`NH_(4)Cl` crystallises in a body centred cubic lattice with a unit cell distance of 387 pm. Calculate
(a) the distance between the oppositely charged ions in the lattice.
(b) the radius of `NH_(4)^(+)` ion if that of `Cl^(-)` ion is 181 pm.

Explanation

(a) In a body centred cubic lattice, oppositely charged ions touch each other along the cross- diagonal of the cube. So, we can write,
`2r^(+)+2r^(-)=sqrt(3)arArrr^(+)+r^(-)=(sqrt3)/(2)a=(sqrt3)/(2)(387"pm")=335.15"pm"`
(b) Now, since `r^(-)=181"pm"`
We have, `r^(+)=(335.15-181)"pm"=154.15"pm"`