In cubic `ZnS (II-VI)` compounds, if the radii of `Zn` and `S` atoms are `0.74 Å` and `1.70 Å`, the lattice parameter of cubic `ZnS` is

Options

A. `11.87 Å`
B. `5.634 Å`
C. `5.14 Å`
D. `2.97 Å`

Explanation

To find the lattice parameter \( a \) of cubic \( ZnS \), we can follow these steps:

### Step 1: Understand the Structure of \( ZnS \)
In cubic \( ZnS \), the sulfide ions \( S^{2-} \) form a face-centered cubic (FCC) lattice, while the zinc ions \( Zn^{2+} \) occupy alternate tetrahedral voids.

### Step 2: Identify the Radii of the Ions
Given:
- Radius of \( Zn^{2+} \) (R_Zn) = \( 0.74 \, \text{Å} \)
- Radius of \( S^{2-} \) (R_S) = \( 1.70 \, \text{Å} \)

### Step 3: Calculate the Distance Between the Ions
The distance between the \( Zn^{2+} \) ion in the tetrahedral void and the \( S^{2-} \) ion can be expressed as:
\[ \text{Distance} = R_{Zn} + R_{S} = 0.74 \, \text{Å} + 1.70 \, \text{Å} = 2.44 \, \text{Å} \]

### Step 4: Relate the Distance to the Lattice Parameter
In an FCC lattice, the tetrahedral voids are located at a distance of \( \frac{\sqrt{3}}{4} a \) from the corner of the unit cell. Therefore, we can set up the equation:
\[ \frac{\sqrt{3}}{4} a = R_{Zn} + R_{S} \]
Substituting the values we have:
\[ \frac{\sqrt{3}}{4} a = 2.44 \, \text{Å} \]

### Step 5: Solve for the Lattice Parameter \( a \)
To find \( a \), we rearrange the equation:
\[ a = \frac{4 \times 2.44 \, \text{Å}}{\sqrt{3}} \]
Calculating this gives:
\[ a \approx \frac{9.76 \, \text{Å}}{1.732} \approx 5.634 \, \text{Å} \]

### Conclusion
The lattice parameter \( a \) of cubic \( ZnS \) is approximately \( 5.634 \, \text{Å} \).

### Final Answer
The correct option is **B: 5.634 Å**.

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