In `CsCl` with cubic structure, `Cl^(-)` ions are located at each corner and `Cs^(+)` ions at the centre of the unit cell. If `r_(Cs^(+))` and `r_(Cl^(-))` and `1.69Å` and `1.81Å` respectively. Find the value of edge length of cube.

Answer

`4.04 Å`

Explanation

We assume that the closest `Cs^(+)` to `Cl^(-)` ion distance is the sum of ionic radii of `Cs^(+)` and `Cl^(-)` ions.
Nearest neighbour distance = `1.69 + 1.81 = 3.50 Å`
This distance is one half of the body diagonal of the cube
Body diagonal = `sqrt(3)a`
`:. (sqrt(3)a)/(2) = 3.50`
or `a = (3.50 xx 2 )/(sqrt3) = 4.04 Å`.