`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.

Answer

(a) 335.15 pm, (b) 154.15 pm

Explanation

(a) In a body cented cubic lattice
Charged ions each other along the cross-diagonal of the cube. Therefore
`2r^(+)+2r^(-)=3sqrta," "r^(+)+r^(-)=(sqrt3)/(2)a," "r^(+)+r^(-)=(sqrt3)/(2)("387pm")=335.15"pm"`
(b) Now, `r^(+)(NH_(4)^(+))=335.15-r^(-)(Cl^(-))=335.15-r^(-)(181)=154.15"pm."`