Question

Rhodium crystallizes in a face-centered cubic unit cell that has an edge length of \(0.381 \mathrm{nm}\). (a) Calculate the atomic radius of a rhodium atom. (b) Calculate the density of rhodium metal.

Short answer

The atomic radius of a rhodium atom is approximately 0.13493 nm. The density of rhodium metal is approximately 12.40 g/cm³.

Step-by-step solution

Find the lattice constant (a)

The edge length of the unit cell is equal to the lattice constant, which we know to be 0.381 nm. We convert this value to meters: \[a = 0.381 * 10^{-9} \mathrm{m}\]

Geometric relationship between atomic radius and lattice constant in FCC unit cell

In an fcc unit cell, we can form an isosceles right triangle using the atomic radius (r), edge length (a), and two similar triangular faces along a diagonal in one of the faces. Using the Pythagorean Theorem, we can find the relationship between a and r for an fcc unit cell: \[a^2 + a^2 = (4r)^2\]

Calculate the atomic radius (r)

We will solve the relationship equation for r and then substitute the value of a that we have: \[r = \frac{a\sqrt{2}}{4}\] Now substitute the value of a, \[r = \frac{0.381 * 10^{-9}\sqrt{2}}{4} = 0.13493 * 10^{-9} \mathrm{m}\] Thus, the atomic radius of a rhodium atom is approximately 0.13493 nm.

Calculate the density of Rhodium

To find the density of Rhodium, we will use the formula for density: \[\rho = \frac{mass}{volume}\] Volume of the Rhodium unit cell can be calculated by \(V = a^3\). Furthermore, there are four Rhodium atoms per fcc unit cell and there are \(N_A = 6.022 * 10^{23}\) atoms per mole. The molar mass of Rhodium is \(M_R = 102.91 \frac{g}{mol}\). We can calculate the mass of the Rhodium atoms in the unit cell using the molar mass and the number of atoms. First, let's calculate the volume of one Rhodium unit cell: \[V = (0.381*10^{-9})^3 = 5.5289*10^{-29}\,m^3\] Now, find the mass of four Rhodium atoms: \[mass = \frac{4 \,atoms * 102.91\frac{g}{mol}}{6.022 * 10^{23}\,atoms/mol} = 6.8473 * 10^{-22}\,g\] Now we can calculate the density: \[\rho = \frac{6.8473 * 10^{-22} \,g}{5.5289*10^{-29}\,m^3} = 12.40 \frac{g}{cm^3}\] The density of Rhodium metal is approximately 12.40 g/cm³.

Key concepts

Face-Centered Cubic Lattice

The face-centered cubic (FCC) lattice is a common arrangement in crystal structures, where each atom is situated at the corners and the centers of all the cube faces of the unit cell. This arrangement is highly efficient, as it allows the atoms to pack closely together. In an FCC lattice:

• Each corner atom is shared among eight adjoining unit cells.
• Each face-centered atom is shared between two adjacent unit cells.

This results in each FCC unit cell effectively containing four whole atoms—each contributing to the compact and stable nature of the structure. The use of an FCC lattice for rhodium demonstrates its dense atomic packing in its crystal form. Understanding this structure is crucial for calculations related to the material's properties, like atomic radius and density.

Atomic Radius Calculation

To calculate the atomic radius of rhodium within an FCC lattice, we rely on the geometric relationship that emerges from the arrangement of atoms in the unit cell. This involves an understanding of a unique triangle formed within the lattice:An edge of the cubic cell forms the triangle's two equal sides, and the diagonal across a face forms the hypotenuse. Here, the face diagonal equals four times the atomic radius (since it spans across the entirety of two touching radius lines of atoms).The calculation is guiding by the Pythagorean theorem:\[ a^2 + a^2 = (4r)^2 \]Solving for the atomic radius \(r\), we find the formula:\[ r = \frac{a\sqrt{2}}{4} \]Given the edge length \(a\) of 0.381 nm, substituting this into the formula results in an atomic radius of approximately 0.13493 nm for rhodium. Breaking down such derivations allows us to grasp the intimate linkage between lattice shapes and physical dimensions.

Density Calculation

Density is a fundamental property of materials, defined as the mass per unit volume. For rhodium within a face-centered cubic lattice, calculating density involves both the atomic mass and the dimensions of the unit cell.First, the unit cell volume is calculated from the edge length \(a\) using the formula:\[ V = a^3 \]With four atoms per FCC unit cell, the total atomic mass within the cell is derived from the molar mass of rhodium (102.91 g/mol), and Avogadro's number \(6.022 \times 10^{23}\) atoms/mol.The mass of the rhodium atoms in one unit cell, and the calculated volume allow determination of the density:\[ \rho = \frac{\text{mass}}{\text{volume}} \]Using these values, the density of rhodium works out to approximately 12.40 g/cm³. This process illustrates the importance of understanding not just the numerical answer, but also the underlying concepts that link atomic arrangement to macro scale properties.