If the crystal structure of KCl is the same a | vedclass

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(A) For an $NaCl$ type structure,the edge length $a$ is given by $a = 2(r_{cation} + r_{anion})$.For $NaCl$: $a_{NaCl} = 2(r_{Na^+} + r_{Cl^-}) = 2r_{

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$1.123$

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(A) For an $NaCl$ type structure,the edge length $a$ is given by $a = 2(r_{cation} + r_{anion})$.For $NaCl$: $a_{NaCl} = 2(r_{Na^+} + r_{Cl^-}) = 2r_{Cl^-} \left( \frac{r_{Na^+}}{r_{Cl^-}} + 1 \right) = 2r_{Cl^-} (0.55 + 1) = 2r_{Cl^-} (1.55)$.For $KCl$: $a_{KCl} = 2(r_{K^+} + r_{Cl^-}) = 2r_{Cl^-} \left( \frac{r_{K^+}}{r_{Cl^-}} + 1 \right) = 2r_{Cl^-} (0.74 + 1) = 2r_{Cl^-} (1.74)$.The ratio of edge lengths is $\frac{a_{KCl}}{a_{NaCl}} = \frac{2r_{Cl^-} (1.74)}{2r_{Cl^-} (1.55)} = \frac{1.74}{1.55} \approx 1.1225 \approx 1.123$.

If the crystal structure of $KCl$ is the same as that of $NaCl$,and the radius ratios are $r_{Na^+}/r_{Cl^-} = 0.55$ and $r_{K^+}/r_{Cl^-} = 0.74$,calculate the ratio of the edge lengths of the unit cells of $KCl$ and $NaCl$.

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