Lead metal has a density of `11.34 g//cm^(3)` and crystallizes in a face-centered lattice. Choose the correct alternatives

Options

A. the volume of one unit cell is `1.214 xx 10^(-22) cm^(3)`
B. the volume of one unit cell is `1.214 xx 10^(-19) cm^(3)`
C. the atomic radius of lead is 175 pm
D. the atomic radius of lead is 155.1 pm

Explanation

To solve the problem regarding lead metal's density and its crystallization in a face-centered cubic (FCC) lattice, we will follow these steps:

### Step 1: Understand the Density Formula
The density (D) of a substance can be expressed using the formula:
\[ D = \frac{Z \times M}{N_A \times V} \]
where:
- \(D\) = density of the substance
- \(Z\) = number of atoms per unit cell
- \(M\) = molar mass of the substance
- \(N_A\) = Avogadro's number (\(6.02 \times 10^{23}\) mol\(^{-1}\))
- \(V\) = volume of the unit cell

### Step 2: Determine the Number of Atoms in FCC
In a face-centered cubic (FCC) lattice:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.

Calculating the total number of atoms (\(Z\)):
\[ Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Step 3: Use Given Data
Given:
- Density \(D = 11.34 \, \text{g/cm}^3\)
- Molar mass of lead \(M = 207 \, \text{g/mol}\)

### Step 4: Calculate the Volume of the Unit Cell
Rearranging the density formula to find volume \(V\):
\[ V = \frac{Z \times M}{D \times N_A} \]
Substituting the values:
\[ V = \frac{4 \times 207}{11.34 \times 6.02 \times 10^{23}} \]
Calculating \(V\):
\[ V = \frac{828}{68.19768 \times 10^{23}} \approx 1.214 \times 10^{-22} \, \text{cm}^3 \]

### Step 5: Calculate the Edge Length of the Unit Cell
The volume of the cubic unit cell is given by:
\[ V = a^3 \]
where \(a\) is the edge length. Thus,
\[ a = V^{1/3} = (1.214 \times 10^{-22})^{1/3} \]
Calculating \(a\):
\[ a \approx 4.95 \times 10^{-8} \, \text{cm} \]

### Step 6: Relate Atomic Radius to Edge Length
In an FCC lattice, the relationship between the atomic radius \(r\) and the edge length \(a\) is given by:
\[ 4r = \sqrt{2}a \]
Thus,
\[ r = \frac{\sqrt{2}a}{4} \]
Substituting the value of \(a\):
\[ r = \frac{\sqrt{2} \times 4.95 \times 10^{-8}}{4} \]
Calculating \(r\):
\[ r \approx 1.75 \times 10^{-8} \, \text{cm} \]

### Step 7: Convert Radius to Picometers
Since \(1 \, \text{cm} = 10^{10} \, \text{pm}\):
\[ r \approx 1.75 \times 10^{-8} \, \text{cm} \times 10^{10} \, \text{pm/cm} = 175 \, \text{pm} \]

### Conclusion
The radius of the lead atom in the face-centered cubic lattice is approximately \(173 \, \text{pm}\).