`MgAl_(2)O_(4)`, is found in the Spinal structure in which `O^(2-)` ions consititute CCP lattice, `Mg^(2+)` ions occupy 1/8th of the Tetrahedral voids and `Al^(3+)` ions occupy 1.2 of the Octahedral voids. Find the total +ve charge contained in one unit cell

Options

A. `+7//4` electronic charge
B. `+6` electronic charge
C. `+2` electronic charge
D. `+8` electronic charge

Explanation

To find the total positive charge contained in one unit cell of `MgAl2O4` in the spinal structure, we can follow these steps:

### Step 1: Identify the charges of the ions
- Magnesium ions (`Mg^(2+)`) have a charge of +2.
- Aluminum ions (`Al^(3+)`) have a charge of +3.
- Oxygen ions (`O^(2-)`) have a charge of -2, but we will not include their charge in our calculation of total positive charge.

### Step 2: Determine the number of tetrahedral voids in the unit cell
In a cubic close-packed (CCP) lattice, which is also known as face-centered cubic (FCC):
- There are a total of 8 tetrahedral voids.

### Step 3: Calculate the contribution of `Mg^(2+)` ions
- According to the problem, `Mg^(2+)` ions occupy 1/8th of the tetrahedral voids.
- Therefore, the number of `Mg^(2+)` ions in one unit cell is:
\[ \text{Number of } Mg^{2+} = 8 \text{ (total tetrahedral voids)} \times \frac{1}{8} = 1 \]
- The total positive charge from `Mg^(2+)` ions in one unit cell is:
\[ \text{Charge from } Mg^{2+} = 1 \text{ (ion)} \times +2 \text{ (charge)} = +2 \]

### Step 4: Determine the number of octahedral voids in the unit cell
In a cubic close-packed (CCP) lattice:
- There are a total of 4 octahedral voids.

### Step 5: Calculate the contribution of `Al^(3+)` ions
- According to the problem, `Al^(3+)` ions occupy 1/2 of the octahedral voids.
- Therefore, the number of `Al^(3+)` ions in one unit cell is:
\[ \text{Number of } Al^{3+} = 4 \text{ (total octahedral voids)} \times \frac{1}{2} = 2 \]
- The total positive charge from `Al^(3+)` ions in one unit cell is:
\[ \text{Charge from } Al^{3+} = 2 \text{ (ions)} \times +3 \text{ (charge)} = +6 \]

### Step 6: Calculate the total positive charge in the unit cell
Now, we can sum up the positive charges from both `Mg^(2+)` and `Al^(3+)` ions:
\[ \text{Total positive charge} = +2 \text{ (from Mg)} + +6 \text{ (from Al)} = +8 \]

### Final Answer
The total positive charge contained in one unit cell of `MgAl2O4` is **+8**.

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