What is the oxidation state of Cu and Fe in chalcopyrite ($\ce{CuFeS2}$)? Some websites tell that it is $\ce{Cu(I)Fe(III)S2}$, and others tell that it is $\ce{Cu(II)Fe(II)S2}$. Which is right?
2 Answers
A number of papers report that the formal valency states of chalcopyrite are best considered as $\ce{Cu+Fe^3+S2}$, based on various computational and spectroscopic evidence.
References:
Klekovkina, V. V.; Gainov, R. R.; Vagizov, F. G.; Dooglav, A. V.; Golovanevskiy, V. A.; Pen’kov, I. N. Oxidation and magnetic states of chalcopyrite CuFeS2: A first principles calculation. Opt. Spectrosc. 2014, 116 (6), 885–888. DOI: 10.1134/S0030400X14060149.
Pearce, C.; Pattrick, R.; Vaughan, D.; Henderson, C.; van der laan, G. Copper oxidation state in chalcopyrite: Mixed Cu d9 and d10 characteristics. Geochim. Cosmochim. Acta 2006, 70 (18), 4635–4642. DOI: 10.1016/j.gca.2006.05.017.
Goh, S. W.; Buckley, A. N.; Lamb, R. N.; Rosenberg, R. A.; Moran, D. The oxidation states of copper and iron in mineral sulfides, and the oxides formed on initial exposure of chalcopyrite and bornite to air. Geochim. Cosmochim. Acta 2006, 70 (9), 2210–2228. DOI: 10.1016/j.gca.2006.02.007.
One clue to the likely oxidation state is offered by some experiments I was doing in the chemistry laboratory when I was in graduate school. Looking for a clue to how thiourea might alter copper ekectrodeposition, I tried adding this stuff in bulk to copper(II) salt solutions.
Copper sulfate gave a loss of color, blue -> colorless.
Copper chloride gave a white precipitate and then it dissolved, forming a colorless solution, when more thiourea was added.
Given that white or colorless copper compounds are typically copper(I), it appeared that thiourea reduces and complexes the copper, and in the sulfur-ligand environment provided by thiourea copper favors the +1 oxidation state.
A Cu(I)-Fe(III) model for chalcopyrite, where the metals are bonded to sulfur, would be quite consistent with such findings.