
The coordination number of cation and anion in $Ca{F_2}$ and $CsCl$ are respectively:
A. $8:4$ and $6:3$
B. $6:3$ and $4:4$
C. $8:4$ and $8:8$
D. $4:2$ and $2:4$
Answer
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Hint: Here to solve this question one needs to know the arrangement of cations and anions in the structures which will develop an approach to the solution. In the structure, $Ca{F_2}$ the calcium cations are present at the cubic close pack site while the fluoride anions are present at the tetrahedral voids. In the structure, $CsCl$ the caesium cations are present at corners of cubes while the chloride anions are present at the central cubic void. By using this data one can find the coordination number and the ratio.
Complete step by step answer:
-In the molecular structure, $Ca{F_2}$ the calcium cations are present at the cubic close pack site while the fluoride anions are present at the tetrahedral voids which shows the coordination number for cation as $8$ and for anion as $4$ which will give the ratio for these as $8:4$ for the structure.
-In the structure, $CsCl$ the caesium cations are present at corners of cubes while the chloride anions are present at the central cubic void which shows the coordination number for the cations as $8$ and for anion as $8$ which will give the ratio for these as $8:8$ for the structure.
-Therefore, the ratio of cations and anion in the molecule $Ca{F_2}$ is $8:4$ and for the molecule $CsCl$ the ratio of cations and anions is $8:8$
Therefore, the above combination of the ratio is found out in option C, which shows the option C as the correct choice.
Note: In the molecular structure $Ca{F_2}$ the four cations of calcium i.e. $C{a^{2 + }}$ is surrounded by the eight anions of the fluoride i.e. ${F^ - }$ which can also be said as for one calcium cation there are four fluoride anions. Which gives the coordination number ratio as $8:4$. In the molecular structure, $CsCl$ the molecular ratio is $1:1$ which is similar to the ratio $8:8$. This is also an alternative method to find out the coordination number ratio in a molecule.
Complete step by step answer:
-In the molecular structure, $Ca{F_2}$ the calcium cations are present at the cubic close pack site while the fluoride anions are present at the tetrahedral voids which shows the coordination number for cation as $8$ and for anion as $4$ which will give the ratio for these as $8:4$ for the structure.
-In the structure, $CsCl$ the caesium cations are present at corners of cubes while the chloride anions are present at the central cubic void which shows the coordination number for the cations as $8$ and for anion as $8$ which will give the ratio for these as $8:8$ for the structure.
-Therefore, the ratio of cations and anion in the molecule $Ca{F_2}$ is $8:4$ and for the molecule $CsCl$ the ratio of cations and anions is $8:8$
Therefore, the above combination of the ratio is found out in option C, which shows the option C as the correct choice.
Note: In the molecular structure $Ca{F_2}$ the four cations of calcium i.e. $C{a^{2 + }}$ is surrounded by the eight anions of the fluoride i.e. ${F^ - }$ which can also be said as for one calcium cation there are four fluoride anions. Which gives the coordination number ratio as $8:4$. In the molecular structure, $CsCl$ the molecular ratio is $1:1$ which is similar to the ratio $8:8$. This is also an alternative method to find out the coordination number ratio in a molecule.
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