Question

In the rock salt structure, the number of formula units per unit cell is equal to:
A.2
B.3
C.8
D.4

Answer 1

A $NaCl$ structure is known as the rock salt structure. In this structure, the sodium cations are present at the octahedral voids of an FCC lattice created by the chlorine ions. To solve this question, you must recall the formula used to find the total number of atoms present in a unit cell.
Formula used:
$n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4}$
Where, n = effective number of atoms present in the unit cell
${n_c}$ is number of atoms present at the corners of the unit cell
${n_f}$ is number of atoms present at the six faces of the unit cell
${n_i}$ is number of atoms present at the body center of the unit cell
${n_e}$ is number of atoms present at the edge centres of the unit cell

Complete step by step answer:
The number of chlorine anions in the unit cell is given by,
$n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4}$
Substituting the values, we get
$n = \dfrac{8}{8} + \dfrac{6}{2} + \dfrac{0}{1} + \dfrac{0}{4}{\text{ }}$
$n = 1 + 3 + 0 + 0{\text{ }}$
Thus, $n = 4$. It means there are four chlorine anions.
The number of sodium cations in the unit cell is given by,
$n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4}$
Substituting the values, we get
$n = \dfrac{0}{8} + \dfrac{0}{2} + \dfrac{1}{1} + \dfrac{{12}}{4}{\text{ }}$
$n = 0 + 0 + 1 + 3$
Thus, $n = 4$. It means there are four sodium cations.
Therefore the number of formula units in one unit cell is 4.

Thus, the correct answer is D.

Note:
$NaCl$ has a $1:1$ ratio of the cations to the anions. So the number of sodium ions present must be equal to the number of chlorine ions which will give us the number of formula units of the salt present in one unit cell.
We know that the chlorine anions form a Face- Centred Cubic lattice, that is, chlorine is present at the corners and at the face centers of the unit cell, while the sodium cations occupy the octahedral voids, that is, sodium is present at the edge centres and the body center.