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

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?

• Redox, maybe?$\!$ – Ivan Neretin Mar 29 '18 at 6:09
• Have you tried to filter the stuff? When I see black and am using reagents with sulfur in the -2 oxidation state like thiocyanate, I think there could be a sulfide precipitate. – Oscar Lanzi Mar 29 '18 at 12:53

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.

• If I recall that correctly the problem with copper is that the copper is right at the border between thiocyanate and isothiocyanate, and thiocyanate is a pseudohalogen which can also reduce the Cu(II) to Cu(I). So you have a mixture of ionic, coordinated, oxidized and reduced linkage isomers which all have different colors. – Justanotherchemist Mar 29 '18 at 8:18
• Yes, this answer is thoroughly wrong, even maybe not an answer at all. – Mithoron Mar 29 '18 at 16:04

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).