# 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 $(d(\ce{Rb^{+}})/2) + (d(\ce{I^{-}})/2) = 3.76 \times 10^{-10} \mathrm{m}$

• Thank you, can you please clarify the solution? How can we explaint that ions do not touch each other? Why is there a gap between them? – user5070 Apr 3 '14 at 14:29
• There is not really a gap between the ions since they are closely packed in the structure. The calculated distance is the distance between the "centers" of the ions when you imagine them as spheres, that means, the distance between the nuclei. – Jannis Andreska Apr 3 '14 at 14:37
• Thank you. I guess it was my interpretation of the question. – user5070 Apr 3 '14 at 14:41
• Unfortunately this is wrong. The distance between the centres of the ions is the sum of the radii, not only half of the radii. – Martin - マーチン May 21 '14 at 3:04
• @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