In an ionic solid `r_((+))=1.6A` and `r_((-))=1.864A`. Use the radius ratio to determine the edge length of the cubic unit cell in `A`.

Answer

4

Explanation

`(r_(+))/(r_(-)) = (1.6)/(1.864) = 0.858`
So, it is CsCl type unit cell
So `sqrt3 a = 2 (r_(+) + r_(-))`
So `a = (2(1.864 + 1.6))/(sqrt3) Å = 2 xx 2 Å = 4Å`