Questions: Crystal Structures and Unit Cells

Question 1

A face-centered cubic (FCC) unit cell contains how many atoms?

• A. 1
• B. 2
• C. 4
• D. 8

Answer: (C)

Each corner atom is shared among 8 unit cells (8 corners x 1/8 = 1 atom), and each face-center atom is shared between 2 unit cells (6 faces x 1/2 = 3 atoms), giving 1 + 3 = 4 atoms per unit cell. This counting method — fractional contribution per atom — is the standard way to determine unit cell contents for any crystal structure.

Question 2

The 14 Bravais lattices represent all possible ways to arrange points in three-dimensional space such that every point has an identical environment.

• T. True
• F. False

Answer: True

Auguste Bravais proved in 1848 that there are exactly 14 distinct lattice types in 3D that satisfy translational symmetry — every lattice point sees the same arrangement of neighbors in the same orientations. These 14 lattices span the 7 crystal systems and differ by the placement of additional lattice points (body-centered, face-centered, base-centered) beyond the primitive corners. Any crystal structure, no matter how complex, maps onto one of these 14 lattice types.

Question 3

In the rock salt structure, both the cation and anion sublattices independently form FCC arrangements. Why does this produce 6-fold coordination for both ions?

Model answer: Each Na+ ion sits in an octahedral hole of the Cl- FCC sublattice, surrounded by 6 nearest-neighbor Cl- ions at the vertices of an octahedron. By symmetry, each Cl- is similarly surrounded by 6 Na+ ions. The interpenetrating FCC arrangement places every ion at the center of an octahedron formed by the other sublattice, giving coordination number 6 for both species.

The rock salt structure can be visualized as two interpenetrating FCC lattices offset by half a unit cell edge along each axis. The octahedral holes in an FCC lattice are exactly the right size for ions with radius ratios between about 0.414 and 0.732, which is why NaCl (ratio ~0.56) adopts this structure. If the cation were much smaller, it would rattle in the octahedral hole and the structure would become unstable — the system would prefer tetrahedral coordination instead (zinc blende).

Question 4

A body-centered cubic metal has a unit cell edge length of 2.87 Angstroms and an atomic mass of 55.85 g/mol. What is the density of this metal?

Model answer: BCC has 2 atoms per unit cell. Density = (2 atoms x 55.85 g/mol) / (6.022 x 10^23 atoms/mol x (2.87 x 10^-8 cm)^3) = 7.87 g/cm^3. This is iron (alpha-Fe), confirming that the BCC structure with these parameters matches the known density of iron.

The density calculation from unit cell parameters is fundamental in crystallography: rho = nM/(N_A x a^3), where n is atoms per unit cell, M is molar mass, N_A is Avogadro's number, and a is the edge length. This formula works because the unit cell, by definition, tiles to fill all space — so the density of the unit cell equals the macroscopic density. Matching calculated density to known values is a standard check on proposed crystal structures.