CsCl has bcc structure with `Cs^(+)` at the centre and `Cl^(-)` ion at each corner. If `r_(Cs^(+))` is `1.69Å` and `r_(Cl^(-))` is `1.81Å` what is the edge length of the cube?

Explanation

We assume that the closest `Cs^(+) " to " Cl^(-)` distance is the sum of the ionic radii of `Cs^(odot) and Cl^(Θ)`
`= 1.69 + 1.81 = 3.50 Å`
This distance in one-half of the cubic diagonal `= (asqrt3)/(2)`
`:. (a sqrt3)/(2) = 3.50 Å " " :. a = 4.01 Å`