If iridium has a density of 23.3 g/cm³ and forms a face-centered cubic lattice, what is the atomic radius of the iridium atom?

Question

Quizplus · Accepted Answer

The atomic radius of an iridium atom can be calculated using the density and the crystal structure information provided. Iridium forms a face-centered cubic (fcc) lattice, where the relationship between the atomic radius (r) and the edge length of the unit cell (a) is given by $a = 2\sqrt{2}r$. The density ($\rho$) can be related to the mass of a single atom (m), the Avogadro number (N_A), and the volume of the unit cell (a^3) by the formula $\rho = \frac{mN_A}{a^3}$. Given the density of iridium and knowing it forms an fcc lattice, we can solve for the atomic radius. The correct calculation leads to an atomic radius of approximately 135.7 pm, making option A the correct answer.

If Iridium Has a Density of 23