The ratio of the side of the unit cell for KCl to that for NaCl is — Solid State Chemistry Question

💡 Solution & Explanation

$a_{NaCl} = 2(0.5r_{Cl^-} + r_{Cl^-}) = 3r_{Cl^-}$. $r_{K^+} = 0.5r_{Cl^-} / 0.7 = \frac{5}{7}r_{Cl^-}$. $a_{KCl} = 2(\frac{5}{7}r_{Cl^-} + r_{Cl^-}) = \frac{24}{7}r_{Cl^-}$. Ratio $= (\frac{24}{7}) / 3 = \frac{8}{7} \approx 1.143$. Therefore, correct answer is A.