Question

The unit cell cube length for $LiCl(NaCl$ structure) is $514\stackrel{\circ }{A}$. Assuming anion-anion contact, calculate the ionic radius for chloride ion.

Answer 1

In a face-centred cubic lattice, anions touch each other along the face diagonal of the cube.
$4r_{C{l}^{-}}=\sqrt{2}a$
$r_{C{l}^{-}}=\frac{\sqrt{2}}{4}a$
$=\frac{\sqrt{2}}{4}\times 5.14=1.82\stackrel{\circ }{A}$
Alternative: Distance between $L{i}^{+}$ and $C{l}^{-}$ ions
$=\frac{5.14}{2}=2.57\stackrel{\circ }{A}$
Thus, distance between two chloride ions
$=\sqrt{(2.57{)}^2+(2.57{)}^2}$
$=3.63\stackrel{\circ }{A}$
Hence,
radius of $C{l}^{-}=\frac{3.63}{2}=1.82\stackrel{\circ }{A}$.