Question

$AB$ has $ZnS$ type structure. What will be interionic distance between $A^{+}$ and $B^{-}$if lattice constant of $AB$ is $200pm$?

• A. $50\sqrt{3}$
• B. $100\sqrt{3}$
• C. $100$
• D. $\frac{100}{\sqrt{2}}$

Answer 1

Answer: (A)

$50\sqrt{3}$

$ZnS$ has cubic packing (ccp) crystal lattice structure which is analogous to the face-centered cubic lattice (FCC)

In $ZnS$, anions form FCC and cations occupy alternate tetrahedral voids.

Given, lattice constant $=$ edge length of a cube $=a=200pm$

Now, tetrahedral voids are present on the body diagonal at ${\left(\frac{1}{4}\right)}^{th}$ of the distance from each corner.

So, distance between $A^{+}$ and $B^{-}=\frac{1}{4}\times \sqrt{3}a$

$=\frac{\sqrt{3}}{4}\times 200pm=50\sqrt{3}pm$