A crystal of lead (II) sulphide has NaCl strcuture . In this crystal the shorest distance between a `Pb^(2+)` ion and `S^(2-)` ion is 297 pm . What is the volume the of unit cell in lead sulphide ?

Options

A. `209.6 xx 10^(-24) cm `
B. `207.8 xx 10^(-23) cm `
C. `22.3 xx 10^(-23) cm`
D. `209.8 xx 10 ^(-23) cm `

Answer

To find the volume of the unit cell in lead (II) sulfide (PbS) with a NaCl structure, we can follow these steps:

### Step 1: Understand the Structure
Lead (II) sulfide has a NaCl structure, which means:
- The sulfide ions (S²⁻) are located at the face-centered cubic (FCC) positions (corners and face centers).
- The lead ions (Pb²⁺) occupy the octahedral sites.

### Step 2: Identify the Shortest Distance
The shortest distance between a Pb²⁺ ion and an S²⁻ ion is given as 297 pm. This distance represents the sum of the radii of the Pb²⁺ and S²⁻ ions.

### Step 3: Relate Shortest Distance to Edge Length
In the NaCl structure, the shortest distance can be expressed as:
\[ r_{S^{2-}} + r_{Pb^{2+}} = \frac{a}{2} \]
where \( a \) is the edge length of the unit cell.

Given that the shortest distance is 297 pm, we can express this as:
\[ r_{S^{2-}} + r_{Pb^{2+}} = 297 \text{ pm} \]

### Step 4: Calculate Edge Length
Since the shortest distance between the ions is equal to the sum of their radii, we can express the edge length \( a \) as:
\[ a = 2 \times (r_{S^{2-}} + r_{Pb^{2+}}) \]
Substituting the given distance:
\[ a = 2 \times 297 \text{ pm} = 594 \text{ pm} \]

### Step 5: Convert Edge Length to Centimeters
To calculate the volume, we need the edge length in centimeters:
\[ 594 \text{ pm} = 594 \times 10^{-12} \text{ m} = 5.94 \times 10^{-8} \text{ cm} \]

### Step 6: Calculate the Volume of the Unit Cell
The volume \( V \) of the cubic unit cell is given by:
\[ V = a^3 \]
Substituting the edge length:
\[ V = (5.94 \times 10^{-8} \text{ cm})^3 \]
Calculating this gives:
\[ V = 2.08 \times 10^{-23} \text{ cm}^3 \]

### Step 7: Final Result
Thus, the volume of the unit cell in lead (II) sulfide is approximately:
\[ V \approx 2.08 \times 10^{-23} \text{ cm}^3 \]