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

We must remember that the zinc blende is described as a cubic close packing (ccp), otherwise called face-centered cubic, structure. The FCC unit cell is made of four cubic layers which are closely packed. In zinc sulfide, Face centered cubic cells are formed by anions and the alternate tetrahedral voids are occupied by cations.

Complete step by step answer:
We need to know that the Zinc sulfide $\left( {ZnS} \right)$ is a one of a kind compound that structures two sorts of glasslike structures. These two polymorphs are wurtzite and zinc blende (otherwise called sphalerite). Wurtzite has a hexagonal structure, whereas zinc blende is cubic. It is portrayed by single bonds between every molecule and upkeep of a $1:1$zinc to sulfur proportion.
In zinc sulfide, Face centered cubic cells are formed by anions and the alternate tetrahedral voids are occupied by cations.
Tetrahedral voids are found on the diagonal of the body at ${\left( {\dfrac{1}{4}} \right)^{th}}$ of the distance from every corner. So, distance between ${A^ + }$ and ${B^ - }$ is calculated as,
Distance between ${A^ + }$ and ${B^ - }$ is given by the formula $\dfrac{1}{4} \times \sqrt 3 a$
Here, a is the lattice constant which is the edge length of a cube. The value of the edge length of a cube is $200pm$. The edge length of a cube is given as a.
Let us now substitute the value of edge length of a cube in the formula.
Distance between ${A^ + }$ and ${B^ - }$ = $\dfrac{1}{4} \times \sqrt 3 a$
Distance between ${A^ + }$ and ${B^ - }$= $\dfrac{{\sqrt 3 }}{4} \times 200pm$
Distance between ${A^ + }$ and ${B^ - }$= $50\sqrt 3 pm$
The distance between ${A^ + }$ and ${B^ - }$ is $50\sqrt 3 pm$.

So, the correct answer is Option A.

Note: One should know that in a face centered cubic cell, the atoms present on the edge are not in bonds with each other whereas the atoms present on diagonal are touching each other and so they make up the interionic distance too. In a face centered cubic, the spherical atoms take $74\%$ of the whole volume.