Question

If a solid $AB$ having $ZnS$ structure is heated so that the ions along two of the axis passing through the face centre particles are lost and bivalent ion $\left( Z \right)$ enters here to maintain the electrical neutrality so that the new formula unit becomes ${A_x}{B_y}{Z_c}.$ Find the value of $x + y - c.$

Answer 1

The smallest indivisible particle of solids is known as a unit cell. Consisting of these units cells, there are three structures:- Primitive unit cell, body centered unit cell (BCC) and face centred unit cell (FCC).

Complete step by step answer:
We know that $ZnS$ is a type of face-centered cubic unit cell. A face centered cubic unit cell contains atoms at all the $8$ corners and the centre of all the $6$ faces of the cube. The atoms located at the corners of a cubic unit cell are shared by $8$ unit cells. Thus, only $\dfrac{1}{8}$ th of an atom actually belongs to a particular unit cell. Moreover each atom located at the face-centre is shared between two adjacent unit cells and only $\dfrac{1}{2}$ of each atom belongs to a unit cell. Thus the total number of atoms per unit cell is $8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = 4{\text{ atoms}}{\text{.}}$
There are two other types of unit cell; Primitive cubic unit cell and body centered-cubic unit cell. In a primitive cubic unit cell, the atoms are only present at the corners of the unit cell and thus the total number of atoms per unit cell is $1$ atom. In a body centered cubic (BCC) unit cell, atoms are present at the corners of the unit cell and at the centre of the unit cell which is not shared with any other unit cell. Therefore total number of atoms per unit cell is $\Rightarrow$ $\dfrac{1}{8} \times 8 + 1 \times 1 = 2{\text{ atoms}}{\text{.}}$
Now we got to know that $ZnS$ is fcc structure and the atoms present per unit cell is $4$.
Therefore, number of anions present $=$ $4/unit{\text{ cell}}$
[ because anions are present on the corners and at the face centres]
So, number. of ${A^ + }$ ions present $= 4/{\text{unit cell}}$
$4\left[ {{\text{Two from each face centre}}} \right] \times \dfrac{1}{2}\left[ {{\text{Per face centre share}}} \right] = 2$
Therefore, No. of ${B^ - }$ left $= 4 - 2 = 2/{\text{Unit cell}}$
Now to maintain electrical neutrality, ${Z^{ - 2}}$ is added. Now $2{B^ - }$ are removed/lost so the loss of charge is $2\left( { - 1} \right) = - 2$. To overcome this charge, one ${z^{ - 2}}$ is added to maintain electrical neutrality.
Thus the formula of compound become $=$ ${A_4}{\text{ }}{{\text{B}}_2}{\text{ }}{{\text{Z}}_1}$
Comparing this with ${A_{x\;}}\;{B_{y\;}}\;{C_{z\;}},$ we get
$x = 4,\;y = 2\;\& \,\;C = 1$
Therefore, value of $x + y - c\; = \;4 + 2 - 1 = 5$
Hence, $5$ is the required answer.

Note: The empty space between the atoms in a unit cell is known as void. Types of voids present in a unit cell are :- tetrahedral voids and octahedral voids. If the no. of close packed particles are $N$
Then, the no. of octahedral voids $=$ $N$
and no. of tetrahedral voids $=$ $2N$
The tetrahedral voids are double of octahedral voids.