# What is the minimum distance of Rubidium and Iod ions in a centered cube?

I have a problem to visualize the following assignment:

What is the minimum distance between $\ce{Rb^+}$ and $\ce{I^-}$ ions if radius of $\ce{Rb^+}$ is ${1.49\cdot10^{-10}}~\mathrm{m}$ and radius of $\ce{I^-}$ is ${2.17\cdot10^{-10}}~\mathrm{m}$ if they are arranged in centered cube?

• I guess you have a problem visualizing the "centered cube". The problem is that the expression is ambiguous: You can either have a face-centered cube or a body-centered cube. – Philipp Apr 3 '14 at 16:35

The distance between $\ce{Rb^{+}}$ and $\ce{I^{-}}$ in the fcc lattice would be $(d(\ce{Rb^{+}})/2) + (d(\ce{I^{-}})/2) = 3.76 \times 10^{-10} \mathrm{m}$
• @Martin You are right. In the formula above, the ion diameter $d$ should be used instead of $r$. I have edited this accordingly. – Jannis Andreska May 21 '14 at 13:57