why is scandium(2) virtually unknown
The electronic configuration of Sc is [Ar]3d14s2. Due to which it exist only in +3 oxidation state.
When Sc loses its two electron, it forms Sc+2.
Sc+2 has a configuration of [Ar]3d1
As only single electron in d orbital is quite unstable, therefore scandium readily losses the last electron in order to attain noble gas configuration i.e. Sc+3 = [Ar]