Question

In LiI Crystal, $I^{-}$ ions form a cubical closest packed arrangement and $L{i}^{+}$ ions occupy octahedral holes. What is the relationship between the edge-length of the unit cells and the radii of the $I^{–}$ ions? Calculate the limiting ionic radii of $L{i}^{+}$ and $I^{-}$ if a = 600pm.

Answer 1

From question
$2r^{+}+2r^{-}=600pm\Rightarrow r^{+}+r^{-}=300pm----\left(1\right)$
Since, $L{i}^{+}$ ions occupy octahedral hole
So, for octahedral hole
$\frac{r^{+}}{r^{-}}=0.414-----\left(ii\right)$
From equation (I)and equation (II)
$r^{-}(\frac{r^{+}}{r^{-}}+1)=300\Rightarrow r^{-}\times (0.414+1)=300\Rightarrow r^{-}=\frac{300}{1.414}=212.164(\because \frac{r^{+}}{r^{-}}=0.414)\therefore r^{+}=212.164\times 0.414=r^{+}=87.84pm$