Question

# Number of $Na$ atoms present in the unit cell of $NaCl$ crystal is : A. $1$B. $2$C. $3$D. $4$

Hint: Firstly, A unit cell is referred to as the smallest repeating unit in a crystal lattice. Unit cells occur in many different varieties. And the cubic crystal system is one of them which has three types of unit cells and one of them is face-centered unit cell and $NaCl$ is an example of fcc lattice.
In $NaCl$ the $C{l^ - }$ ions are present at the corners of the unit cell and at the face-centers of the unit cell and the Na+ ions are present at the octahedral voids of the unit cell. Imagining a unit cell as a cube there are $8$ corners and $6$ face-centers. The octahedral voids present in a unit cell are “$n$”, where ”$n$” is the number of atoms present in a unit cell.
Before calculating the number of carbon atoms in the unit cell one must know about the contribution of the atoms towards a unit cell in the lattice. An atom present at the corner contributes $\dfrac{1}{8}$to a unit cell as it is shared to $8$ different unit cells attached together, whereas the one present at the face-center contributes $\dfrac{1}{2}$to a unit cell, and the contribution of atoms present at octahedral voids is $1$.
Now, let’s calculate the number of $Na$ atoms present in $NaCl$:
As there are $4$ atoms present at octahedral voids therefore,
$1 \times 4 = 4$,where $n = 4$ for fcc lattice.
So, the total number of $Na$ atoms is $4$.
A student can be confused about the number of octahedral (“ $n$ ”) and tetrahedral voids (“ $2n$ ”) present in a lattice so a student should keep the values in mind. And when it comes to the contribution of the octahedral and the tetrahedral voids, they have a contribution of $1$, as they are totally a part of the lattice.