Platinum crystallize in a face centered cube crystal with a unit cell length of `3.9231 Å`. The density and atomic radius of platinum are respectively: [Atomic mass of Pt = 195]

Options

A. `45.25 g cm^(-3), 2.516 Å`
B. `21.86 g cm^(-3), 1.387 Å`
C. `29.46 g cm^(-3), 1.48 Å`
D. None of these

Explanation

To find the density and atomic radius of platinum (Pt) crystallizing in a face-centered cubic (FCC) structure with a unit cell length of \(3.9231 \, \text{Å}\), we can follow these steps:

### Step 1: Determine the Atomic Radius
In a face-centered cubic (FCC) unit cell, the relationship between the atomic radius \(r\) and the unit cell length \(a\) is given by the formula:

\[ 4r = a\sqrt{2} \]

Where:
- \(r\) = atomic radius
- \(a\) = unit cell length

Given \(a = 3.9231 \, \text{Å}\), we can rearrange the formula to solve for \(r\):

\[ r = \frac{a\sqrt{2}}{4} \]

### Step 2: Calculate the Atomic Radius
Substituting the value of \(a\):

\[ r = \frac{3.9231 \times \sqrt{2}}{4} \]

Calculating this gives:

\[ r = \frac{3.9231 \times 1.4142}{4} \approx \frac{5.548}{4} \approx 1.387 \, \text{Å} \]

### Step 3: Determine the Density
The density \(\rho\) of a substance can be calculated using the formula:

\[ \rho = \frac{Z \cdot M}{N_A \cdot a^3} \]

Where:
- \(Z\) = number of atoms per unit cell (for FCC, \(Z = 4\))
- \(M\) = molar mass (atomic mass of Pt = 195 g/mol)
- \(N_A\) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))
- \(a\) = unit cell length in cm (convert \(3.9231 \, \text{Å}\) to cm)

### Step 4: Convert Unit Cell Length to cm
\[ a = 3.9231 \, \text{Å} = 3.9231 \times 10^{-8} \, \text{cm} \]

### Step 5: Calculate the Volume of the Unit Cell
\[ a^3 = (3.9231 \times 10^{-8})^3 \approx 6.048 \times 10^{-23} \, \text{cm}^3 \]

### Step 6: Substitute Values into the Density Formula
Now substituting the values into the density formula:

\[ \rho = \frac{4 \cdot 195}{6.022 \times 10^{23} \cdot 6.048 \times 10^{-23}} \]

Calculating this gives:

\[ \rho = \frac{780}{3.628 \approx 221.86 \, \text{g/cm}^3} \]

### Final Answers
- Atomic radius of platinum: \(1.387 \, \text{Å}\)
- Density of platinum: \(221.86 \, \text{g/cm}^3\)