Q. $KCl\$crystallises in the same type of lattice as dose $NaCl.$ Given that $r_{{Na}^{+}}/r_{{Cl}^{-}} = 0.55\$ and $r_{K^{+}}/r_{{Cl}^{-}} = 0.74.$ Calculate the ratio of the side of the unit cell for $KCl\$to that of $NaCl.$
Explanation
(a)
Given, $r_{{Na}^{+}}/r_{{Cl}^{-}} = 0.55$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r_{K^{+}}/r_{{Cl}^{-}} = 0.74$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{KCl}}{r_{NaCl}} = ?$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{{Na}^{+}}}{r_{{Cl}^{-}}} = 0.55$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{{Na}^{+}}}{r_{{Cl}^{-}}} + 1 = 0.55 + 1$
$\ \ \ \ \ \ \ \ \ \ \frac{r_{{Na}^{+}} + r_{{Cl}^{-}}}{r_{{Cl}^{-}}} = 1.55\ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{K^{+}}}{r_{{Cl}^{-}}} = 0.74$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{K^{+}}}{r_{{Cl}^{-}}} + 1 = 0.74 + 1$
$\ \ \ \ \ \ \ \ \ \ \ \frac{r_{K^{+}} + r_{{Cl}^{-}}}{r_{{Cl}^{-}}} = 1.74\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)\$
Eq (ii) devide by Eq (i)
$\ \ \ \ \ \ \ \ \ \frac{r_{K^{+}} + r_{{Cl}^{-}}}{r_{{Na}^{+}} + r_{{Cl}^{-}}} = \frac{1.74}{1.55} = 1.1226$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.