The unit cube length for `LiCl` (`NaCl` structure) is `5.14 Å`. Assuming anion-anion contact, calculate the ionic radius for chloride ion.

Answer

1.81 Å

Explanation

Interionic distance of LiCl=`5.14/2`=2.57 Å
`therefore BC=sqrt(AB^2+AC^2)=sqrt((2.57)^2+(2.57)^2)=3.63`
Radius of `Cl^-` ion=`1/2` x 3.63 Å =1.81 Å