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`.

Options

A. 4
B. `2sqrt(3)`
C. `3sqrt(3)`
D. `(4)/(sqrt(3))`

Answer

1

Explanation

`(r_(+))/(r_(_))=(1.6)/(1.864)=0.858`
So, it is `CsCl` type of unit cell
So `sqrt(3)a=2(r_(+)+r_(_))`
So, `a=(2(1.864+1.6))/(sqrt(3))Å=2xx2Å=4Å`