In a crystal of LiI, I^{-} ions are arranged | vedclass

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(D) The structure of $LiI$ is similar to $NaCl$ (rock salt structure).In this structure, the edge length $a$ is related to the ionic radii $r_{+}$ $(L

Vedclass · Accepted Answer

$85$

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(D) The structure of $LiI$ is similar to $NaCl$ (rock salt structure).In this structure, the edge length $a$ is related to the ionic radii $r_{+}$ $(Li^{+})$ and $r_{-}$ $(I^{-})$ by the relation: $a = 2(r_{+} + r_{-})$.For $ccp$ arrangement of anions, the anions touch along the face diagonal: $4r_{-} = \sqrt{2}a$.Thus, $r_{-} = \frac{\sqrt{2}a}{4} = \frac{1.414 	imes 600}{4} = 212.1 \ pm$.Using the relation $a = 2(r_{+} + r_{-})$:$600 = 2(r_{+} + 212.1)$$300 = r_{+} + 212.1$$r_{+} = 300 - 212.1 = 87.9 \ pm \approx 88 \ pm$.The closest option is $85 \ pm$.

In a crystal of $LiI$, $I^{-}$ ions are arranged in a $ccp$ structure and $Li^{+}$ ions occupy octahedral voids. If $a = 600 \ pm$, what is the radius of the $Li^{+}$ ion in $pm$?

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