Question

The coordination number of calcium fluoride type structure is (cation : anion):
(a) 1:2
(b) 4:4
(c) 4:8
(d) 8:4

Answer 1

Hint: Calcium fluoride has a fluorite type of structure in which the cation forms the FCC unit cell while the anions occupy the tetrahedral sites.

Complete step by step solution:
For solving this question, we need to understand the structure of the face centred cubic unit cell. The diagram of Face centred cubic unit cell is given below:

Here atoms are present at the corners as well as at the face centres. The atoms present at the corners are shared among 8 unit cells while the atoms present at face centres are shared among 2 unit cells. Therefore the total number of atoms per unit cell is = $8\times \frac{1}{8}+6\times \frac{1}{2}=4$
Let us calculate the packing efficiency for this unit cell. For that we need to establish a relationship between r and a where r is the radius of the atoms and a is the side of the cube.
Since the atoms present at the face centers are touching the atoms present at the corners of that face, therefore:
$a=\sqrt{8}r$.

According to the Pythagoras equation:
$(4r{)}^2=a^2+a^2$
Therefore, $a=\sqrt{8}r$.
The volume occupied by the atoms present in the unit cell is = $4\times {\displaystyle \frac{4}{3}}\pi r^3$
 The volume of the unit cell will be:
 Since $a=\sqrt{8}r$, therefore
$a^3=8\sqrt{8}{r}^3$
 The packing efficiency of the unit cell is = $\frac{4\times \frac{4}{3}\pi r^3}{8\sqrt{8}{r}^3}=0.74$
Hence a face centred cubic unit cell has some voids present in its structure. These voids are octahedral voids and tetrahedral voids. The octahedral voids have a coordination number of six while the tetrahedral voids have a coordination number of four. If there are N atoms per unit cell, then there will be N octahedral voids and 2N tetrahedral voids.
In ${CaF}_2$, the ${Ca}^{2+}$ ions occupy all the corner and the face centres positions in the face centred cubic unit cell while the $F^{-}$ ions occupy all the tetrahedral voids. Therefore the coordination number of calcium fluoride type structure will be 8:4.

Hence the correct answer is (d) 8:4

Note: ${CaF}_2$ has a fluorite type of structure. In a fluorite type of structure, the cations form the FCC unit cell while the anions occupy the tetrahedral sites. There are some compounds that have an anti-fluorite type of structure. In these compounds the anions form the FCC structure while the cations occupy the tetrahedral sites. For example ${Mg}_{2}Si$.