Nickel has a face-centred cubic structure and has a density of 8.90 g cm^-3.What is its atomic radius? A)125 pm B)249 pm C)353 pm D)997 pm

Question

Quizplus · Accepted Answer

The atomic radius of nickel in a face-centered cubic (FCC) structure can be calculated using the formula for the density of a crystal structure, $\rho = \frac{ZM}{a^3N_A}$, where $\rho$ is the density, $Z$ is the number of atoms per unit cell (which is 4 for FCC), $M$ is the molar mass of nickel (58.69 g/mol), $a$ is the lattice parameter, and $N_A$ is Avogadro's number ($6.022 \times 10^{23} \text{mol}^{-1}$). The atomic radius $r$ is related to the lattice parameter in an FCC structure by $a = 2\sqrt{2}r$. Solving for $r$ gives an atomic radius close to 125 pm.

Nickel Has a Face-Centred Cubic Structure and Has a Density