Question

The unit cell cube length of $LiCl$ ($NaCl$ structure) is $5.14\stackrel{o}{A}$. Assuming anion-anion contact, calculate the ionic radius for chloride ion.

• A. $2.05$
• B. $1.82$
• C. $2.92$
• D. $3.92$

Answer 1

Answer: (B)

$1.82$

In NaCl structure, anions have a face-centered cubic arrangement.

In such a unit cell face diagonally in $4$ times the radius of the anion(as shown in the figure).

Face diagona$[=\sqrt{2}.a=\sqrt{2}\times 5.14A^{\circ }$

Face diagonal$=4r$.

$4r=\sqrt{2}\times 5.14$

Implies that

$r=\frac{\sqrt{2}\times 5.14}{4}=1.82A^{\circ }$