1. Ch 7 The Crystalline Solid State
Formulas and Structures
Simple Structures
Unit Cell = simplest repeating unit of the regular crystalline array
Bravais Lattices = 14 possible basic crystal structure unit cell types

2. Atoms on corners and edges are shared between unit cells
Rectangular corners shared by 8 unit cells (8 x 1/8 = 1 total atom in cell)
Other corners shared unequally, but still contribute 1 atom to unit cell
Edges shared by 4 cells, contribute (4 x ¼ = 1 atom to unit cell)
Faces shared by 2 cells, contribute ½ atom to each
Angles and Dimensions can vary: triclinic has all different lengths and angles
Lattice Points = positions of atoms needed to generate the whole crystal
Body Centered Cubic: (0,0,0) = origin and (½ , ½, ½) = center
All other atoms can be generated from these 2 by moving them exactly one cell length
Cubic Structures
Primitive Cubic is the simplest type
To fully describe: length of side, 90o angles, (0,0,0) lattice point
8 x 1/8 = 1 atom in the unit cell
each atom surrounded by 6 others (Coordination Number = CN = 6)
Not efficiently packed; only 52.4% of volume is occupied (74.1% max)
Vacant space at center with CN = r sphere would fit here

3. Body-Centered Cubic (bcc)
One more atom is added to the center of the cube
Size of unit cell must increase over simple cubic
Diagonal across center = 4r (r = radius of one atom)
Corner atoms not in contact with each other due to cell size expansion
Side = 2.31r (Calculate this in Ex. 7-2)
Unit cell contains 1(1) + 8(1/8) = 2 atoms
Lattice Points: (0,0,0) and ( ½ , ½ , ½ )
Close-Packed Structures
Spheres will arrange to take up the least space, not cubic or bcc
Two structures have almost identical packing efficiency (74.1%)
Hexagonal Close Packing = (hcp)
Cubic Close Packing = (ccp) = Face-Centered Cubic = (fcc)
Both have CN = 12 for each atom: 6 in its layer, 3 above, and 3 below
HCP has 3rd layer in identical position as first ABA; second layer above holes; CCP has 3rd layer displaced so above holes in first layer ABC
Both have 2 tetrahedral (Td) holes per atom (CN = 4) formed by 3 atoms in one layer and one atom above/below
Both have 1 octahedral (Oh) hole per atom (CN = 6) formed by 3 atoms in one layer and 3 atoms above/below 4r l l

5. HCP unit cell is smaller than the hexagonal prism
Take four touching atoms in 1st layer and extend lines up to 3rd layer
8(1/8) = 2 atoms in unit cell
Dimensions = 2r, 2r, and 2.83r
Angles = 120o, 90o, and 90o
Lattice Points = (0,0,0,) and ( 1/3, 2/3, 1/2)

6. CCP unit cell
One corner from layer 1, opposite corner from layer 4
6 atoms each from layers 2 and 3 in the unit cell
Unit cell is fcc
8(1/8) + 6(1/2) = 4 atoms in the unit cell
Lattice Points = (0,0,0), ( ½ , 0 , ½), ( ½ , ½, 0), (0 , ½ , ½)

7. Holes
2 Td and 1 Oh hole per atom in both (hcp) and (ccp)
If ionic compound, small cations can fill these holes
Td hole = 0.225r Oh hole = 0.414r
NaCl has Cl- in (ccp) with Na+ (ccp) but also in the Oh holes
Na+ = 0.695r, so it forces the Cl- to be farther apart, but still CN = 6
Metallic Crystals
Most metals crystallize in (bcc), (ccp), or (hcp) structures
Changing pressure or temperature can interchange these forms for a metal
Must consider bonding, not just geometric packing only
Soft and malleable metals usually have (ccp) structure (copper)
Harder and more brittle metals usually have the (hcp) structure (zinc)
Most metals can be bent due to non-directional bonding
Weak bonding to all neighbors, not strong bonding to any single one
Atoms can slide past each other and then realign into crystal form
Dislocations = imperfections in lattice; make it easier to bend
Impurities = other elements; allow slippage of layers
Work Hardening = hammer until impurities are together
Heat: can soften by dispersing impurities or harden if control cooling

8. Diamond
Each atom is tetrahedrally bonded to 4 other carbon atoms
Directional single bonds, unlike metals, cause hardness
Binary Compound Structures
Simplest structures just have 2nd element in holes of the first element’s lattice
Small cations can fit in Td/Oh holes of large anions
Large cations may only be able to fit in Oh holes
Even larger cations may cause changes in structure if they don’t fit in holes
Relative Number of Cations/Anions
M2X won’t allow close-packing of anion lattice with cations in Oh holes because there are more cations than holes
Alternatives: cations in Td holes, vacancies, not close-packed

9. Sodium Chloride Structure: NaCl
Na+ in (fcc) and Cl- in (fcc)
Offset by ½ unit cell length
Na+ are centered in edges of Cl- lattice (or vice versa)
Most alkali halides have this same structure
Large size difference of ions facilitate this structure
Each Cl- has CN = 6 Na+ Each Na+ has CN = 6 Cl-
Cesium Chloride Structure: CsCl
Cs in simple cubic structure with Cl- in center (or vice versa)
Cl- = 0.83Cs+ size (0.73r in center is ideal)
Rare structure, need big cation (Cs, Tl only cations known with this structure)

10. Zinc Blende Structure: ZnS
Same as diamond structure with alternating Zn and S atoms
Alternate: Zn and S each in (fcc) lattices combined so each ion is in a Td hole of the other lattice
Stoichiometry: only ½ of the Td holes are occupied and ½ are vacant
Wurtzite Structure: ZnS
Rarer than Zinc Blende structure for ZnS; formed at higher Temperatures
Zn and S each in (hcp) lattices combined so each ion is in a Td hole of the other lattice
Again, ½ of the Td holes are vacant

11. Fluorite Structure: CaF2
Ca2+ in (ccp) lattice with 8 F- surrounding each and occupying all Td holes
Alternate: F- in simple cubic lattice with Ca2+ in alternate body centers
Nearly perfect radius fits for this structure
Antifluorite Structure
Reverse stoichiometry compounds like Na2O
Every Td hole in the anion lattice is occupied by a cation

12. Nickel Arsenide Structure: NiAs
As atoms in close packed layers exactly above each other
Ni atoms in all the Oh holes
Both Ni and As have CN = 6
Alternate: Ni atoms occupy all Oh holes of (hcp) As lattice
Usual for MX compounds where X = Sn, As, Bi, S, Se, Te
Rutile Structure: TiO2
Distorted TiO6 octahedra forming columns by sharing edges
Ti CN = 6; O CN = 3
Adjacent columns connected by sharing corners of octahedra
Unit cell has Ti at corners and in the body center, 4 O in the faces, and 2 O in the plane of the body center Ti
MgF2, ZnF2 are other examples

13. Complex Structures
Substitution of cation or anion in a given structure is possible, giving a range of compounds with essentially the same structure.
When charge or size is different in the substitution, the structure must change
Balance charge by leaving vacancies
Lattice can spread/contract out to accommodate larger/smaller ions
Complex, non-spherical anions may distort the basic lattice shape
Large cations may require increased coordination number (> 6)
Calcite
Stucture: CN = 6
LiNO3
NaNO3
MgCO3
CaCO3
FeCO3
InBO3
YBO3
Aragonite Stucture: CN = 9
CaCO3
KNO3
SrCO3
LaBO3
Hexagonal, CN = 6 O per metal
Orthorhombic, CN = 9 O per metal

14. The Radius Ratio
We can crudely predict CN by using the ratio r+/r-
This assumes atoms are just packing as hard spheres (not really all that occurs)
Examples:
NaCl r+/r- = 113/167 (CN = 4) = 0.667
r+/r- = 116/167 (CN = 6) = Fits best with CN = 6
ZnS r+/r- = 74/170 (CN = 4) = Fits best with CN = 4
r+/r- = 88/170 (CN = 6) = CN = 4 in actual structure
4) Example: CaF2 has fluoride ions in a simple cubic array and calcium ions in alternate body centers, with r+/r- = What are the coordination numbers of the two ions predicted by r+/r- ? What are the coordination numbers observed? Predict coordination numbers of Ca2+ in CaCl2 and CaBr2.