Salt AB has a zinc blend structure. The radius of `A^(2+)` and `B^(2-)` ion are `0.7Å and 1.8Å` respectively. The edge length of AB unit cell is:

Options

A. (a) `2.5Å`
B. (b) `5.09Å`
C. (c) `5 Å`
D. (d) `5.77Å`

Answer

To find the edge length of the unit cell of salt AB with a zinc blend structure, we can follow these steps:

### Step 1: Understand the Structure
The zinc blend structure is a face-centered cubic (FCC) lattice where cations occupy tetrahedral voids. In this case, A is the cation (A²⁺) and B is the anion (B²⁻).

### Step 2: Calculate the Radius Ratio
We are given the ionic radii:
- Radius of A²⁺ (r_A) = 0.7 Å
- Radius of B²⁻ (r_B) = 1.8 Å

Now, we calculate the radius ratio (r_A/r_B):
\[ \text{Radius Ratio} = \frac{r_A}{r_B} = \frac{0.7}{1.8} \approx 0.388 \]

### Step 3: Determine the Type of Void Occupied
The radius ratio of 0.388 falls between 0.225 and 0.414, which indicates that the cation A²⁺ occupies the tetrahedral voids in the zinc blend structure.

### Step 4: Use the Relationship for Edge Length
In a zinc blend structure, the relationship between the edge length (a) of the unit cell and the ionic radii is given by:
\[ \frac{\sqrt{3}}{4} a = r_A + r_B \]
Substituting the values of r_A and r_B:
\[ \frac{\sqrt{3}}{4} a = 0.7 + 1.8 = 2.5 \, \text{Å} \]

### Step 5: Solve for Edge Length (a)
Now, we can solve for a:
\[ a = \frac{4}{\sqrt{3}} \times 2.5 \]
Calculating this gives:
\[ a \approx \frac{10}{\sqrt{3}} \approx 5.77 \, \text{Å} \]

### Conclusion
The edge length of the AB unit cell is approximately **5.77 Å**.

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