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# KCl crystallizes in the same type of lattice as does $NaCl$ . Given that $\large\frac{r_{Na^{+}}}{r_{cl^{-}}}$$=0.5 and \large\frac{r_{Na^{+}}}{r_{K^{+}}}$$=0.5$ calcute the ratio of the unit cell for $KCl$ to that for $NaCl$

$\begin{array}{1 1} 103.5 \\ 1.143 \\ 8.4 \\ 9.5 \end{array}$

Answer : $1.143$
Nacl Crystallizes in the face- centered cubic unit cell, Such that $r_{Na} +r_{cl-}=\large\frac{a}{2}$
Where a is the edge length of unit cell.
Now Since $r_{Na^{+}}/r_{Cl^-}=0.5$ and $r_{Na}/r_{K^{+}}=0.7$ we will have
$\large\frac{r_{Na^+}+r_{Cl^-}}{r_{Cl^{-}}}$$=1.5 and \large\frac{r_{K^{+}}}{r_{Cl^{-}}}=\frac{r_{K^+}}{r_{Cl^{-}}}=\frac{r_{K^+}}{r_{Na^+}/0.5} \qquad= \large\frac{0.5}{r_{Na^+}/r_{K^+}}=\frac{0.5}{0.7} \Large\frac{r_{K^+}+r_{Cl^{-}}}{r_{Na^+}+r_{Cl^{-}}}$$=\large\frac{1.2}{0.7 } \times \frac{1}{1.5}$
$\large\frac{a_{KCl}}{a_{NaCl}}$$= \frac{1.2}{1.05}$$=1.143$