What is the minimum distance of Rubidium and Iod ions in a centered cube?
The distance between $\ce{Rb^{+}}$ and $\ce{I^{-}}$ in the fcc lattice would be $(d(\ce{Rb^{+}})/2) + (d(\ce{I^{-}})/2) = 3.76 \times 10^{-10} \mathrm{m}$