Question

The no of ion pairs that Constitute one unit cell of CsCl

Answer 1

Answer:

Contribution to each unit cell is 18. Therefore, 1 cation and 1 anion per unit cell.

Answer 2

We know that,

$Cs^{+}$ is present at the cubic hole and$CI^{-}$ is present at the corners of the cube forms a simple cubic lattice.

Cation is at the center of the cube, its contribution to the unit cell is 1.

Therefore the number of cations per unit cell is    

And anions are at 8 corners of the cube, and each anion is shared by 8 unit cells.

Therefore contribution to each unit cell is

$\frac{1}{8}$

So, the number of anions per unit cell is

$=8 corners*\frac{1}{8} \\=1$

Hence,  The number of ion pairs that Constitute one unit cell of CsCl is 1 cation and 1 anion per unit cell.