Question

$B{r}^{-}$ions form a close packed structure. If the radius of $B{r}^{-}$ ions is $195pm$, calculate the radius of the cation that just fits into the teteahedral hole. Can a cation $A^{+}$ having a radius of $82pm$ be shipped into be octathedral hole of the crystal $A^{+}B{r}^{-}$?

Answer 1

For tetrahedral voids, radius ratio of cation vs anion is,

$0.225<r<0.414$

$\frac{r+}{r-}=0.414$

$r-=195pm$

$r+=0.414\times 195pm$

$r+=80.73pm$

This is the highest radius cation which fit in tetrahedral void. On increasing the radius the cation will fit into octahedral void as radius ratio for octahedral void is $0.414<r<0.732$

Thus, $82pm>80.73pm$ will be shipped into octahedral void.