A crystal is made of particle \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\). A forms FCC packing, B occupies all octahedral voids of A and C occupies all tetrahedral voids of \(\mathrm{A}\), if all the particles along one body diagonal are removed then the formula of the crystal would be: (a) \(\mathrm{A}_{5} \mathrm{BC}_{8}\) (b) \(\mathrm{A}_{5} \mathrm{~B}_{4} \mathrm{C}_{8}\) (c) \(\mathrm{A}_{8} \mathrm{~B}_{4} \mathrm{C}_{5}\) (d) \(\mathrm{A}_{5} \mathrm{~B}_{2} \mathrm{C}_{8}\)

Face-centered cubic (FCC) packing is a highly efficient way of arranging spherical particles in a crystal lattice.
In this structure, particles are located at each of the corners and the centers of all the faces of the cube, creating a repeating pattern.

One unit cell of an FCC structure contains particles that contribute fractions of their volume to the cell. Specifically, there are 8 corner particles, each contributing 1/8th of its volume, and 6 face-centered particles each contributing half of its volume. After accounting for these overlaps, the unit cell effectively contains 4 whole particles in total.

The efficiency of this packing results in a densely packed structure with little empty space, and it's commonly found in many metal crystals. It also determines the number of voids present within the structure, which are spaces where other, smaller particles can reside, leading us to the concept of octahedral and tetrahedral voids.

Octahedral voids are spaces within a crystal lattice where a smaller particle can fit in between the larger particles without disrupting their arrangement.
In an FCC crystal, these voids are formed by the space bounded by six particles: four from the top and bottom face centers and two from the opposite edges of the unit cell.

The number of octahedral voids in any FCC lattice is equal to the number of particles making up the FCC structure. Thus, in a unit cell of an FCC lattice, there are 4 octahedral voids.

These voids play a crucial role in determining the arrangements of different atoms in a compound. In the case of the given problem, particle B is occupying these octahedral voids, which significantly affects the stoichiometry of the final crystal structure. Each octahedral void is shared among multiple unit cells, hence why removing particles along the body diagonal decreases the count of B atoms in the structure.

Tetrahedral voids in a crystal structure are another type of void space that can host smaller particles.
As the name suggests, the space is in the shape of a tetrahedron, a pyramid with a triangular base formed by four atoms. In an FCC lattice, these voids are found in the gaps where four spheres from two adjacent layers meet.

There are twice as many tetrahedral voids as there are FCC particles, resulting in 8 tetrahedral voids per unit cell. This is because each tetrahedral void is surrounded by only one-fourth of each of the four spheres that form its corners.

In our problem, particle C occupies the tetrahedral voids. Due to their larger quantity compared to octahedral voids, the stoichiometry of C will be higher relative to B in the composition of the crystal. As with octahedral voids, when particles are removed along a body diagonal, the count of C atoms in the final formula will change proportionally, illustrating the interconnected nature of crystal structures and their voids.