The radius of K^{+} is 133 pm and that of Cl | vedclass

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(C) $KCl$ crystallizes in a face-centered cubic $(FCC)$ structure,where the edge length $a$ is related to the ionic radii by the formula: $a = 2(r_{K^

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$2.48$

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(C) $KCl$ crystallizes in a face-centered cubic $(FCC)$ structure,where the edge length $a$ is related to the ionic radii by the formula: $a = 2(r_{K^{+}} + r_{Cl^{-}})$.Given,$r_{K^{+}} = 133 \ pm$ and $r_{Cl^{-}} = 181 \ pm$.$a = 2(133 + 181) = 2(314) = 628 \ pm$.Converting edge length to centimeters: $a = 628 	imes 10^{-10} \ cm = 6.28 	imes 10^{-8} \ cm$.The volume of the unit cell is $V = a^{3}$.$V = (6.28 	imes 10^{-8} \ cm)^{3} = 247.97 	imes 10^{-24} \ cm^{3} \approx 2.48 	imes 10^{-22} \ cm^{3}$.

The radius of $K^{+}$ is $133 \ pm$ and that of $Cl^{-}$ is $181 \ pm$. The volume of the unit cell of $KCl$ expressed in $10^{-22} \ cm^{3}$ is

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