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Question: A binary solid \(\left( {{A}^{+}}{{B}^{-}} \right)\) has zinc blende structure with \({{B}^{-}}\) io...

A binary solid (A+B−)\left( {{A}^{+}}{{B}^{-}} \right) has zinc blende structure with B−{{B}^{-}} ions constituting the lattice and A+{{A}^{+}} occupying 25% of the tetrahedral holes. The formula of the solid is:
[A] AB
[B] A2B{{A}_{2}}B
[C] AB2A{{B}_{2}}
[D] AB4A{{B}_{4}}

Explanation

Solution

To solve this firstly identify the zinc blende lattice. Find the voids in that lattice and the arrangement of ions in it. Compare the binary solid with that structure. Do not forget to consider only 25% of the tetrahedral voids.

Complete step by step answer:
To solve this firstly we have to understand the zinc blende structure and also find out the ions in its lattice and the voids.
Zinc blende is a structure that ZnS i.e. zinc sulphide can take. In this structure, the structure is cubic close packed and the crystals are in a face-centred cubic lattice.
Here, the anion i.e. the sulphide ion occupies the corners and the face centres. So let us calculate its contribution.

We know there are 8 corners in a cube and each corner is shared by 8 other corners.
So, the total contribution from the corner atoms is 8×188\times \dfrac{1}{8} i.e. 1.
Also, we know that there are 6 face centres. So in every face centre there is one sulphide ions and each face centre is shared with another face.
So, the total contribution from the face centred ions is 6×12=36\times \dfrac{1}{2}=3.
Therefore, the total number of sulphide ions is 4.
Now, let us discuss the contribution from the zinc ions.
We know in fcc there are 8 tetrahedral voids. In zinc blende structure, Zn2+Z{{n}^{2+}} occupies alternate tetrahedral voids i.e. it occupies 4 tetrahedral voids.

So, we have equal number of zinc and sulphide ions (which is also obvious from the formula ZnS)
Now, the given binary compound given to us has a zinc blende structure i.e. the B−{{B}^{-}} ions occupy the lattice positions i.e. similar to the sulphide ions. So, there are 4 B−{{B}^{-}} ions.
And A+{{A}^{+}} occupy 25% of the tetrahedral voids. There are 8 tetrahedral voids so 25% of 8 voids is 2.
Therefore, the formula of the binary solid is A2B4 or AB2{{A}_{2}}{{B}_{4}}\text{ or A}{{\text{B}}_{2}}.
So, the correct answer is “Option C”.

Note: Zinc sulphide comes in two structures. One we have discussed above and the other is wurtzite. In both the forms zinc and sulphur ratio is 1:1 and maintains tetrahedral arrangement. The above discussed structure is also known as sphalerite. Sphalerite is thermodynamically stable compared to wurtzite but both of them are found.