Question
In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion $\ce{(Ca^{2+})}$ and fluoride ion $(F^−)$ are:
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Answer
Calcium fluoride crystallizes in a Face $-$ Centered Cubic unit cell $\ce{(FCC)}$ having an edge length of $5.463$ Angstroms.
The cell is displayed here with calcium cations $($in blue$)$ defining $\ce{FCC}$ lattice sites, and fluoride anions $($in green$)$ occupying all tetrahedral sites.
Fluoride anions have $4$ neighbors of opposite charge arranged at vertices of an tetrahedron.
Calcium cations have $\text{EIGHT}$ neighbors of opposite charge arrange at corners that outline a smaller cube.
So the $Ca: F$ coordination ratio is $8 : 4$ or $2 : 1$.
In $\ce{4CaF_2}$ the the co ordination number of $\ce{Ca^{+2}}$ is $\ce{8 F^−}$ is $4$.
The cell is displayed here with calcium cations $($in blue$)$ defining $\ce{FCC}$ lattice sites, and fluoride anions $($in green$)$ occupying all tetrahedral sites.
Fluoride anions have $4$ neighbors of opposite charge arranged at vertices of an tetrahedron.
Calcium cations have $\text{EIGHT}$ neighbors of opposite charge arrange at corners that outline a smaller cube.
So the $Ca: F$ coordination ratio is $8 : 4$ or $2 : 1$.
In $\ce{4CaF_2}$ the the co ordination number of $\ce{Ca^{+2}}$ is $\ce{8 F^−}$ is $4$.
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