Question

$KCl$ crystallizes in the same type of lattice as $NaCl$. Given that $\frac{r_{N{a}^{+}}}{r_{C{l}^{-}}}=0.50$ and $\frac{r_{K^{+}}}{r_{Cl-}}=0.70$. Calculate the ratio of the side of the unit cell for $KCl$ to that for $NaCl$
(Give your answer upto two decimal value)

Answer 1

For $NaCl$ type structure,
$r^{+}+r^{-}=\frac{a}{2},$
where $a$ is the edge length of the unit cell.
$r^{+}\text{and}{r}^{-}\text{is the radius of cation and anion}$
Given,
$\frac{r_{N{a}^{+}}}{r_{C{l}^{-}}}=0.50,\frac{r_{K^{+}}}{r_{C{l}^{-}}}=0.7$
So, $\frac{a_{KCl}}{a_{NaCl}}=\frac{r_{K^{+}}+r_{C{l}^{-}}}{r_{N{a}^{+}}+r_{C{l}^{-}}}=\frac{\frac{r_{K^{+}}}{r_{C{l}^{-}}}+1}{\frac{r_{N{a}^{+}}}{r_{C{l}^{-}}}+1}=\frac{1.7}{1.5}=1.13$