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I know that usually more electronegative anions would displace less electronegative anions but is there a way to get different anions to "share" a cation?

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A cursory search showed some interesting cases.

The oldest reported examples seemingly are mercury halides $\ce{HgXY}$.

  • $\ce{HgClI}$ was prepared by iodination of solid $\ce{HgCl2}$
  • $\ce{HgBrCl}$ was obtained from the chlorination of an ethanolic suspension of $\ce{HgBr2}$
  • $\ce{HgBrI}$ was prepared by mixing ethanolic solutions of $\ce{HgBr2}$ and $\ce{HgI2}$

The crystal structure of $\ce{Cs6Cl3[Fe(CN)6].H2O}$ has been reported.

Passing $\ce{H2S}$ through solutions of oxidosulfido molybdates and tungstates let to various mixed salts with halide, hydrogen sulfide and thiosulfate as additional anions beside the trisulfidometalates $\ce{[MoOS3]^{2−}}$ and $\ce{[WOS3]^{2−}}$.

But the concept isn't limited to inorganic chemistry.

The crystal structures of mixed salts of amino acids with different anions were reported.

From the field of organic conductor research, salts of BEDT-TTF (bis(ethylenedithio)tetrathiafulvalene) with mixed anions of $\ce{[CuBr2]-}$ and $\ce{[CuCl2]-}$ are known.

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    $\begingroup$ Mixed anion salts are more interesting than I thought! I like how the Hg salts have different properties depending on the order that anions are switched. $\endgroup$ – gsurfer04 Feb 21 '14 at 12:08
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Sure, why not? In the bad old days geology claiming isomorphous substitution had a hissy fit with chemistry claiming exact stoichiometry. Rochelle salt is potassium sodium tartrate. Potassium aluminum sulfate (alum) is often doped with colored transition metal ions substituting for some of the aluminum (chrome alum). Ruby, where Cr(III) substitutes for some (Al(III).

If size is about the same for identical charge, the alien ion goes into the crystal lattice.

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  • $\begingroup$ This answer discusses salts with varying cations, while the question asks for anions. $\endgroup$ – cbeleites Feb 22 '14 at 16:55
  • $\begingroup$ It works the same way either way. $\endgroup$ – Uncle Al Feb 23 '14 at 21:49

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