Question

What is the total number of ions present in one unit cell of sodium chloride lattice?

• A. $4$
• B. $6$
• C. $12$
• D. $8$

Answer 1

Answer: (A)

$4$

The total number of ions present in one unit cell of sodium chloride lattice is $8$. It includes $4$ $N{a}^{+}$ ions and $4$ $C{l}^{-}$ ions.
There are $8$ $C{l}^{-}$ ions at $8$ corners of fcc unit cell (each one contributes one eigth to unit cell) and $6$ $C{l}^{-}$ ions at 6 face centres (each one contributes one half to unit cell). Total number of $C{l}^{-}$ ions in the unit cell $=8\times \frac{1}{8}+6\times \frac{1}{2}=1+3=4$
$N{a}^{+}$ ions occupies octahedral voids. There are 4 octahedral voids in $fcc$ unit cell. Thus, there are 4 $N{a}^{+}$ ions in $fcc$ unit cell.