Question

In the $ZnS$ (Zinc Blende) structure, what would be the distance for the nearest cation $\left(Z{n}^{2+}\right)$ from an anion $\left(S^{2-}\right)$ ?
Here $a$ is the edge cell length.

• A. $\frac{\sqrt{3}a}{4}$
• B. $\sqrt{2}a$
• C. $\frac{a}{4}$
• D. $a$

Answer 1

The correct option is A $\frac{\sqrt{3}a}{4}$

$ZnS$ (Zinc Blende) structure :
$S^{2-}$ ions form the FCC lattice,while $Z{n}^{2+}$ ions occupy alternate (nonadjacent) tetrahedral voids.


For an anion $S^{2-}$,
Distance of the nearest cation $Z{n}^{2+}$ will be $\frac{\sqrt{3}a}{4}$.
This is the distance from the corner atom $S^{2-}$ to the atom at tetrahedral void $Z{n}^{2+}$

Option (a) is correct.
Number of nearest cations are 4
Theory :

Number of $S^{2-}$ ions in Zinc blended structure = 4
Number of $Z{n}^{2+}$ ions in Zinc blended structure = 4
Total $ZnS$ units in one unit cell = 4

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