1
$\begingroup$

What is the oxidation state of iron in $\ce{CuFeS2}$? And also could you please tell it's proper representation: $\ce{Cu2S.FeS.FeS2}$ or $\ce{Cu(I)FeS2}$?

$\endgroup$

closed as off-topic by Todd Minehardt, airhuff, Tyberius, Mithoron, Jan Dec 22 '17 at 0:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ $\ce{Cu2S.FeS.FeS2}$ and $\ce{CuFeS2}$ are fine, but $\ce{Cu(I)FeS2}$ is wrong. Oxidation state in parentheses used in written names; if you want to denote monovalent copper, then $\ce{Cu^IFeS2}$ $\endgroup$ – andselisk Dec 21 '17 at 11:13
  • $\begingroup$ Can you please also comment on their oxidation state ie. Is iron in +2 and copper in +1? $\endgroup$ – user45903 Dec 21 '17 at 11:14
  • $\begingroup$ But you've already answered this part by typing $\ce{Cu^I}$, haven't you? :) The structure contains $\ce{Cu+}$ and $\ce{Fe^3+}$ cations. $\endgroup$ – andselisk Dec 21 '17 at 11:16
  • $\begingroup$ In the formula Cu2S.FeS.FeS2, the 1st iron is in +2 and the second one is also in +2 by me, ie it is polysulphide ion ( S2 (2- ) ion ), so their avg will also be 2 right? $\endgroup$ – user45903 Dec 21 '17 at 11:24
  • $\begingroup$ No. $\ce{Cu2S.FeS.FeS2}$ is just a convenient representation, nothing else. In reality the cubic structure consists of array of vertex-linked $\ce{CuS4}$ and $\ce{FeS4}$ tetrahedra, that's it. There is no separate standalone metal sulfides as one would imagine. $\endgroup$ – andselisk Dec 21 '17 at 11:31