Clear Notion
Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 03 / 07Solid State

Crystal Lattices, Unit Cells and Counting Atoms per Unit Cell

Crystal lattice and lattice points, unit cell and its parameters, primitive and centred unit cells, the seven crystal systems and fourteen Bravais lattices, and calculation of the number of atoms in simple cubic, body-centred and face-centred cubic unit cells, including formula determination from positions.

Figure 1
Crystal lattice. A regular three-dimensional arrangement of points (lattice points) in space, each representing a particle (atom, ion or molecule) or a group of particles. Joining the points gives the lattice. Unit cell. The smallest portion of the lattice that, repeated in three dimensions, generates the whole crystal. It is described by six parameters: edge lengths aa, bb, cc and the angles α\alpha (between bb and cc), β\beta (between aa and cc), γ\gamma (between aa and bb). Primitive (simple) unit cell: particles only at the corners. Centred unit cells: body-centred (one extra particle at the body centre), face-centred (one at the centre of each face), end-centred (at the centres of two opposite faces). Seven crystal systems and fourteen Bravais lattices. Cubic: a=b=ca = b = c, α=β=γ=90∘\alpha = \beta = \gamma = 90^{\circ}; primitive, body-centred, face-centred (NaCl, Cu, ZnS). Tetragonal: a=b≠ca = b \ne c, all 90∘90^{\circ}; primitive, body-centred (white tin, SnO2\mathrm{SnO_{2}}, TiO2\mathrm{TiO_{2}}). Orthorhombic: a≠b≠ca \ne b \ne c, all 90∘90^{\circ}; primitive, body-, face- and end-centred (rhombic sulfur, KNO3\mathrm{KNO_{3}}, BaSO4\mathrm{BaSO_{4}}). Hexagonal: a=b≠ca = b \ne c, α=β=90∘\alpha = \beta = 90^{\circ}, γ=120∘\gamma = 120^{\circ}; primitive (graphite, ZnO, CdS). Rhombohedral (trigonal): a=b=ca = b = c, α=β=γ≠90∘\alpha = \beta = \gamma \ne 90^{\circ}; primitive (calcite, cinnabar HgS). Monoclinic: a≠b≠ca \ne b \ne c, α=γ=90∘\alpha = \gamma = 90^{\circ}, β≠90∘\beta \ne 90^{\circ}; primitive, end-centred (monoclinic sulfur, Na2SO4⋅10H2O\mathrm{Na_{2}SO_{4} \cdot 10H_{2}O}). Triclinic: a≠b≠ca \ne b \ne c, α≠β≠γ≠90∘\alpha \ne \beta \ne \gamma \ne 90^{\circ}; primitive (K2Cr2O7\mathrm{K_{2}Cr_{2}O_{7}}, CuSO4⋅5H2O\mathrm{CuSO_{4} \cdot 5H_{2}O}, H3BO3\mathrm{H_{3}BO_{3}}). In total, 3+2+4+1+1+2+1=143 + 2 + 4 + 1 + 1 + 2 + 1 = 14 Bravais lattices. Counting particles per unit cell (see figure). A particle at a corner is shared by 88 cells (contributes 18\frac{1}{8}), on an edge by 44 (14\frac{1}{4}), on a face by 22 (12\frac{1}{2}), and at the body centre by 11. Simple cubic: 8×18=18 \times \frac{1}{8} = 1. Body-centred cubic (bcc): 1+1=21 + 1 = 2. Face-centred cubic (fcc): 1+6×12=41 + 6 \times \frac{1}{2} = 4. For a hexagonal close-packed (hcp) unit cell (hexagonal prism): 12×16+2×12+3=612 \times \frac{1}{6} + 2 \times \frac{1}{2} + 3 = 6. The effective number zz allows formulas to be deduced from positions of different atoms. Key Takeaways: - Seven crystal systems, fourteen Bravais lattices. - Corner 18\frac{1}{8}, edge 14\frac{1}{4}, face 12\frac{1}{2}, body 11. - zz: sc 11, bcc 22, fcc 44, hcp prism 66. - Pitfall: in a hexagonal prism, a corner atom is shared by 66 cells, not 88. Brain-Teaser 1: In a cubic solid, atoms A occupy the corners and atoms B occupy the face centres. What is the formula? What if one corner atom is missing from every unit cell? Solution: A: 8×18=18 \times \frac{1}{8} = 1; B: 6×12=36 \times \frac{1}{2} = 3; formula AB3\mathrm{AB_{3}}. With one corner missing: A =7×18=78= 7 \times \frac{1}{8} = \frac{7}{8}, so A : B =78:3=7:24= \frac{7}{8} : 3 = 7 : 24, formula A7B24\mathrm{A_{7}B_{24}}. Brain-Teaser 2: A compound has atoms X at the corners, Y at the body centre and Z at the edge centres of a cube. Find its formula. Solution: X =1= 1, Y =1= 1, Z =12×14=3= 12 \times \frac{1}{4} = 3: XYZ3\mathrm{XYZ_{3}} (the perovskite-like arrangement, as in CaTiO3\mathrm{CaTiO_{3}}-type formulas). Brain-Teaser 3: Which crystal system has a=b≠ca = b \ne c and γ=120∘\gamma = 120^{\circ}, and which has all edges equal but angles not 90∘90^{\circ}? Solution: Hexagonal (a=b≠ca = b \ne c, α=β=90∘\alpha = \beta = 90^{\circ}, γ=120∘\gamma = 120^{\circ}). Rhombohedral (a=b=ca = b = c, α=β=γ≠90∘\alpha = \beta = \gamma \ne 90^{\circ}).

Practice this concept

Free · no account required · instant solutions

Crystal Lattices, Unit Cells and Counting Atoms per Unit Cell — DPP

10 questionsSee the solution the moment you answer

Practice