Why is a [Cu(SCN)2] complex black?

Question

I've been creating various Copper(II) complexes using different ligands and predicting their relative hues using Crystal Field Theory by using the spectrochemical series to predict the extent of d orbital splitting on the Copper(II) ion and the resultant colour of the copper(II) complex.

When I added potassium thiocyanate ($\ce{KSCN}$) to a solution containing hexaaquacopper(II) the entire solution turned deep black. Doing some research online and knowing that $\ce{SCN-}$ tends to form complexes with planar geometry I'm fairly certain that the resultant copper(II) complex was $\ce{[Cu(SCN)4]}$.

Because of the relative strength of the $\ce{SCN-}$ ligand compared to the pale blue complex formed with $\ce{H2O}$ I assumed the solution would appear redder in hue as higher frequency wavelengths would be absorbed, however, I fail to see why the entire visible spectrum would suddenly be absorbed. Can this phenomenon be explained with Crystal Field Theory?

Also perhaps my solution was not dilute enough?

Answer 1

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).

Answer 2

When you add $\ce{SCN^-}$ ions to aqueous solution of $\ce{Cu^{2+}}$, the thiocyanate ligands will start to substitute the water molecules from the octahedral $\ce{[Cu(H_2O)_6]^{2+}}$ complex.

Though $\ce{SCN^-}$ is a weak field ligand, It can substitute relatively stronger field ligand $\ce{H_2O}$ from $\ce{Cu^{2+}}$ complexes, which can be explained by Pearson's HSAB Theory. $\ce{Cu^{2+}}$ is a very soft acid, as it has less positive charge on it, more no. of $\ce{d}$ electrons etc. which matches with the properties of the soft acid. Between, $\ce{H2O}$ and $\ce{SCN-}$, $\ce{H2O}$ has its ligand site as $\ce{O}$ which is a hard base centre, but in $\ce{SCN-}$, the ligand site $\ce{S}$ is a relatively softer base centre, as it has a larger size, lesser electronegativity, and lesser electron density than $\ce{O}$,and these properties make $\ce{SCN-}$ a soft base. Now, according to HSAB theory, soft acids prefer to bind with soft bases. Thus $\ce{Cu^2+}$ has more affinity towards $\ce{SCN-}$ rather than $\ce{H2O}$.

Thus when you have a aqueous solution of $\ce{Cu^2+}$, and add $\ce{SCN-}$ gradually by little amounts, the following complexes will start to form.

$$\ce{[Cu(H2O)6]^2 ->[SCN-] [Cu(H2O)4(SCN)2]}\text{( apple-green colour)}$$ $$\ce{[Cu(H2O)4(SCN)2]^2 ->[SCN-] [Cu(H2O)2(SCN)4]^2-}\text{ (pale -yellow colour)}$$

This shift in the colours of the new complexes can be explained by CFT. As more weak field ligands ($\ce{SCN-}$) is introduced in the complex, the octahedral crystal field splitting energy ($\Delta_\mathrm{o}$) of the complexes decreases and thus the wavelength absorbed shifts to higher wavelengths and complementary colours also become higher in wavelengths.

Now, if you add sufficiently higher amounts of $\ce{SCN-}$, all the complexes will be destroyed and you will get only a compound of $\ce{Cu^{2+}}$ and $\ce{SCN-}$, which is a normal ionic compound, and no complex is left thereafter.

$$\ce{[Cu(H2O)2(SCN)4]^2- ->[high SCN-] [Cu(SCN)6]^4- (unstable) -> Cu(SCN)2(black)}$$

Thus the final compound is not at all any complex and just a black coloured compound, which might have formed if you have added significantly higher amount of $\ce{SCN-}$, in aqueous solution of $\ce{Cu^2+}$. So, it is irrelevant to judge the colour of final product i.e. $\ce{Cu(SCN)2}$ through CFT as it is now no longer a complex at all.