Question

In the fluorite structure if the radius ratio is $(\sqrt{\frac{3}{2}}-1)$, how many ions does each cation touch?

• A. 4 anions
• B. 12 cations
• C. 8 anions
• D. No cations

Answer 1

Answer: (B), (C)

The correct options are
B 12 cations
C 8 anions

Fluorite formula is $Ca{F}_2$ which occupies in FCC.

$8F^{-}$ ions occupies $8$ tetrahedral holes and $4C{a}^{+2}$ is present in an FCC unit cell.

$\frac{r^{-}}{r^{+}}=0.225\approx \sqrt{\frac{3}{2}}-1\to$ which means that $f^{-}$ ion is in the tetrahedral void.

Here, $r^{-}\to$ Radius of anion

$r^{+}\to$ Radius of cation

So, $A$ cation can touch $12$ cations and $8$ anions.