Aluminium metal has a density of `2.72 g cm^(-3)` and crystallizes in a cubic lattice with an edge of `404` pm. Which is//are correct?

Options

A. It forms an fcc unit cell.
B. It forms a bcc unit cell.
C. Its coordination number is `8`.
D. Its coordination number is `12`.

Explanation

To solve the problem regarding aluminum metal's density and its crystallization structure, we can follow these steps:

### Step 1: Convert the edge length from picometers to centimeters
Given:
- Edge length (a) = 404 pm = \(404 \times 10^{-10}\) cm

### Step 2: Use the density formula to find the number of formula units (Z)
The formula for density (d) is given by:
\[ d = \frac{Z \cdot M}{N_a \cdot a^3} \]
Where:
- \(d\) = density = 2.72 g/cm³
- \(M\) = molar mass of aluminum = 27 g/mol
- \(N_a\) = Avogadro's number = \(6.022 \times 10^{23}\) mol⁻¹
- \(a\) = edge length in cm = \(404 \times 10^{-10}\) cm
- \(Z\) = number of formula units per unit cell

Rearranging the formula to solve for \(Z\):
\[ Z = \frac{d \cdot N_a \cdot a^3}{M} \]

### Step 3: Calculate \(a^3\)
\[ a^3 = (404 \times 10^{-10})^3 = 6.58 \times 10^{-29} \text{ cm}^3 \]

### Step 4: Substitute the values into the equation for \(Z\)
\[ Z = \frac{2.72 \cdot (6.022 \times 10^{23}) \cdot (6.58 \times 10^{-29})}{27} \]

### Step 5: Perform the calculation
Calculating \(Z\):
\[ Z \approx \frac{2.72 \cdot 6.022 \cdot 6.58}{27} \approx 4 \]

### Step 6: Determine the crystal structure
Since \(Z \approx 4\), aluminum crystallizes in a face-centered cubic (FCC) structure.

### Step 7: Find the coordination number
In an FCC structure, the coordination number is known to be 12.

### Conclusion
Based on the calculations:
1. The number of formula units \(Z\) is approximately 4, indicating an FCC structure.
2. The coordination number for FCC is 12.

Therefore, the correct statements are:
- The structure is FCC (Option 1).
- The coordination number is 12 (Option 4).