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LiI occurs as cubical-closest packing.If the edge length of a unit cell is 624 pm,determine the ionic radii of $Li^+$ and $I^-$ ions.

$\begin{array}{1 1}220.65pm,91.35pm\\120.65pm,81.35pm\\320.65pm,71.35pm\\420.65pm,101.35pm\end{array} $

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Answer : 220.65pm,91.35pm
The cubical closest packing has a face-centred cubic unit cell.$I^-$ ions occupy the corners and the face centres.These ions touch each other along the face diagonal of the cube.Hence
$4r_{I^-}=\sqrt 2 a$
$r_{I^-}=\large\frac{a}{2\sqrt 2}=\frac{624pm}{2(1.414)}=$$220.65$pm
Now along the edge,we will have $I^-Li^+I^-$ arrangement,where $I^-$ are at the corners and $Li^+$ ion at the centre of the edge (octahedral void ).Since in closest packing,they touch each other,we will have
$2r_I^-+2r_{Li^+}=a$
$r_{Li^+}=\large\frac{a}{2}$$-r_{I^-}=\large\frac{624pm}{2}$$-220.65pm=91.35pm$
answered Jul 16, 2014 by sreemathi.v
 

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