Question

The coordination number of cation and anion in flourite $Ca{F}_2$ 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 1

The correct option is C 8 : 4 and 8 : 8

1.
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
$C.N.ofCa{F}_2=8:4$

2.
In $CsCl,C{l}^{-}$ form simple cubic lattice. Thus, $C{l}^{-}$ ions occupies the corners of the cube and the number of $C{l}^{-}$ ions per unit cell is
$\frac{1}{8}\times 8=1$
$C{s}^{+}$ ions occupies the cubical void in unit cell.
This type of structure is formed when the anion and cation have similar sizes.
Here each $C{s}^{+}$ is surrounded by 8 $C{l}^{-}$ and each $C{l}^{-}$ is surrounded by 8 $C{s}^{+}$ions.
Thus, the coordination number of $CsCl$,
$C.N.ofCsCl=8:8$