How to determine the packing geometry of compounds with contradicting coordination number and radius-ratio values? Determine the packing geometry of the following compounds by using the radius ratio rules. For your convenience assume a coordination number of six for both cation and anion.
Be specific in the your answer, &quot;body-centered cubic with the anion occupying the corners of the cell [1 atom] and the cation occupying the central hole [1 atom].

Below is the table we are given, and we're are choosing between geometries of cubic, body-centered, or face-centered.

The compound I'm on is $\ce{LiBr}$
I did $\frac{r^+}{r^-}=\frac{79}{182}=0.43$ for the Radius-Ratio value.
But from here would I choose square planar because it's closer to 0.414 or do I choose octahedral because we're assuming coordination number 6? And how would I determine the locations of the cation/anion?
My best guess is cubic geometry because the cation is too large to fit tetrahedral holes and octahedral holes, with $\ce{Li}$ and $\ce{Br}$ taking occupying complementary corners.

I'm making the assumption that the coordination number 6 is used as a reference to the atomic radii.
I've currently adjusted my answer based off of other examples I could find to

FCC with OH holes occupied by $\ce{Li^+}$ [4 atoms] and the $\ce{Br^-}$ forms the FCC Lattice [4 atoms (by stoichiometry}] for a 100% occupancy rate.

But I still have an issue as to how to conceptually approach the problem.
You've calculated your radius ratio value and you know you're between 0.414 and 0.732, so cubic is off the table, based on your chart.

You're looking at either square planar or octahedral. Your chart also says that ionic compounds do not use square planar packing and the coordination number would have to be 4. LiBr is an ionic compound (like NaCl) with a coordination number of 6, therefore you're looking at octahedral.

From there you can determine your "specific" answer.