Question

ln calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion $\left(C{a}^{2+}\right)$ and fluoride ion $\left(F^{-}\right)$ are

• A. 4 and 8
• B. 4 and 2
• C. 6 and 6
• D. 8 and 4

Answer 1

The correct option is D 8 and 4

In fluorite structure, the $C{a}^{2+}$ forms the FCC lattice.

Thus, $C{a}^{2+}$ occupies the corners and face centres of the cube.
The number of $C{a}^{2+}$ ions in each unit cell is
$\frac{1}{8}\times 8+\frac{6}{2}=4$

$F^{-}$ ion occupies all the tetrahedral voids of the FCC. There are 8 tetrahedral voids in FCC.
Thus, the number of $F^{-}$ ions in each unit cell is 8.

Here, each $F^{-}$ ion in tetrahedral void is surrounded 4 $C{a}^{2+}$ ions and each $C{a}^{2+}$ ion is surrunded by 8 $F^{-}$ ion.

Thus, the coordination number of $F^{-}$ is 4 and coordination number of $C{a}^{2+}$ is 8
$C.N.ofCa{F}_2=8:4$