In a unit cell containing `X^(2+), Y^(3+)` and `Z^(2-)` where `X^(2+)` occupies 1/8 th of tetrahedral voids, `Y^(3+)` occupies 1/2 of octanhedral voids and `Z^(2-)` forms ccp structure. Then formula of compound is

Options

A. `X_(2)Y_(4)Z`
B. `XY_(2)Z_(4)`
C. `XY_(3)Z_(4)`
D. `X_(4)YZ_(2)`

Answer

To find the formula of the compound containing \(X^{2+}\), \(Y^{3+}\), and \(Z^{2-}\) in a unit cell where \(Z^{2-}\) forms a cubic close-packed (CCP) structure, we can follow these steps:

### Step 1: Determine the number of \(Z^{2-}\) ions in the unit cell
In a cubic close-packed (CCP) structure, the number of ions (Z) is 4. This is because in a face-centered cubic (FCC) arrangement, there are:
- 8 corner atoms contributing \( \frac{1}{8} \) each = 1 atom
- 6 face-centered atoms contributing \( \frac{1}{2} \) each = 3 atoms
Thus, total = \(1 + 3 = 4\).

### Step 2: Determine the number of tetrahedral voids
In a unit cell with Z = 4, the number of tetrahedral voids is \(2n\), where \(n\) is the number of atoms in the unit cell. Therefore, for Z = 4:
\[ \text{Number of tetrahedral voids} = 2 \times 4 = 8 \]

### Step 3: Determine the number of \(X^{2+}\) ions
According to the problem, \(X^{2+}\) occupies \( \frac{1}{8} \) of the tetrahedral voids. Since there are 8 tetrahedral voids:
\[ \text{Number of } X^{2+} = \frac{1}{8} \times 8 = 1 \]

### Step 4: Determine the number of octahedral voids
In a unit cell with Z = 4, the number of octahedral voids is equal to the number of atoms, which is 4.

### Step 5: Determine the number of \(Y^{3+}\) ions
The problem states that \(Y^{3+}\) occupies \( \frac{1}{2} \) of the octahedral voids. Since there are 4 octahedral voids:
\[ \text{Number of } Y^{3+} = \frac{1}{2} \times 4 = 2 \]

### Step 6: Compile the formula
Now we can compile the formula based on the number of each type of ion in the unit cell:
- \(X^{2+}\) = 1
- \(Y^{3+}\) = 2
- \(Z^{2-}\) = 4

Thus, the formula of the compound is:
\[ \text{Formula} = X_1Y_2Z_4 \]

### Final Answer
The formula of the compound is \(X_1Y_2Z_4\).

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