Aluminum metal crystallizes in a face-centered cubic unit cell. (a) How many aluminum atoms are in a unit cell? (b) What is the coordination number of each aluminum atom? (c) Estimate the length of the unit cell edge, \(a,\) from the atomic radius of aluminum (143 pm). (d) Calculate the density of aluminum metal.

In a face-centered cubic (FCC) unit cell, there are eight atoms at the corners and six atoms on the face centers. Each corner atom is shared by eight adjacent unit cells, so they contribute 1/8 of an atom to each cell. Since there are eight corners, they contribute a total of 8*(1/8) = 1 atom. Each face-centered atom is shared by two adjacent unit cells, so they contribute 1/2 of an atom to each cell. Since there are six faces, they contribute a total of 6*(1/2) = 3 atoms. Therefore, there are #(1+3)=4# aluminum atoms in a face-centered cubic unit cell.

In an FCC unit cell, each atom is in contact with 12 other atoms. This is due to the fact that each atom: - touches 4 atoms along the edges of the same face - touches 4 atoms on the adjacent face in front - touches 4 atoms on the adjacent face behind So, the coordination number of each aluminum atom in an FCC unit cell is 12.

In an FCC unit cell, the diagonal of each face can be related to the atomic radius (r) and edge length (a) as follows: \(4r = \sqrt{2a^2}\) Given the atomic radius of aluminum, r = 143 pm, we can find the length of the unit cell edge. \(4(143 \,\text{pm}) = \sqrt{2a^2}\) Solve for \(a\): \(a = \frac{\sqrt{2(4)(143\,\text{pm})^2}}{4}\) \(a \approx 404.95\,\text{pm}\) So, the edge length of the unit cell is approximately 404.95 pm.

To find the density of aluminum, we can use the formula: \(\text{Density} = \frac{\text{mass of atoms in the unit cell}}{\text{volume of the unit cell}}\) First, we need to find the mass of Aluminum atoms. Aluminum has an atomic mass of 26.98 g/mol. In one unit cell, there are 4 aluminum atoms, so the mass of atoms in the unit cell is: \(\text{Mass} = 4 \times \frac{26.98\, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} = 1.794 \times 10^{-22}\, \text{g}\) Now, we can calculate the volume of the unit cell: \(\text{Volume} = a^3 = (404.95\, \text{pm})^3 = 6.644 \times 10^{-23} \,\text{cm}^3\) Now, we can find the density: \(\text{Density} = \frac{1.794 \times 10^{-22}\, \text{g}}{6.644 \times 10^{-23} \,\text{cm}^3} \approx 2.70\, \text{g/cm}^3\) So, the density of aluminum metal is approximately 2.70 g/cm³.