Question

AB has ZnS type structure. What will be the interionic distance between \[{A^ + }\] and \[{B^ - }\] if the lattice constant of AB is 200pm?
A. \[50\sqrt 3 \]
B. \[100\sqrt 3 \]
C.100
D. \[\dfrac{{100}}{{\sqrt 2 }}\]

Answer 1

Hint: ZnS has cubic crystal lattice structure which is analogous to the face-centred cubic lattice (FCC) in which anions form FCC and cations occupy alternate tetrahedral voids that are present on the body diagonal at $\dfrac{1}{4}$ th of the distance from each corner.

Complete step by step answer:
Face-centred cubic lattice has lattice points at the eight corners of the unit cell and additional lattice points at the face-centres of the unit cell. Its unit cell vectors are a=b=c and four atoms in its unit cell.
 \[8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = 4\]
Let us suppose the radius of an atom is r and the edge length of the unit cell is E. For FCC lattice, the edge length of the unit cell is \[\dfrac{4}{{\sqrt 3 }}r\] .
We can write it as, \[E = \dfrac{4}{{\sqrt 3 }}r\]
Further on rearranging it, we get \[\sqrt 3 E = 4r\]
We have to calculate the inter-ionic distance between \[{A^ + }\] and \[{B^ - }\] , which is equal to r.
For r, the formula now becomes \[r = \dfrac{1}{4} \times \sqrt 3 E\] which implies that the alternate tetrahedral voids are present on the body diagonal at $\dfrac{1}{4}$ th of the distance from each corner.
Given that E = 200pm, so putting this value in the above formula, we get
  \[r = \dfrac{{\sqrt 3 }}{4} \times 200pm = 50\sqrt 3 \]

Hence, the correct option is (A).

Note:
One must know that in FCC, the atoms on edge are not in contact with each other but atoms on diagonal are touching each other and so they make up the interionic distance too. In FCC, the spherical atoms occupy \[74\% \] of the total volume.