Question

For the fluorite structure $\left(Ca{F}_2\right)$, which of the following is/are correct?

• A. For an anion, $F^{-}$ , number of nearest cations are 6
• B. For an anion $F^{-}$ , number of nearest cations are 4
• C. For a cation $C{a}^{2+}$, distance of the nearest anion is equal to $\frac{\sqrt{3}a}{4}$
• D. For a cation $C{a}^{2+}$, distance of the nearest cation is equal to edge cell length $a$

Answer 1

Answer: (B), (C)

For a cation $C{a}^{2+}$, distance of the nearest anion is equal to $\frac{\sqrt{3}a}{4}$

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

Thus, $C{a}^{2+}$ occupies the corners and face centres of the cube. $F^{-}$ ion occupies all the tetrahedral voids of the fcc.

$\text{Total}Ca{F}_2\text{units in one unit cell}=4$

Nearest distance between two $C{a}^{2+}$ ions is the distance between corner and face centre which is equal to $\frac{a}{\sqrt{2}}$
Option (a) is incorrect.

For a cation $C{a}^{2+}$, distance of the nearest anion is equal to the distance between $C{a}^{2+}$ ion present at corner and $F^{-}$ ion present at tetrahedral void, which is equal to $\frac{\sqrt{3}a}{4}$
Option (b) is correct.

$F^{-}$ is present at tetrahedral voids, the number of nearest cations $C{a}^{2+}$ are 4
Option (c) is correct