Question

# The coordination number of cation and anion in fluorite $(Ca{F_2}){\text{ and CsCl}}$ are respectively:$(a){\text{ 8:4 and 6:3}} \\ (b){\text{ 6:3 and 4:4}} \\ (c){\text{ 8:4 and 8:8}} \\ (d){\text{ 4:2 and 2:4}} \\$

Hint – In this question use the concept that in $Ca{F_2}$ cations are present at ccp sites and anions are present at tetrahedral voids whereas in $CsCl$ cations are present at corners of cubes and anions are present at the central cubic void. This will help getting the ratio of coordination numbers in the respective fluorite.

In $Ca{F_2}$ cations are present at ccp sites and anions are present at tetrahedral voids, and coordination number of cations and anions in $Ca{F_2}$ are 8 and 4 respectively.
In $CsCl$ cations are present at corners of cubes and anions are present at the central cubic void, and coordination number of cations and anions in $CsCl$ are 8 and 8 respectively.
So the ratio of cations and anion in $Ca{F_2}$ is (8:4).
And the ratio of cations and anion in $CsCl$ is (8:8).
Note – Calcium fluoride ($Ca{F_2}$) is an (8, 4) structure, meaning that each cation $C{a^{2 + }}$ is surrounded by eight ${F^ - }$anion neighbors, and each ${F^ - }$ anion by four $C{a^{2 + }}$. So coordination numbers of $C{a^{2 + }}$ and ${F^ - }$ ions in $Ca{F_2}$ crystal are 8 and 4 respectively. The formula unit for cesium chloride is CsCl, also a 1:1 ratio or (8: 8) as each Cl − ion is also surrounded by eight $C{s^ + }$ ions.