Question

In sodium chloride, $C{l^ - }$ ions form ccp arrangement. Which site will $N{a^ + }$ ions occupy in this structure?
A. cubic
B. tetragonal
C. octahedral
D. trigonal bipyramidal

Answer 1

For solving this question we need to understand the arrangement of atoms in a ccp arrangement. In a ccp arrangement, there are a total of 4 spheres per unit cell and has a coordination number of 12. Also, we know that the packing efficiency is the fraction of volume in a crystal structure that is occupied by constituent particles, rather than empty space. Therefore, the packing efficiency of ccp arrangement is 74%.

Complete step by step answer:
In the question we need to define the bond between sodium chloride, where the $C{l^ - }$ ions form a ccp arrangement. The $N{a^ + }$ ions occupy the octahedral voids in NaCl structure. The compound NaCl has coordination number of 6.
We can confirm the coordination number of the compound from the radius ratio. In case of octahedral voids, the radius ratio is $\dfrac{r}{R}$ = 0.414.
We can define the radius ratio as the ratio of the radius (r) of the atom or ion which can exactly fit in the octahedral void formed by spheres of radius R.
So, radius ratio of NaCl = $\dfrac{r}{R} = \dfrac{{95}}{{181}} = 0.524$

So, the correct answer is Option C.

Note:
After solving this question, we need to understand the concept of different arrangements. In a hexagonal closest packed (hcp) the coordination number is 12 and it contains 6 atoms per unit cell. In a face-centered cubic (fcc), the coordination number is 12 and there are 4 atoms per unit cell. And in a body-centered cubic (bcc) the coordination number is 8 and there are 2 atoms per unit cell.