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?

Options

A. `4.04`
B. `3.50`
C. `3.03`
D. `1.95`

Answer

A

Explanation

The closest `Cs^(+)` to `Cl^(-)` distance
= sum of ionic radii of `Cs^(+)` and `Cl^(-)`
`=1.69+1.81`
`=3.50 A^(@)`
The distance in one-half of the cubic diagonal `(asqrt(3))/(2)`
`a=4.04A^(@)`