The problem describes a compound with three types of ions: X, Y, and Z.
In an FCC unit cell, let the number of atoms (Z ions) be denoted by $n$. For FCC, $n=4$.
We can now calculate the number of X and Y ions based on the voids they occupy:
The ratio of ions X:Y:Z is $\frac{4}{3} : \frac{4}{3} : 4$. To find the simplest whole number ratio, we multiply the ratio by 3:
Ratio = $(\frac{4}{3} \times 3) : (\frac{4}{3} \times 3) : (4 \times 3)$
Ratio = $4 : 4 : 12$
Now, we simplify this ratio by dividing by the greatest common divisor, which is 4:
Simplified Ratio = $\frac{4}{4} : \frac{4}{4} : \frac{12}{4}$
Simplified Ratio = $1 : 1 : 3$
Therefore, the chemical formula of the compound is $XYZ_3$.
For the given unit cells of a two dimensional square lattice, which option lists all the primitive cells?

The number of distinct ways the primitive unit cell can be constructed for the two dimensional lattice as shown in the figure is ______.

Consider a three-dimensional crystal of $N$ inert gas atoms. The total energy is given by $U(R) = 2N\epsilon \left[p \left(\frac{\sigma}{R}\right)^{12} - q \left(\frac{\sigma}{R}\right)^6\right]$, where $p = 12.13$, $q = 14.45$, and $R$ is the nearest neighbour distance between two atoms. The two constants, $\epsilon$ and $R$, have the dimensions of energy and length, respectively. The equilibrium separation between two nearest neighbour atoms in units of $\sigma$ (rounded off to two decimal places) is ________