If `Al^(3+)` replaces `Na^(+)` at the edge centre of `NaCl` lattice ,then the cation vacancies in `1` mole of `NaCl` will be

Options

A. `6.022xx10^(23)`
B. `6.775xx10^(23)`
C. `4.517xx10^(23)`
D. `3.01xx10^(23)`

Explanation

To solve the problem of determining the number of cation vacancies in 1 mole of NaCl when Al³⁺ replaces Na⁺ at the edge center of the NaCl lattice, we can follow these steps:

### Step 1: Understand the structure of NaCl
NaCl crystallizes in a face-centered cubic (FCC) lattice structure. In this structure:
- Na⁺ ions are located at the corners and the edge centers of the cube.
- Cl⁻ ions are located at the face centers.

### Step 2: Calculate the number of Na⁺ ions in 1 mole of NaCl
In 1 mole of NaCl, there are Avogadro's number of Na⁺ ions, which is approximately \(6.022 \times 10^{23}\) Na⁺ ions.

### Step 3: Determine the contribution of Na⁺ ions from the edge centers
In the NaCl lattice:
- Each edge of the cube has 4 edge centers.
- Each edge center contributes \( \frac{1}{4} \) of a Na⁺ ion to the unit cell.

Thus, the total contribution of Na⁺ ions from the edge centers in one unit cell is:
\[ \text{Contribution from edge centers} = 4 \times \frac{1}{4} = 1 \text{ Na}^+ \]

### Step 4: Replacement of Na⁺ by Al³⁺
When Al³⁺ replaces Na⁺ at the edge center, we need to consider the charge balance:
- Al³⁺ has a charge of +3.
- Na⁺ has a charge of +1.

If 1 Al³⁺ replaces 3 Na⁺ ions, then for every Al³⁺ that replaces Na⁺, we create vacancies for the remaining Na⁺ ions.

### Step 5: Calculate the number of vacancies created
Since 1 Al³⁺ replaces 3 Na⁺ ions, the replacement leads to the creation of vacancies:
- For every Al³⁺ that replaces 3 Na⁺, we create 2 vacancies (since we have 1 Al³⁺ and we are losing 3 Na⁺).

### Step 6: Calculate the total number of vacancies in 1 mole of NaCl
If we have 1 mole of NaCl, the number of Al³⁺ ions that can replace Na⁺ ions is \( \frac{1}{3} \) mole (since 1 Al³⁺ replaces 3 Na⁺). Thus, the number of vacancies created is:
\[ \text{Vacancies} = \frac{1}{3} \text{ moles of Al}^{3+} \times 2 = \frac{2}{3} \text{ moles of vacancies} \]

### Step 7: Convert moles of vacancies to number of vacancies
To find the total number of vacancies in terms of particles, we multiply the number of moles of vacancies by Avogadro's number:
\[ \text{Total vacancies} = \frac{2}{3} \times 6.022 \times 10^{23} \approx 4.015 \times 10^{23} \text{ vacancies} \]

### Final Answer
The number of cation vacancies in 1 mole of NaCl when Al³⁺ replaces Na⁺ at the edge center is approximately \(4.015 \times 10^{23}\).

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