Question

$KCl$ crystallises in the same type of lattice as $NaCl$. Given that $r_{N{a}^{+}}/r_{C{l}^{-}}=0.50$ and $r_{N{a}^{+}}/r_{K^{+}}=0.70$, calculate the ratio of the side of the unit cell of $KCl$ to that of $NaCl$:

• A. 1.143
• B. 1.224
• C. 1.414
• D. 0.875

Answer 1

The correct option is A 1.143

Rock salt type lattice structure means the edge length will be $a=r_{+}+r_{-}$
Hence, $\frac{a_{KCl}}{a_{NaCl}}=\frac{r_{K^{+}}+r_{C{l}^{-}}}{r_{N{a}^{+}}+r_{C{l}^{-}}}=\frac{\frac{r_{K^{+}}}{r_{C{l}^{-}}}+1}{\frac{r_{N{a}^{+}}}{r_{C{l}^{-}}}+1}=\frac{\frac{0.5}{0.7}+1}{0.5+1}=1.143$