Are binary ionic solids like $\ce{NaCl}$ comprised of a cubic close packed structure of the anions with the cations filling interstital holes or do the cations and the anions contribute to the cubic close packed structure equally? I am really struggling with solid state chemistry - any help would be gratefully received. Also any suggestions for good learning resources on this topic would be great. Thanks.
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$\begingroup$ This might be helpful. When we talk about close packed structures, we're really only describing the positions of atoms relative to one another, not their bonding and there are often multiple equivalent ways to think about them, though which ions are involved do of course influence which crystal structures are stable. $\endgroup$– Michael DM DrydenFeb 3, 2015 at 21:39
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1 Answer
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The relative radii of the ions, and the stoichiometric ratio of the ions needs to be considered.
For NaCl, the radius of Na+ is 0.695 that of Cl-.
Na+ ions occupy octahedral holes in the cubic close packed lattice of Cl-.
If the radii are more equal (ratio greater than 0.732) then each type of ion will form a simple cubic lattice, with each ion of one lattice filing a cubic hole of the other lattice. For example, CsCl.
If the radii are very different (ratio less than 0.414) the smaller ions will fill tetrahedral holes in the cubic close packed lattice of the larger ions. For example ZnS.
See http://depts.washington.edu/chemcrs/bulkdisk/chem162A_sum06/handout_Lecture_11.pdf for more information.
