Question

Polonium is the only element known to crystallize in the simple cubic form. In this structure, the interatomic distance between a Po atom and each of its six nearest neighbors is \(335 \mathrm{pm}\). Use this description of the crystal structure to estimate the density of polonium.

Short answer

The estimated density of polonium is approximately \(9.23 \, g/cm^3\).

Step-by-step solution

- Calculation of Unit Cell Volume

The unit cell of a simple cubic structure is a cube, and the length of each side of the cube is equal to the given interatomic distance. Convert this distance from picometers to centimeters:\[335 \mathrm{pm} * \frac{100 \mathrm{cm}}{1 \mathrm{m}} * \frac{1 \mathrm{m}}{10^{12}\mathrm{pm}} = 3.35 \times 10^{-8} \mathrm{cm}\] Now calculate the volume (\(V\)) of the cube using the formula \(V=a^3\), so \(V=(3.35\times 10^{-8})^3=3.76\times 10^{-23}\) cubic centimeters.

- Calculation of Density

Density (\(\rho\)) is mass divided by volume. In one unit cell, there's one Po atom, so the mass of one unit cell is the molar mass of Po divided by Avogadro's number. Covert the molar mass of Po from g/mol to g: \[209 \, g/mol * \frac{1 \, mol}{6.022 \times 10^{23} \, atoms} = 3.47 \times 10^{-22} \, g\] Now, use the formula \(\rho=m/V\) to find the density and express the result in g/cm^3.\[\rho = \frac{3.47 \times 10^{-22} \, g}{3.76 \times 10^{-23} \, cm^3} = 9.23 \, g/cm^3\]

- Report the Result

After completing the steps above and performing the calculations, report the result for the density of polonium which is found to be around \(9.23 \, g/cm^3\)

Key concepts

Simple Cubic Crystal Structure

The simple cubic crystal structure is one of the seven types of crystal systems that materials can form. This structure is characterized by atoms arranged at the corners of a cube. In total, one atom is shared equally among each corner of eight unit cells. Although the arrangement is simple, it is not densely packed compared to other structures like face-centered cubic or body-centered cubic.

Polonium is unique for adopting this structure as it is the only known element to do so naturally. This arrangement means each Po atom is in contact with six other atoms, forming a basic but crucial lattice. Understanding this arrangement helps in determining properties such as interatomic distances and unit cell volumes, essential for calculating material density.

Interatomic Distance

The interatomic distance in a simple cubic structure is the distance between the centers of two adjacent atoms. In the case of polonium, this distance is given as 335 pm (picometers). This measurement is crucial because it defines the edge length of the unit cell.

To work with densities and cell volumes, this distance needs to be converted into centimeters. The conversion involves the following calculation:

• Convert picometers to meters: \(335 \mathrm{pm} = 3.35 \times 10^{-10} \mathrm{m}\).
• Then convert meters to centimeters: \(3.35 \times 10^{-10} \mathrm{m} = 3.35 \times 10^{-8} \mathrm{cm}\).

Grasping this concept allows you to link the microscopic arrangement with the macroscopic properties of the material.

Unit Cell Volume

The volume of a unit cell is an essential attribute for understanding the density and packing efficiency of a crystalline structure. In a simple cubic structure, the unit cell is a cube with sides equal in length to the interatomic distance.

Therefore, the volume can be calculated using the formula for the volume of a cube: \(V = a^3\).
For example, with polonium:

• The side of the cube (\(a\)) is \(3.35 \times 10^{-8} \mathrm{cm}\).
• Thus, the volume \(V = (3.35 \times 10^{-8})^3 = 3.76 \times 10^{-23} \mathrm{cm}^3\).

Understanding unit cell volume is crucial for determining the density, as it relates to the space the atoms occupy within the structure, affecting how mass is distributed throughout the material.