Rhodium (Rh) crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodium in g·cm–3.

Question

BlackTom AI · Accepted Answer

Before evaluating the options, let’s lay out the problem clearly: Rhodium (Rh) forms a face-centered cubic (FCC) lattice, the atomic radius is 135 pm, and we’re asked for the density in g/cm^3.
Option A: 0.337 g·cm–3. This is far too small for a metal with transition-metal atoms of this size. If you were to estimate density from size alone, you’d expect a value orders of magnitude larger than a few tenths of a g/cm^3 for a dense metal like Rh, so this option seems implausible.
Option B: 278 g·cm–3. This would be an amazingly high density, exceeding most metals by a wide margin. Since the density of Rh is known to be around 12 g/cm^3, 278 g/cm^3 would imply an enormously heavy mass per well-defined volume, which clashes with standard elemental data. This option is inconsistent with typical metallic densities and with what the crystallographic geometry would permit.
Option C: 3.07 g·cm–3. This value is in the right order of magnitude for some metals, but still noticeably lower than the known density of rhodium. If one were to crash-test this by rough reasoning, the volume per atom in a close-packed lattice and the number of atoms per unit cell should yield a density closer to ~12 g/cm^3, not as low as 3 g/cm^3.
Option D: 12.3 g·cm–3. This aligns with the well-established density of rhodium and with the expected result from an FCC lattice where four atoms occupy each unit cell. Now, let’s show how that result arises from the geometry and stoichiometry: In FCC, the number of atoms per unit cell Z = 4. The lattice parameter a relates to the atomic radius r by the face diagonal touching atoms: √2 a = 4r, so a = 2√2 r. With r = 135 pm = 1.35 × 10^−8 cm, we get a ≈ 2.828 × 1.35 × 10^−8 cm ≈ 3.82 × 10^−8 cm. The cell volume is a^3 ≈ (3.82 × 10^−8 cm)^3 ≈ 5.57 × 10^−23 cm^3. The mass per unit cell is Z × (M / N_A) = 4 × (102.91 g/mol) / (6.022 × 10^23 mol^−1) ≈ 6.84 × 10^−22 g. Therefore, density ρ = mass/volume ≈ (6.84 × 10^−22 g) / (5.57 × 10^−23 cm^3) ≈ 12.3 g·cm^−3.
In summary, the first three options deviate either by underestimating the compact metal packing, misapplying the number of atoms per cell, or miscalculating the lattice parameter, while the fourth option matches the crystallographic calculation and known physical data for rhodium.
Hence, the reasoning shows that Option D is consistent with the FCC geometry and the observed density of rhodium, while Options A, B, and C fail to reproduce the correct mass-to-volume relationship given the lattice parameters and atomic mass.

Rhodium (Rh) crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodium in g·cm–3.

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