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}} \\\$

Answer 1

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.

Complete answer:
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).
So this is the required answer.

Hence option (C) is the correct answer.
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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.