Ch 7 The Crystalline Solid State
I.
Formulas and Structures
A.
Simple Structures
1) Unit Cell = simplest repeating unit of the regular crystalline array
2)
Bravais Lattices = 14 possible basic crystal structure unit cell types
3)
a)
Atoms on corners and edges are shared between unit cells
i. Rectangular corners shared by 8 unit cells (8 x 1/8 = 1 total atom in cell)
ii. Other corners shared unequally, but still contribute 1 atom to unit cell
iii. Edges shared by 4 cells, contribute (4 x ¼ = 1 atom to unit cell)
iv. Faces shared by 2 cells, contribute ½ atom to each
b)
Angles and Dimensions can vary: triclinic has all different lengths and angles
c)
Lattice Points = positions of atoms needed to generate the whole crystal
i. Body Centered Cubic: (0,0,0) = origin and (½ , ½, ½) = center
ii. All other atoms can be generated from these 2 by moving them exactly
one cell length
Cubic Structures
a) Primitive Cubic is the simplest type
i. To fully describe: length of side, 90o angles, (0,0,0) lattice point
ii. 8 x 1/8 = 1 atom in the unit cell
iii. each atom surrounded by 6 others (Coordination Number = CN = 6)
iv. Not efficiently packed; only 52.4% of volume is occupied (74.1% max)
v. Vacant space at center with CN = 8
0.73r sphere would fit here
4)
4r
l
2l
5)
Body-Centered Cubic (bcc)
a) One more atom is added to the center of the cube (68% efficient)
b) Size of unit cell must increase over simple cubic
c) Diagonal across center = 4r (r = radius of one atom)
d) Corner atoms not in contact with each other due to cell size expansion
e) Side = 2.31r (Calculate this in Ex. 7-2)
f) Unit cell contains 1(1) + 8(1/8) = 2 atoms
g) Lattice Points: (0,0,0) and ( ½ , ½ , ½ )
Close-Packed Structures
a) Spheres will arrange to take up the least space, not cubic or bcc
b) Two structures have almost identical packing efficiency (74.1%)
i. Hexagonal Close Packing = (hcp)
ii. Cubic Close Packing = (ccp) = Face-Centered Cubic = (fcc)
iii. Both have CN = 12 for each atom: 6 in its layer, 3 above, and 3 below
iv. 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
v. Both have 2 tetrahedral (Td) holes per atom (CN = 4) formed by 3 atoms
in one layer and one atom above/below
vi. Both have 1 octahedral (Oh) hole per atom (CN = 6) formed by 3 atoms
in one layer and 3 atoms above/below
c)
HCP unit cell is smaller than the hexagonal prism
i) Take four touching atoms in 1st layer and extend lines up to 3rd layer
ii) 8(1/8) + 1 = 2 atoms in unit cell
iii) Dimensions = 2r, 2r, and 2.83r
iv) Angles = 120o, 90o, and 90o
v) Lattice Points = (0,0,0,) and ( 1/3, 2/3, 1/2)
d)
CCP unit cell
i) One corner from layer 1, opposite corner from layer 4
ii) 6 atoms each from layers 2 and 3 in the unit cell
iii) Unit cell is fcc
iv) 8(1/8) + 6(1/2) = 4 atoms in the unit cell
v) Lattice Points = (0,0,0), ( ½ , 0 , ½), ( ½ , ½, 0), (0 , ½ , ½)
e)
6)
Holes
i) 2 Td and 1 Oh hole per atom in both (hcp) and (ccp)
ii) If ionic compound, small cations can fill these holes
iii) Td hole = 0.225r
Oh hole = 0.414r
iv) NaCl has Cl- in (ccp) with Na+ (ccp) but also in the Oh holes
v) Na+ = 0.695r, so it forces the Cl- to be farther apart, but still CN = 6
Metallic Crystals
a) Most metals crystallize in (bcc), (ccp), or (hcp) structures
b) Changing pressure or temperature can interchange these forms for a metal
c) Must consider bonding, not just geometric packing only
d) Soft and malleable metals usually have (ccp) structure (copper)
e) Harder and more brittle metals usually have the (hcp) structure (zinc)
f) Most metals can be bent due to non-directional bonding
i. Weak bonding to all neighbors, not strong bonding to any single one
ii. Atoms can slide past each other and then realign into crystal form
iii. Dislocations = imperfections in lattice; make it easier to bend
iv. Impurities = other elements; allow slippage of layers
v. Work Hardening = hammer until impurities are together
vi. Heat: can soften by dispersing impurities or harden if control cooling
7)
B.
Diamond
a) Each atom is tetrahedrally bonded to 4 other carbon atoms
b) Directional single bonds, unlike metals, cause hardness
Binary Compound Structures
1) Simplest structures just have 2nd element in holes of the first element’s lattice
a) Small cations can fit in Td/Oh holes of large anions
b) Large cations may only be able to fit in Oh holes
c) Even larger cations may cause changes in structure if they don’t fit in holes
2) Relative Number of Cations/Anions
a) M2X won’t allow close-packing of anion lattice with cations in Oh holes
because there are more cations than holes
b) Alternatives: cations in Td holes, vacancies, not close-packed
2)
Sodium Chloride Structure: NaCl
a) Na+ in (fcc) and Cl- in (fcc)
b) Offset by ½ unit cell length
c) Na+ are centered in edges of Cl- lattice (or vice versa)
d) Most alkali halides have this same structure
e) Large size difference of ions facilitate this structure (Na+ ~70% of Cl-)
f) Each Cl- has CN = 6 Na+ Each Na+ has CN = 6 Cl-
3)
Cesium Chloride Structure: CsCl
a) Cs in simple cubic structure with Cl- in center (or vice versa)
b) Cl- = 0.83Cs+ size (0.73r in center is ideal)
c) Rare structure, need big cation (Cs, Tl only cations known with this structure)
4)
Zinc Blende Structure: ZnS
a) Same as diamond structure with alternating Zn and S atoms
b) Alternate: Zn and S each in (fcc) lattices combined so each ion is in a Td hole
of the other lattice
c) Stoichiometry: only ½ of the Td holes are occupied and ½ are vacant
5)
Wurtzite Structure: ZnS
a) Rarer than Zinc Blende structure for ZnS; formed at higher Temperatures
b) Zn and S each in (hcp) lattices combined so each ion is in a Td hole of the
other lattice
c) Again, ½ of the Td holes are vacant
6)
Fluorite Structure: CaF2
a) Ca2+ in (ccp) lattice with 8 F- surrounding each and occupying all Td holes
b) Alternate: F- in simple cubic lattice with Ca2+ in alternate body centers
c) Nearly perfect radius fits for this structure
d) Antifluorite Structure
i. Reverse stoichiometry compounds like Na2O
ii. Every Td hole in the anion lattice is occupied by a cation
7)
Nickel Arsenide Structure: NiAs
a) As atoms in close packed layers exactly above each other
b) Ni atoms in all the Oh holes
c) Both Ni and As have CN = 6
d) Alternate: Ni atoms occupy all Oh holes of (hcp) As lattice
e) Usual for MX compounds where X = Sn, As, Bi, S, Se, Te
8)
Rutile Structure: TiO2
a) Distorted TiO6 octahedra forming columns by sharing edges
b) Ti CN = 6; O CN = 3
c) Adjacent columns connected by sharing corners of octahedra
d) 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
e) MgF2, ZnF2 are other examples
9)
Complex Structures
a) Substitution of cation or anion in a given structure is possible, giving a range
of compounds with essentially the same structure.
b) When charge or size is different in the substitution, the structure must change
i. Balance charge by leaving vacancies
ii. Lattice can spread/contract out to accommodate larger/smaller ions
c) Complex, non-spherical anions may distort the basic lattice shape
d) Large cations may require increased coordination number (> 6)
Calcite
Stucture: CN = 6
LiNO3
NaNO3
MgCO3
CaCO3
FeCO3
InBO3
YBO3
Hexagonal, CN = 6 O per metal
Aragonite
Stucture: CN = 9
CaCO3
KNO3
SrCO3
LaBO3
Orthorhombic, CN = 9 O per metal
D) The Radius Ratio
1) We can crudely predict CN by using the ratio r+/r2) This assumes atoms are just packing as hard spheres (not really all that occurs)
3) Examples:
a) NaCl
r+/r- = 113/167 (CN = 4) = 0.667
r+/r- = 116/167 (CN = 6) = 0.695
Fits best with CN = 6
b)
4)
ZnS
r+/r- = 74/170 (CN = 4) = 0.435
r+/r- = 88/170 (CN = 6) = 0.518
Fits best with CN = 4
CN = 4 in actual structure
Example: CaF2 has fluoride ions in a simple cubic array and calcium ions in
alternate body centers, with r+/r- = 0.97. 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.