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Solid-State Chemistry

Definition 1 (Bravais Lattice): An infinite array of discrete points generated by discrete translation operations. There are 14 Bravais lattices in 3D: 1 triclinic, 2 monoclinic, 4 orthorhombic, 2 tetragonal, 1 rhombohedral, 1 hexagonal, 3 cubic.

Simple Cubic (SC):

  • Atoms at the 8 corners of a cube.
  • Coordination number (CN) = 6.
  • Atoms per unit cell: 8×18=1\frac{8 \times 1}{8} = 1.
  • Packing fraction: π652.4%\frac{\pi}{6} \approx 52.4\%.

Body-Centered Cubic (BCC):

  • Atoms at 8 corners + 1 at the center.
  • CN = 8.
  • Atoms per unit cell: 8×18+1=2\frac{8 \times 1}{8} + 1 = 2.
  • Packing fraction: 3π868.0%\frac{\sqrt{3}\,\pi}{8} \approx 68.0\%.
  • Examples: Fe (α\alpha), Cr, W, Mo.

Face-Centered Cubic (FCC / Cubic Close-Packed, CCP):

  • Atoms at 8 corners + 1 at the center of each face.
  • CN = 12.
  • Atoms per unit cell: 8×18+6×12=4\frac{8 \times 1}{8} + \frac{6 \times 1}{2} = 4.
  • Packing fraction: π3274.0%\frac{\pi}{3\sqrt{2}} \approx 74.0\% (maximum for equal spheres).
  • Examples: Cu, Ag, Au, Al, Ni, Pt.

Theorem 1 (HCP Structure): ABAB stacking sequence. Each atom has CN = 12. Same packing fraction as FCC (74.0%).

Atoms per unit cell: 6 (in the conventional cell).

Examples: Mg, Zn, Ti, Co, Cd.

1.4 Relationship Between Lattice Parameters and Atomic Radius

Section titled “1.4 Relationship Between Lattice Parameters and Atomic Radius”
StructureRelationshipRadius in Terms of aa
SCa=2ra = 2rr=a/2r = a/2
BCC4r=3a4r = \sqrt{3}\,ar=3a/4r = \sqrt{3}\,a/4
FCC4r=2a4r = \sqrt{2}\,ar=2a/4=a/(22)r = \sqrt{2}\,a/4 = a/(2\sqrt{2})
HCPa=2ra = 2r, c/a=1.633c/a = 1.633 (ideal)r=a/2r = a/2

Example 1: Iron has a BCC structure with a=286.6a = 286.6 pm. Calculate the atomic radius.

r=3×286.64=124.1 pmr = \frac{\sqrt{3} \times 286.6}{4} = 124.1 \text{ pm}

\blacksquare

  • FCC arrangement of anions with cations in octahedral holes.
  • CN = 6 for both ions.
  • Formula: MX (1:1 stoichiometry).
  • Examples: NaCl, KBr, MgO, CaO.
  • Simple cubic arrangement of anions with cation at the body center.
  • CN = 8 for both ions.
  • Formula: MX (1:1 stoichiometry).
  • Examples: CsCl, CsBr, TlCl.
  • FCC arrangement of S2^{2-} with Zn2+^{2+} in half the tetrahedral holes.
  • CN = 4 for both ions.
  • Formula: MX.
  • Examples: ZnS, CuCl, GaAs.
  • FCC arrangement of Ca2+^{2+} with F^- in all tetrahedral holes.
  • CN: Ca2+^{2+} = 8, F^- = 4.
  • Formula: MX2_2.
  • Examples: CaF2_2, UO2_2, ZrO2_2.

Theorem 2 (Radius Ratio Rules): The ratio r+/rr_+/r_- determines the coordination geometry:

r+/rr_+/r_-CoordinationStructure
0.225–0.4144 (tetrahedral)ZnS (zinc blende)
0.414–0.7326 (octahedral)NaCl (rock salt)
0.732–1.08 (cubic)CsCl

Definition 2 (Lattice Energy, ΔU\Delta U): The energy released when 1 mol of an ionic solid is formed from its gaseous ions. Always exothermic.

Theorem 3 (Born-Haber Cycle): Lattice energy can be calculated thermodynamically:

ΔU=ΔfHΔHatomIE12BDEEA\Delta U = \Delta_f H^\circ - \Delta H_{\text{atom}} - \text{IE} - \frac{1}{2}\text{BDE} - \text{EA}

For NaCl: ΔU(NaCl)=ΔfH(NaCl)ΔHsub(Na)12D(Cl2)IE1(Na)EA(Cl)\Delta U(\text{NaCl}) = \Delta_f H^\circ(\text{NaCl}) - \Delta H_{\text{sub}}(\text{Na}) - \frac{1}{2}D(\text{Cl}_2) - \text{IE}_1(\text{Na}) - \text{EA}(\text{Cl})

Theorem 4 (Born-Lande Equation):

ΔU=NAMz+ze24πε0r0(11n)\Delta U = -\frac{N_A M z^+ z^- e^2}{4\pi\varepsilon_0 r_0}\left(1 - \frac{1}{n}\right)

where:

  • MM is the Madelung constant (depends on structure: NaCl = 1.748, CsCl = 1.763, ZnS = 1.638).
  • z+z^+, zz^- are ionic charges.
  • r0r_0 is the distance of closest approach.
  • nn is the Born exponent (8–12, related to the compressibility).

Example 2: Calculate the lattice energy of NaCl with r0=282r_0 = 282 pm, n=8n = 8, M=1.748M = 1.748.

ΔU=6.022×1023×1.748×1×1×(1.602×1019)24π×8.854×1012×282×1012(118)\Delta U = -\frac{6.022 \times 10^{23} \times 1.748 \times 1 \times 1 \times (1.602 \times 10^{-19})^2}{4\pi \times 8.854 \times 10^{-12} \times 282 \times 10^{-12}}\left(1 - \frac{1}{8}\right)

=1.748×(96.485 kJ/mol)282×1012×0.875=787 kJ/mol= -\frac{1.748 \times (96.485 \text{ kJ/mol})}{282 \times 10^{-12}} \times 0.875 = -787 \text{ kJ/mol}

\blacksquare

Definition 3 (Energy Band): When NN atoms are brought close together, their atomic orbitals overlap and split into NN closely spaced energy levels forming a continuous band.

  • Valence band: Highest occupied band at T=0T = 0.
  • Conduction band: Lowest unoccupied band at T=0T = 0.
  • Band gap (EgE_g): Energy difference between the top of the valence band and the bottom of the conduction band.
TypeBand GapConductivity (TT)Examples
ConductorEg=0E_g = 0DecreasesCu, Al, Na, Au
SemiconductorEg0.1E_g \sim 0.144 eVIncreases exponentiallySi (1.1 eV), Ge (0.67 eV)
InsulatorEg>4E_g > 4 eVVery lowDiamond (5.5 eV), SiO2_2

Intrinsic semiconductor: Conductivity due to thermally excited electrons across the band gap:

ni=ne=nh=NcNveEg/2kBTn_i = n_e = n_h = \sqrt{N_c N_v}\,e^{-E_g/2k_BT}

where nen_e is the electron concentration, nhn_h is the hole concentration, and NcN_c, NvN_v are the effective density of states.

Extrinsic semiconductors:

  • n-type: Doped with donors (Group 15 in Si, e.g., P, As) — extra electrons in the conduction band.
  • p-type: Doped with acceptors (Group 13 in Si, e.g., B, Al) — holes in the valence band.

Theorem 5 (pn Junction): At the interface of p-type and n-type material:

  • Depletion region forms (no free carriers).
  • Forward bias: Current flows; reverse bias: Current blocked.
  • Basis of diodes, transistors, and solar cells.

Definition 4 (Effective Mass): The curvature of the band determines the effective mass:

m=2(d2Edk2)1m^* = \hbar^2\left(\frac{d^2E}{dk^2}\right)^{-1}

Electrons near the bottom of the conduction band have positive mm^*; holes near the top of the valence band have negative mm^* (positive effective mass in the opposite direction).

Definition 5 (Schottky Defect): A cation-anion pair vacancy. Maintains electrical neutrality and approximately constant stoichiometry.

Vacancy concentration: nsNeΔHs/2kBT\text{Vacancy concentration: } n_s \approx N\,e^{-\Delta H_s/2k_BT}

Common in NaCl, CsCl (high CN, similar ionic sizes).

Definition 6 (Frenkel Defect): An ion displaced from its lattice site to an interstitial position. Common when one ion is much smaller (e.g., AgCl, AgBr).

Frenkel concentration: nfNNieΔHf/2kBT\text{Frenkel concentration: } n_f \approx \sqrt{N\,N_i}\,e^{-\Delta H_f/2k_BT}

where NiN_i is the number of interstitial sites.

Metal excess defects:

  • Anion vacancies with trapped electrons (F-centers, color centers). Examples: NaCl heated in Na vapor turns yellow (F-centers absorb blue light).
  • Interstitial cations.

Metal deficiency defects:

  • Cation vacancies with compensating charge (e.g., Fe1x_{1-x}O, where some Fe2+^{2+} is replaced by Fe3+^{3+} and vacancies maintain charge balance).
  • Dislocations: Edge dislocations (extra half-plane) and screw dislocations.
  • Grain boundaries: Boundaries between crystalline domains with different orientations.
  • Stacking faults: Errors in the stacking sequence (e.g., ABCABABC instead of ABCABC).

Theorem 6 (Bragg’s Law): Constructive interference occurs when:

nλ=2dsinθn\lambda = 2d\sin\theta

where nn is the order of reflection, λ\lambda is the X-ray wavelength, dd is the interplanar spacing, and θ\theta is the angle of incidence.

Definition 7 (Miller Indices): A set of integers (hkl)(hkl) that describe the orientation of a plane in a crystal lattice.

For a plane intercepting the crystallographic axes at (a/h,b/k,c/l)(a/h, b/k, c/l):

  • (100)(100): plane perpendicular to the aa axis.
  • (110)(110): plane bisecting aa and bb axes.
  • (111)(111): plane bisecting all three axes.

For a cubic crystal:

dhkl=ah2+k2+l2d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}

Example 3: For NaCl (a=564a = 564 pm) with Cu Kα\alpha radiation (λ=154.2\lambda = 154.2 pm), find the first-order Bragg angle for the (200) reflection.

d200=5644=282 pmd_{200} = \frac{564}{\sqrt{4}} = 282 \text{ pm}

sinθ=λ2d=154.22×282=0.2734\sin\theta = \frac{\lambda}{2d} = \frac{154.2}{2 \times 282} = 0.2734

θ=15.9°\theta = 15.9°

\blacksquare

Theorem 7 (Systematic Absences): Certain reflections are absent due to the symmetry of the unit cell (glide planes, screw axes, centering).

Lattice TypeAbsent When
SCNone
BCCh+k+lh + k + l = odd
FCCh,k,lh, k, l mixed (not all odd or all even)

Definition 8 (Polymorphism): The ability of a solid to exist in more than one crystal structure.

Examples:

  • Carbon: diamond (cubic), graphite (hexagonal).
  • Iron: α\alpha-Fe (BCC, ferromagnetic) → γ\gamma-Fe (FCC, paramagnetic) at 912°C → δ\delta-Fe (BCC).
  • Ti: α\alpha-Ti (HCP) → β\beta-Ti (BCC) at 882°C.

Solid solution: Atoms of different elements share the same lattice.

  • Substitutional: Similar-sized atoms (e.g., Cu–Ni).
  • Interstitial: Small atoms in the voids of a metal lattice (e.g., C in Fe → steel).

Theorem 8 (Quantum Confinement): When a semiconductor particle has a size comparable to the exciton Bohr radius, the band gap increases (blue shift in absorption/emission).

Eg(nanoparticle)=Eg(bulk)+2π22R2(1me+1mh)1.8e24πε0εrRE_g(\text{nanoparticle}) = E_g(\text{bulk}) + \frac{\hbar^2\pi^2}{2R^2}\left(\frac{1}{m_e^*} + \frac{1}{m_h^*}\right) - \frac{1.8e^2}{4\pi\varepsilon_0\varepsilon_r R}

where RR is the particle radius.

Definition 9 (Surface-to-Volume Ratio): For a nanoparticle of radius RR:

AV=3R\frac{A}{V} = \frac{3}{R}

As R0R \to 0, surface atoms become a larger fraction of total atoms, leading to:

  • Enhanced reactivity.
  • Lower melting points (Tm1/RT_m \propto 1/R for very small particles).
  • Different mechanical properties.
TypeDimensionsExample
Nanoparticles0DAu, Ag, CdSe quantum dots
Nanotubes1DCarbon nanotubes, BN nanotubes
Nanowires1DSi nanowires, Ag nanowires
Nanosheets2DGraphene, MoS2_2
Nanocomposites3DNanoparticle-polymer blends

Definition 10 (Zeolite): Crystalline aluminosilicates with a 3D framework of SiO4_4 and AlO4_4 tetrahedra, creating pores and channels of molecular dimensions.

General formula: Mx/n[(AlO2)x(SiO2)y]mH2O\text{M}_{x/n}[(\text{AlO}_2)_x(\text{SiO}_2)_y]\cdot m\text{H}_2\text{O}

where M is the cation (e.g., Na+^+, K+^+, Ca2+^{2+}) balancing the negative charge from Al3+^{3+} substitution in the framework.

  • Ion exchange: Water softening (Na+^+ replaces Ca2+^{2+}, Mg2+^{2+}).
  • Molecular sieves: Size-selective adsorption based on pore dimensions.
  • Catalysis: Shape-selective catalysis in petrochemical cracking.
  • Gas separation: Separation of gases by molecular size.

Common zeolite structures: Linde Type A (LTA), Faujasite (FAU, includes X and Y zeolites), MFI (ZSM-5), Mordenite (MOR).

Definition 11 (Superconductor): A material with zero electrical resistance below a critical temperature TcT_c.

Theorem 9 (BCS Theory): Below TcT_c, electrons form Cooper pairs via phonon-mediated attraction:

2Δ=3.53kBTc2\Delta = 3.53\,k_B\,T_c

where Δ\Delta is the superconducting energy gap.

Meissner effect: Superconductors expel magnetic fields below TcT_c and HcH_c (critical field).

Cuprate superconductors (e.g., YBa2_2Cu3_3O7_7, Tc=92T_c = 92 K):

  • Layered perovskite structures with CuO2_2 planes.
  • TcT_c depends on oxygen stoichiometry.
  • Iron-based superconductors (e.g., LaFeAsO, Tc26T_c \sim 26 K).

Definition 12 (Non-Stoichiometry): Many transition metal oxides have variable stoichiometry due to mixed oxidation states and defects.

Fe1x_{1-x}O: Some Fe2+^{2+} is oxidized to Fe3+^{3+}, with vacancies maintaining charge balance.

Definition 13 (Superionic Conductor): Solids with exceptionally high ionic conductivity due to mobile ions in a rigid framework.

Examples:

  • β\beta-alumina (Na+^+ conductivity).
  • AgI above 147°C (Ag+^+ mobility).
  • Yttria-stabilized zirconia (YSZ, O2^{2-} conductivity — used in solid oxide fuel cells).
  1. Confusing SC, BCC, and FCC packing fractions. SC = 52.4%, BCC = 68.0%, FCC/HCP = 74.0%. Fix: Calculate: SC=π/6\text{SC} = \pi/6, BCC=3π/8\text{BCC} = \sqrt{3}\pi/8, FCC=π/(32)\text{FCC} = \pi/(3\sqrt{2}).
  2. Wrong atoms per unit cell count. Corner atoms count as 1/8, face atoms as 1/2, edge atoms as 1/4, body atoms as 1. Fix: Always use the correct fractional count for each position.
  3. Using the Born-Lande equation for covalent solids. The equation assumes purely ionic bonding. Fix: Use the Born-Haber cycle with thermodynamic data for more accurate values.
  4. Confusing intrinsic and extrinsic semiconductors. Intrinsic carriers come from thermal excitation; extrinsic carriers come from dopants. Fix: At room temperature, ne=nh=nin_e = n_h = n_i (intrinsic) or neNDn_e \approx N_D (n-type), nhNAn_h \approx N_A (p-type).
  5. Wrong Miller indices for planes. Miller indices (hkl)(hkl) are the reciprocals of the axis intercepts, not the intercepts themselves. Fix: Take reciprocals, clear fractions, reduce to smallest integers.
  6. Confusing Schottky and Frenkel defects. Schottky = vacancy pair (cation + anion); Frenkel = displacement (ion moves to interstitial). Fix: Schottky is favored when ions are similar in size (alkali halides); Frenkel when one ion is much smaller (AgCl).
  7. Ignoring the quantum size effect for nanoparticles. The band gap depends on particle size; bulk properties don’t apply to nanomaterials. Fix: Use the quantum confinement formula or experimental data for nanoparticle-specific properties.
  • Crystal structures: SC (CN=6), BCC (CN=8), FCC/HCP (CN=12); packing fractions 52%, 68%, 74%.
  • Ionic structures: NaCl (6:6), CsCl (8:8), ZnS (4:4), CaF2_2 (8:4); radius ratio rules.
  • Born-Haber cycle: Lattice energy from thermodynamic cycle; Born-Lande equation.
  • Band theory: Conductors (Eg=0E_g = 0), semiconductors (Eg0.1E_g \sim 0.1–4 eV), insulators (Eg>4E_g > 4 eV).
  • Defects: Schottky (vacancy pairs), Frenkel (displacement); non-stoichiometry.
  • X-ray diffraction: Bragg’s law nλ=2dsinθn\lambda = 2d\sin\theta; Miller indices; systematic absences.
  • Nanomaterials: Quantum confinement; high surface-to-volume ratio; quantum dots, nanotubes.
  • Zeolites: Porous aluminosilicates; ion exchange, molecular sieves, catalysis.

Example 1: Calculating Density from Unit Cell Parameters

Section titled “Example 1: Calculating Density from Unit Cell Parameters”

Problem: Sodium chloride crystallises in a face-centred cubic structure with a = 564 pm. Calculate the density of NaCl. (M_Na = 23.0, M_Cl = 35.5 g/mol, N_A = 6.022 x 10^23). Solution: NaCl unit cell contains 4 Na+ and 4 Cl- ions (FCC arrangement). Molar mass of NaCl = 58.5 g/mol. Mass of unit cell = (4 x 58.5) / (6.022 x 10^23) = 3.886 x 10^-22 g. Volume = a^3 = (564 x 10^-10 cm)^3 = 1.795 x 10^-22 cm^3. Density = 3.886 x 10^-22 / 1.795 x 10^-22 = 2.17 g/cm^3. Literature value: 2.16 g/cm^3.

Example 2: Predicting Stoichiometry from Radius Ratio

Section titled “Example 2: Predicting Stoichiometry from Radius Ratio”

Problem: NaCl has r(Na+) = 102 pm and r(Cl-) = 181 pm. Determine the expected coordination geometry using the radius ratio rule. Solution: Radius ratio = r+/r- = 102/181 = 0.564. For 0.414 < r+/r- < 0.732, the predicted coordination number is 6 (octahedral), matching the observed NaCl (rock salt) structure. If the ratio were below 0.414, tetrahedral (ZnS) coordination would be expected. If above 0.732, cubic (CsCl) coordination.

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