital: the overlap between two orbitals (which will later form a bond) is typically best if the orbitals have similar shapes and sizes. Thus, because the HOMO has a rather diffuse lobe on carbon’s side it will prefer binding to other compounds with rather diffuse orbitals in a Lewis acid-base reaction.
That should be all that is required to explain cyanide’s behaviour under Pearson’s theory.