Question

Consider the ionic compound CaF_{2}CaF ​2 ​​ . Given that r_{Ca}=0.197 nmr ​Ca ​​ =0.197nm, r_{Ca}^{+2} =0.106 nmr ​Ca ​+2 ​​ =0.106nm, rF=0.06 nm, and rF^- 0.133 nm, estimate the coordination numbers for each of the elements present in the compound.

Answer 1

The question is a bit unclear; CaF₂ is the only ionic compound present. CaF is hypothetical and doesn’t exist. We need to find out the radius ratio for CaF₂ which is defined as

Radius Ratio = rC/rA where rC and rA are the radii of the cation and the anion.

Radius Ration = (0.106 nm)/(0.133 nm) = 0.79699 ≈ 0.797

The value of radius ratio falls in the range 0.732-1.000 and hence, the co-ordination number of both the cation and the anion will be 8. The void will be cubic in nature and hence the compound must assume cubic geometry (ans).