What is the edge length of ZnS, given the radii of Zn and S ions? [closed]








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What is the edge length of $\ce{ZnS}$, given the radii of $\ce{Zn^2+}$ and $\ce{S^2-}$ ions?




In ZnS, $Zn^{2+}$ is located in alternate tetrahedral voids and $S^{2-}$ at the vertices of cube of ZnS crystal. If $a$ is the length of cube, then the tetrahedral void is present at a distance $√3a/4$ from the ion at the nearest vertex on that diagonal of cube,i.e.,$S^{2-}$.
So, $Zn^{2+}$
 is at a distance
 $√3a/4$ from $ S^{2-}$.


Tetrahedral void is present between centre of cube and it's vertex on the diagonal of the cube.


So $√3a/4$ = radius of $Zn^{2+}$+ radius of $S^{2-}$=$d$ say.
Hence, $a$=($4/√3a$)×$d$


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