Why is a [Cu(SCN)2] complex black?
As the user above states, if you mix concentrated solutions of $\ce{Cu(II)}$ salts and $\ce{NCS}$ you get $\ce{Cu(NCS)_2}$ which is a black solid. I'd disagree with them that it's not longer a complex because it 100% is - and you are correct to assume that the local coordination is $\ce{Cu(NCS)_4}$. To be more precise, each copper is coordinated by $\ce{Cu(NCS)_2(SCN)_2}$, and there's actually a large Jahn Teller distortion so it could also be described as $\ce{Cu(NCS)_2(SCN)_4}$. They're right that the colour can't be easily explained by $d-d$ transitions (i.e. the kind you're thinking of w.r.t. ligand field splitting) though.

$\ce{Cu(NCS)_2}$ is black probably because of ligand to metal charge transfer (the same reason $\ce{Fe(III)NCS}$ is blood red) - i.e. on absorption you transiently form $\ce{Cu(I)}$ and $\ce{NCS}$ from $\ce{Cu(II)}$ and $\ce{NCS–}$. It's more complex that that for sure, but I think it's fair to say no one knows yet, because we only worked out the structure two years ago.

If you want a lot more information about $\ce{Cu(NCS)_2}$ we published on it here https://journals.aps.org/prb/abstract/10.1103/PhysRevB.97.144421 (on the arxiv at https://arxiv.org/abs/1710.04889).