Question

KCl crystallizes in the same type of lattice as does NACl. Given that $\frac{r_{N{a}^{+}}}{r_{C{l}^{-}}}=0.55$ and $\frac{r_{K^{+}}}{r_{C{l}^{-}}}=0.74$. Calculate the ratio of the side of the unit cell for KCl to that of NaCl.

• A. 1.123
• B. 0.891
• C. 1.414
• D. 0.414

Answer 1

The correct option is A

1.123

$\to$ NaCl cyrstallises in fcc.
$\Rightarrow r_{N{a}^{+}C{l}^{-}}=\frac{a}{2}$
Given, $\frac{r_{N{a}^{+}}}{r_{C{l}^{-}}}=0.55$; Adding 1 to both sides
$\frac{r_{N{a}^{+}}+r_{C{l}^{-}}}{r_{C{l}^{-}}}=1.55\dots \dots \left(1\right)$
Now, $\frac{r_{K^{+}}}{r_{C{l}^{-}}}=0.74$; Adding 1 to both sides
$\frac{r_{K^{+}}+r_{C{l}^{-}}}{r_{C{l}^{-}}}=1.74\dots \dots \left(2\right)$

$\frac{2}{1}\Rightarrow \frac{r_{K^{+}}+r_{C{l}^{-}}}{r_{N{a}^{+}}+r_{C{l}^{-}}}=\frac{1.74}{1.55}\Rightarrow \frac{a_{KCl}}{a_{NaCl}}=\frac{1.74}{1.55}=1.123$