Zinc sulfide, copper iodide and copper bromide all adopt the zincblende structure. My lecture notes suggest that this structure is adopted preferentially over the wurtzite structure if the bonding is more covalent (why is this?); thus, it’s adopted where there is a polarizing cation and a polarizable anion.

However, surely a polarizing cation should be small and highly charged (i.e. not ions like $\ce{Cu+}$ surely? It has a low charge and large-ish radius). What makes this polarizing? I’ve also heard this about $\ce{Ag+}$.


1 Answer 1


Generally when you get a preference between two structures due to directional bonding, the first order environment of the cation is different. In this case however both are tetrahedrally coordinated. The Wurtzite structure has a marginally greater Madelung constant (1.641 versus 1.638).

Of the top of my head, I think the dominant factor in favour of the blende structure is stronger directional cation-cation and anion-anion interactions. In the case of Cu(I), the d-orbitals are going to be fairly diffuse due to the low charge, potentially allowing some covalent interaction between adjacent sites. A much better example would be GaAs, where both cation and anion are realtively soft/polarizable, which adopts blende, in contrast to AlN, with hard ionic interactions, which adopts wurzite. I don't think it's really as simple as this because the difference in energy between the two structures is often very small; you can see this in the relative stability of both ZnS polymorphs.

  • $\begingroup$ Any thoughts on the issue raised in the last paragraph of my question? $\endgroup$
    – RobChem
    Apr 14, 2015 at 18:48
  • $\begingroup$ The relative polarisability of both species is the relevent factor here. I don't think Ag(I) is a "polarizing" species at all but I guess it might be in that it forms compounds with a high degree of covalency with polarisable anions e.g. Ag(I)I. I think looking at it in terms of hard-hard/soft-soft interactions is much more intuitive. $\endgroup$
    – J. LS
    Apr 14, 2015 at 19:04
  • $\begingroup$ Why does it form compounds with a high degree of covalency? $\endgroup$
    – RobChem
    Apr 14, 2015 at 19:09
  • $\begingroup$ Like most 4d metals, it has radially diffuse valence orbitals which will significantly overlap with those of the counterion. Significant covalency is acheived if there is reasonable orbital energy match too. $\endgroup$
    – J. LS
    Apr 14, 2015 at 19:33

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