Question

How many sodium ions and chloride ions are present in a unit cell of sodium chloride crystal?

Answer 1

Hint: To solve this we need to know that sodium chloride crystallizes in a cubic crystal lattice. It has an fcc lattice. The chlorine ions are arranged in the fcc lattice and the sodium ions in the octahedral sites. To solve this calculate the contribution from each of these atoms in a unit cell. The number of sodium ions will be equal to the number of chloride ions.

Complete step by step solution:
We know that the sodium chloride crystallizes in a cubic crystal lattice. In the cubic lattice, the chloride ions are arranged in a cubic close-packed arrangement and the sodium ions are present in the octahedral sites.
There are sodium cations and chloride anions present it. They are present in equal proportions as we can understand from its formula i.e. NaCl.
NaCl crystallises in a face-centred cubic lattice. In a face-centred cubic lattice, there are atoms present on each corner of the cube and also in the face centre of each face.
We know that there are 8 corners in a cube and 6 faces. Therefore, there are eight chloride ions on the corners and six chloride ions at the face centre.
Firstly, let us discuss the contribution of the chlorine atoms.
Each of the corner atoms is shared with eight other corners. Therefore contribution from each corner atom is one-eighth. Each face centre atom is shared with another face centre. Therefore, the contribution from each face centre atom is half.
As we have 8 corner and 6 face-centre ions, therefore, we can write that the total contribution of the chloride ion in one unit cell is $8\times \dfrac{1}{8}+6\times \dfrac{1}{2}=4$ . So, there are four chlorine ions in the unit cell of sodium chloride.
Now, let us discuss the contribution from the sodium atoms.
The sodium ions are present on the edge and in the body-centre. The contribution from the body-centre ion is 1 as it is not shared with any other unit cell. There are 12 edges in a cube, so there are 12 atoms on each edge. Each atom on the edge is shared with 4 other unit cells. Therefore, the contribution from each atom on the edge is one-fourth.
As we have 12 edge atoms and 1 body centred atom so we can write that the total contribution of the sodium ions in one unit cell is $12\times \dfrac{1}{4}+1=4$. So, there are four sodium ions in the unit cell of sodium chloride.


We can understand from the above discussion that the required answer is 4 sodium cations and 4 chlorine anions are present in a unit cell of sodium chloride.

Note: We know that a lattice is a network which gives us the three-dimensional arrangement of atoms. There are 14 different Bravais lattices that we study in the case of solids. The body centred and the cubic close-packed structures are also Bravais lattice. BCC is one of the primitive lattices and CCP is just like the face centred Bravais lattice.