A spinal is an important class of oxides consisting of two types of metal ions with oxides ions arranged in CCP layer. The normal spinal has one-eighth of the tetrahedral holes occupied by one type of metal ion and one-half of the octahedral holes occupied by another type of metal ion. such a spinal is formed by `Zn^(2+), Al^(3+) and O^(2-)` in the tetrahedral holes. if formula of the compound is `Zn_(x)Al_(y) O_(z)`, then find the value of `(x + y + z)` ?

Explanation

To solve the question, we need to determine the values of \(x\), \(y\), and \(z\) in the formula \(Zn_xAl_yO_z\) based on the information provided about the spinal structure.

### Step-by-Step Solution:

1. **Identify the number of oxide ions in the unit cell**:
- In a cubic close-packed (CCP) structure, the number of oxide ions (\(O^{2-}\)) per unit cell is 4. Therefore, \(z = 4\).

2. **Determine the number of tetrahedral and octahedral voids**:
- In a CCP structure, the number of tetrahedral voids is \(2n\) and the number of octahedral voids is \(n\), where \(n\) is the number of atoms per unit cell.
- Since \(n = 4\) (the number of oxide ions), the number of tetrahedral voids is \(2 \times 4 = 8\) and the number of octahedral voids is \(4\).

3. **Calculate the number of metal ions in tetrahedral and octahedral voids**:
- According to the problem, one-eighth of the tetrahedral holes are occupied by one type of metal ion (\(Zn^{2+}\)):
\[ \text{Number of } Zn^{2+} = \frac{1}{8} \times 8 = 1 \]
Thus, \(x = 1\).
- The other type of metal ion (\(Al^{3+}\)) occupies half of the octahedral holes:
\[ \text{Number of } Al^{3+} = \frac{1}{2} \times 4 = 2 \]
Thus, \(y = 2\).

4. **Summarize the values**:
- We have determined:
- \(x = 1\)
- \(y = 2\)
- \(z = 4\)

5. **Calculate \(x + y + z\)**:
\[ x + y + z = 1 + 2 + 4 = 7 \]

### Final Answer:
The value of \(x + y + z\) is \(7\).

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