Question

In zinc blende structure, zinc atom filled up
(A) All octahedral voids
(B) All tetrahedral voids
(C) Half number of octahedral voids
(D) Half number of tetrahedral voids

Answer 1

Hint: The chemical formula for zinc blende is $ ZnS $ and hence ions are in the ratio $ 1:1$ , meaning that in a unit cell, the number of zinc ions and sulphate ions will be equal. As the sulphate ions are anions and present on the lattice points, we need to find the total number of atoms in one unit cell using the lattice points, and the number of voids filled up by zinc ion will be equal.

Complete answer:
The chemical formula for Zinc Blende, also known as sphalerite is $ZnS $ and its constituent ions are written as $ Z{{n}^{2+}} $ and $ {{S}^{2-}} $
From the chemical formula, we can say that the amount of zinc and sulphur will be equal. With respect to a unit cell, the number of ions of sulphur will be equal to the number of ions of zinc.
Now, we know that a void is formed in the space left empty between the lattice atoms. Hence, naturally the size of the void is less than the lattice sites. Also, we know that for ions, the size of an anion is larger than the size of the cation, given the size of neutral atoms following the same trend.
Thus, anions, in this case $ {{S}^{2-}} $ will occupy the lattice points and the cations, in this case $ Z{{n}^{2+}} $ will occupy the voids.
It has been practically observed and established that Zinc Blende possesses face centred cubic structure. The lattice points of the FCC structure are the eight corners of the unit cell that contribute $ \dfrac{1}{8} $ in a particular unit cell and the six face centres that contribute $ \dfrac{1}{2} $ in the particular unit cell.
Hence, the total number of atoms present in a single unit cell is $ =8\times \dfrac{1}{8}+6\times \dfrac{1}{2} $
 $ \therefore Total=4 $
Thus, in a single unit cell there are four atoms present, or say for the particular case, four sulphate ions present in a single unit cell.
As the bonding ratio is $ 1:1\; $ , the number of zinc ions present will also be equal to four.
Now, there are two types of void – tetrahedral void and octahedral void.
A tetrahedral void is formed between four atoms, and can be located by dividing the unit cell into eight equal parts and marking the centre of the tetrahedral formed by joining three adjacent face centres and one common corner point. Hence, there are a total of eight tetrahedral voids in a unit cell.
An octahedral void is formed between six atoms and can be located at the body centre of the unit cell and edge centres of the unit cell obtained by joining two adjacent corners, two face centres related to the corners and two face centres of adjacent unit cells. Hence, considering $ \dfrac{1}{4} $ of the edge centres, there are four octahedral voids in a unit cell
From the above explanation, it is clear that the size of the octahedral void is greater than the tetrahedral void.
From practical observation, it is observed that the ratio of atomic radii of zinc and sulphate ions is $ \;0.39 $ . Hence, as the size of zinc ions is very small, it will be arranged in the tetrahedral voids.
As there are eight tetrahedral voids and only four zinc ions, only half of the tetrahedral voids will be filled.
Hence, the correct answer is Option $ (D) $ .

Note:
To fact to remember is that Zinc Blende makes a cubic close packing or face centred cubic structure, because with the help of the structure, the lattice points can be found out. Also, by practical calculations, it can be found that the radius of zinc ions is very less compared to sulphate ions. Hence, zinc will rather fill tetrahedral voids than octahedral voids, as octahedral voids are bigger in size.