The edge length of an FCC cell is $\pu{508 pm}$. If radius of cation is $\pu{110 pm}$, the radius of anion is? Diagram of FCC cell edge length

My approach: $$(2R_{\text{cation}}+2R_{\text{anion}})^2=2a^2$$ Which on simplification gives: $$\sqrt 2(R_{\text{cation}}+R_{\text{anion}})=a$$

On putting the values, the answer comes out to be $\pu{249.21 pm}$ which seems correct but here the answer is something else Please help me about it.

  • $\begingroup$ I think your calculation is correct. $\endgroup$
    – Ed V
    Sep 8, 2021 at 20:09
  • $\begingroup$ Based on the figure, which shows a square of a = 508 pm length on a side, the diagonal, from point A to point B, is a times the square root of 2. This equals the sum of the two ion diameters. And this is what your second equation gives, if you simply multiply both sides by the square root of 2. $\endgroup$
    – Ed V
    Sep 8, 2021 at 20:41
  • 1
    $\begingroup$ To nit-pick: It is actually not fcc, since fcc means that the center of the face is equivalent to the corner. Since at the corner is an anion and at the center of the face is a cation, the cell could be bcc or primitive cubic lattice but not fcc. But with this said. I also think that the answer from the link is wrong. $\endgroup$
    – Fabian
    Sep 9, 2021 at 13:57
  • $\begingroup$ The lattice is not fcc. And the drawing is not perfect. The rectangle should be a square, as both edge lengths are equal. $\endgroup$
    – Maurice
    Sep 9, 2021 at 15:54
  • $\begingroup$ @Maurice The lattice is indeed fcc but the problem was that I m considering the cations to be placed at the face centers but the truth is the anions make the whole fcc and cations are present at the octahedral voids $\endgroup$ Sep 11, 2021 at 14:04


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