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In LiI structure, the two successive iodide ions touch each other and $L i^{+}$ ions are present in the interstitial positions. If $L i I$ has a $N a C l$ structure, calculate the radius of $I^{-}$ ion (In nano meters). Given the inter nuclear distance between $L i^{+}$ and $I^{-}$ ions is $0.3 n m$

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Detailed Solution

IN $N a C l$ type structure
$Z r^{+} + Z r^{-} =$ edge length
And $4 r^{-} = \sqrt{2} a r^{-} = \frac{1.414 \times 6 n m}{4} = 2.121 n m$

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