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

Last updated date: 11th Aug 2024
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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 = $8\times \cfrac { 1 }{ 8 } +6\times \cfrac { 1 }{ 2 } =4$
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 = $4\times \dfrac { 4 }{ 3 } \pi { r }^{ 3 }$
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 = $\cfrac { 4\times \cfrac { 4 }{ 3 } \pi { r }^{ 3 } }{ 8\sqrt { 8 } { r }^{ 3 } } =0.74$
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 ${ Mg }_{ 2 }Si$.