Chromium metal crystallizes with a body-centred cubic lattice. The length of the unit cell edge is found to be `287`pm. Calculate the atomic radius. What woulds be the density of chromium in `g cm^(-3)`?

Options

A. 124.27 pm
B. 287 pm
C. 574 pm
D. 143.5 pm

Explanation

To solve the problem of calculating the atomic radius and density of chromium metal, which crystallizes in a body-centered cubic (BCC) lattice, we can follow these steps:

### Step 1: Identify the given data
- The edge length of the unit cell (A) = 287 pm = 2.87 x 10^-8 cm (since 1 pm = 10^-12 m and 1 cm = 10^-2 m).
- The molecular mass of chromium (MA) = 52 g/mol.
- The coordination number (Z) for BCC = 2.

### Step 2: Calculate the atomic radius (r)
In a body-centered cubic lattice, the relationship between the edge length (A) and the atomic radius (r) is given by the formula:
\[ A = 4r / \sqrt{3} \]

Rearranging this formula to find r:
\[ r = \frac{A \cdot \sqrt{3}}{4} \]

Substituting the value of A:
\[ r = \frac{2.87 \times 10^{-8} \, \text{cm} \cdot \sqrt{3}}{4} \]
\[ r = \frac{2.87 \times 10^{-8} \, \text{cm} \cdot 1.732}{4} \]
\[ r = \frac{4.97364 \times 10^{-8} \, \text{cm}}{4} \]
\[ r = 1.24341 \times 10^{-8} \, \text{cm} \]
\[ r = 124.34 \, \text{pm} \]

### Step 3: Calculate the density (ρ)
The formula for density is given by:
\[ \rho = \frac{Z \cdot MA}{N_A \cdot A^3} \]
Where:
- Z = number of atoms per unit cell (for BCC, Z = 2)
- MA = molar mass of chromium = 52 g/mol
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- \( A \) = edge length in cm = \( 2.87 \times 10^{-8} \, \text{cm} \)

Calculating \( A^3 \):
\[ A^3 = (2.87 \times 10^{-8} \, \text{cm})^3 = 2.36 \times 10^{-23} \, \text{cm}^3 \]

Now substituting the values into the density formula:
\[ \rho = \frac{2 \cdot 52 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 2.36 \times 10^{-23} \, \text{cm}^3} \]
\[ \rho = \frac{104 \, \text{g/mol}}{1.419 \, \text{g/cm}^3} \]
\[ \rho \approx 7.34 \, \text{g/cm}^3 \]

### Final Results
- Atomic radius \( r \approx 124.34 \, \text{pm} \)
- Density \( \rho \approx 7.34 \, \text{g/cm}^3 \)