# 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.

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

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)}=$$220.65pm Now along the edge,we will have I^-Li^+I^- arrangement,where I^- are at the corners and Li^+ ion at the centre of the edge (octahedral void ).Since in closest packing,they touch each other,we will have 2r_I^-+2r_{Li^+}=a r_{Li^+}=\large\frac{a}{2}$$-r_{I^-}=\large\frac{624pm}{2}$$-220.65pm=91.35pm$