In the zinc blende structure, zinc ions occupy alternate tetrahedral voids and `S^(2-)` ions exist as ccp the radii of `Zn^(2+)` and `S^(2-)` ions are 0.83Å and 1.74 Å respectively.The edge length of the ZnS unit cell is :

Options

A. 2.57 Å
B. 5.14 Å
C. 2.96 Å
D. 5.94 Å

Explanation

To solve the problem of finding the edge length of the ZnS unit cell in the zinc blende structure, we can follow these steps:

### Step 1: Understand the Structure
In the zinc blende structure:
- Zinc ions (Zn²⁺) occupy alternate tetrahedral voids.
- Sulfide ions (S²⁻) form a cubic close-packed (CCP) structure.

### Step 2: Identify the Ion Radii
Given:
- Radius of Zn²⁺ = 0.83 Å
- Radius of S²⁻ = 1.74 Å

### Step 3: Determine the Body Diagonal Relationship
In a cubic unit cell, the body diagonal can be expressed in terms of the edge length (a):
\[ \text{Body Diagonal} = a\sqrt{3} \]

### Step 4: Calculate the Distance Between Ions
In the zinc blende structure, the distance between the Zn²⁺ ion located in the tetrahedral void and the S²⁻ ions can be described as:
\[ \text{Distance} = \text{Radius of Zn}^{2+} + \text{Radius of S}^{2-} \]
\[ \text{Distance} = 0.83 \, \text{Å} + 1.74 \, \text{Å} = 2.57 \, \text{Å} \]

### Step 5: Relate the Distance to the Body Diagonal
The distance between the Zn²⁺ ion and the S²⁻ ion along the body diagonal is given by:
\[ \frac{a\sqrt{3}}{4} = \text{Distance} \]
This is because the tetrahedral voids are located at a quarter of the body diagonal.

### Step 6: Set Up the Equation
Substituting the distance into the equation:
\[ \frac{a\sqrt{3}}{4} = 2.57 \]

### Step 7: Solve for Edge Length (a)
Rearranging the equation gives:
\[ a\sqrt{3} = 2.57 \times 4 \]
\[ a\sqrt{3} = 10.28 \]
\[ a = \frac{10.28}{\sqrt{3}} \]

### Step 8: Calculate the Value of a
Calculating the value:
\[ a = \frac{10.28}{1.732} \approx 5.94 \, \text{Å} \]

### Final Answer
The edge length of the ZnS unit cell is approximately **5.94 Å**.

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