A certain form of lead has a cubic closest packed structure with an edge length of 492 \(\mathrm{pm}\) . Calculate the value of the atomic radius and the density of lead.

In a ccp structure, the face diagonal length (\(a\sqrt{2}\)) is 4 times the atomic radius (4r). \[\begin{aligned} 4r &= a\sqrt{2} \\ r &= \frac{a\sqrt{2}}{4} \end{aligned}\] Now, we can substitute the given edge length (492 pm) to find the atomic radius.

Using the given edge length (a = 492 pm) and the relationship we just derived, we can solve for the atomic radius (r). \(r = \frac{a\sqrt{2}}{4}\) \(r = \frac{492\, \text{pm}\ \cdot\sqrt{2}}{4}\) \(r \approx 174\, \text{pm}\) The atomic radius of lead in this ccp structure is approximately 174 pm.

First, let's find the mass of a single lead atom. The atomic mass of lead is 207.2 g/mol, and we can use Avogadro's number (6.022 x 10^23 atoms/mol) to find the mass of a single atom. \[\text{mass of a single Pb atom} = \frac{207.2\, \text{g/mol}}{6.022 \times 10^{23}\, \text{atoms/mol}} \approx 3.44 \times 10^{-22}\, \text{g}\] Since there are 4 atoms in a ccp unit cell, we can find the mass of the atoms in the unit cell: \[\text{mass in unit cell} = 4 \times 3.44 \times 10^{-22}\, \text{g} \approx 1.38 \times 10^{-21}\, \text{g}\] Now let's find the volume of the unit cell using the given edge length (a = 492 pm): \[\text{volume of unit cell} = a^3 = (492 \, \text{pm})^3 \times \frac{1\, \text{cm}^3}{10^{24}\, \text{pm}^3} = 1.186 \times 10^{-22}\, \text{cm}^3\] Finally, we can calculate the density of lead by dividing the mass by the volume: \[\rho = \frac{\text{mass in unit cell}}{\text{volume of unit cell}}\] \[\rho \approx \frac{1.38 \times 10^{-21}\, \text{g}}{1.186 \times 10^{-22}\, \text{cm}^3}\] \[\rho \approx 11.6\, \text{g/cm}^3\] The density of this form of lead with a cubic closest-packed structure is approximately 11.6 g/cm³.