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. 6.8
B. 7.6
C. 6.6
D. 7.23

Answer

To solve the problem, we need to calculate the atomic radius and the density of chromium metal, which crystallizes in a body-centered cubic (BCC) lattice. Let's break down the steps:

### Step 1: Calculate the Atomic Radius
In a body-centered cubic lattice, the relationship between the edge length (a) and the atomic radius (R) is given by the formula:

\[ \sqrt{3}a = 4R \]

From this, we can express the atomic radius (R) as:

\[ R = \frac{\sqrt{3}}{4}a \]

Given that the edge length \( a = 287 \, \text{pm} \):

\[ R = \frac{\sqrt{3}}{4} \times 287 \, \text{pm} \]

Calculating this:

1. Calculate \( \sqrt{3} \approx 1.732 \).
2. Substitute \( a \):

\[ R = \frac{1.732}{4} \times 287 \approx 0.433 \times 287 \approx 124.2 \, \text{pm} \]

### Step 2: Calculate the Density
The formula for density (\( \rho \)) is given by:

\[ \rho = \frac{n \cdot M}{V \cdot N_A} \]

Where:
- \( n \) = number of atoms per unit cell (for BCC, \( n = 2 \))
- \( M \) = molar mass of chromium (approximately \( 52 \, \text{g/mol} \))
- \( V \) = volume of the unit cell (\( a^3 \))
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \, \text{mol}^{-1} \))

#### Step 2.1: Calculate the Volume of the Unit Cell
Convert \( a \) from picometers to centimeters:

\[ a = 287 \, \text{pm} = 287 \times 10^{-10} \, \text{cm} \]

Now calculate the volume \( V \):

\[ V = a^3 = (287 \times 10^{-10})^3 \approx 2.36 \times 10^{-28} \, \text{cm}^3 \]

#### Step 2.2: Substitute Values into the Density Formula
Now substitute the values into the density formula:

\[ \rho = \frac{2 \cdot 52}{2.36 \times 10^{-28} \cdot 6.022 \times 10^{23}} \]

Calculating the numerator:

\[ 2 \cdot 52 = 104 \, \text{g} \]

Calculating the denominator:

\[ 2.36 \times 10^{-28} \cdot 6.022 \times 10^{23} \approx 1.42 \times 10^{-4} \, \text{g} \]

Now calculate the density:

\[ \rho = \frac{104}{1.42 \times 10^{-4}} \approx 7.3 \, \text{g/cm}^3 \]

### Final Answers
- Atomic Radius \( R \approx 124.2 \, \text{pm} \)
- Density \( \rho \approx 7.3 \, \text{g/cm}^3 \)