Question

How many $N{a}^{+}$ ions in each unit cell is in a cubic ideal crystal of $NaCl$?

Answer 1

$N{a}^{+}$ ions are present in the edge centre and at the body centre.

The number of $N{a}^{+}$ ions present in one unit cell $=12\times \frac{1}{4}+1=4atoms$

$C{l}^{-}$ ions are present at the corner and at the centre of the faces.

The number of $C{l}^{-}$ ions present in one unit cell $=6\times \frac{1}{4}+8\times \frac{1}{8}=4atoms$

So, each unit cell has 4 atoms of each $N{a}^{+}$ and $C{l}^{-}$.

Hence, there are total $4$ formula units of $NaCl$ in each unit cell.