JEE Chemistry The Solid State Notes Set 03

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Crystalline and Amorphous Solids

 

In a crystalline solid the particles follow a regular pattern that repeats over long distances (long-range order). Such solids have a sharp melting point, a definite heat of fusion, flat cleavage faces and are anisotropic, meaning some properties such as refractive index or conductivity depend on direction.
In an amorphous solid (glass, rubber, plastics) the order is only short range. They soften over a range of temperature, break into irregular pieces and are isotropic. They are often called super-cooled liquids or pseudo solids.

 

Types of Crystalline Solids

 

TypeParticlesBinding forceExamplesProperties
MolecularMoleculesVan der Waals, hydrogen bondsIce, dry ice, I2Soft, low melting point, insulators
IonicCations and anionsElectrostaticNaCl, CaF2Hard, brittle, high melting point, conduct when molten
Covalent (network)AtomsCovalent bondsDiamond, SiC, quartzVery hard, very high melting point, insulators (graphite conducts)
MetallicKernels in electron seaMetallic bondCu, Fe, NaMalleable, ductile, good conductors

 

Crystal Systems and Bravais Lattices

 

The smallest repeating unit of a lattice is the unit cell, described by edges a, b, c and angles α, β, γ. Seven crystal systems exist and together they give 14 Bravais lattices.

 

SystemEdgesAnglesLattices
Cubica = b = cα = β = γ = 90°sc, bcc, fcc
Tetragonala = b ≠ cα = β = γ = 90°simple, body centred
Orthorhombica ≠ b ≠ cα = β = γ = 90°simple, end, body, face centred
Rhombohedral (trigonal)a = b = cα = β = γ ≠ 90°simple
Hexagonala = b ≠ cα = β = 90°, γ = 120°simple
Monoclinica ≠ b ≠ cα = γ = 90°, β ≠ 90°simple, end centred
Triclinica ≠ b ≠ cα ≠ β ≠ γ ≠ 90°simple

 

Cubic Unit Cells

 

An atom at a corner is shared by 8 cells (contributes 1/8), on a face by 2 cells (1/2), on an edge by 4 cells (1/4) and at the body centre it belongs wholly to one cell (1).
The three cubic cells are drawn below. Black-outlined circles are corner atoms and red-outlined circles are the extra face-centred or body-centred atoms.

 

Simple cubic (Z = 1) Body centred cubic (Z = 2) Face centred cubic (Z = 4)

 

PropertySimple cubicbccfcc
Atoms per cell Z124
Coordination number6812
Touching directionedge: a = 2rbody diagonal: √3a = 4rface diagonal: √2a = 4r
Radiusa/2√3a/4a/(2√2)
Packing efficiency52.4%68%74%

 

Density of a Unit Cell

 

\( \rho = \dfrac{Z \times M}{N_A \times a^3} \), where M is the molar mass in g mol−1 and a is the edge in cm (1 pm = 10−10 cm, 1 Å = 10−8 cm).
The same relation can be used to find Z, M or NA when the other quantities are known. Z must come out as a whole number.

 

Worked example 1. Copper is fcc with edge 361 pm and molar mass 63.5 g mol−1. Find its density.
Answer:
\( a^3 = (3.61\times10^{-8})^3 = 4.70\times10^{-23}\ \text{cm}^3 \)
\( \rho = \dfrac{4\times63.5}{6.022\times10^{23}\times4.70\times10^{-23}} = 8.97\ \text{g cm}^{-3} \)

 

Close Packing of Identical Spheres

 

In one layer, spheres touch six neighbours (two-dimensional hexagonal packing). A second layer sits in the dips of the first. For the third layer there are two choices.
Placing it directly above the first layer gives ABAB... which is hexagonal close packing (hcp). Placing it over the uncovered voids gives ABCABC... which is cubic close packing (ccp), the same as fcc.
Both have coordination number 12 and packing efficiency 74%, leaving 26% empty space.

 

Voids in Close Packing

 

For N spheres in a close-packed array there are N octahedral voids (surrounded by 6 spheres) and 2N tetrahedral voids (surrounded by 4 spheres).
Radius of the void in terms of sphere radius R: tetrahedral 0.225R, octahedral 0.414R. In an fcc cell: 4 octahedral voids and 8 tetrahedral voids.

 

Ionic Crystals and Radius Ratio

 

r+/r−Coordination numberShape of void
0.155 to 0.2253Triangular
0.225 to 0.4144Tetrahedral
0.414 to 0.7326Octahedral
0.732 to 18Cubic

 

StructureDescriptionCoordinationZExamples
Rock salt (NaCl)Cl− ccp, Na+ in all octahedral voids6 : 64NaCl, KCl, MgO
Zinc blende (ZnS)S2− ccp, Zn2+ in half of tetrahedral voids4 : 44ZnS, CuCl, AgI
Fluorite (CaF2)Ca2+ fcc, F− in all tetrahedral voids8 : 44CaF2, BaCl2
AntifluoriteAnion fcc, cations in all tetrahedral voids4 : 84Na2O, K2O
Caesium chlorideCl− simple cubic, Cs+ at body centre8 : 81CsCl, CsBr, TlCl

 

Distance relations: rock salt a = 2(r+ + r−); CsCl type √3a = 2(r+ + r−).

 

Worked example 2. A solid has anions in ccp and cations in all octahedral voids. Find the formula.
Answer:
Per 4 anions there are 4 octahedral voids, so cations = 4.
Ratio 1 : 1, so the formula is MX (rock salt type).

 

Imperfections in Solids

 

Point defects arise above 0 K. Stoichiometric defects keep the cation : anion ratio unchanged.
Schottky defect: equal numbers of cations and anions are missing; found when ions are similar in size (NaCl, KCl, CsCl, AgBr). Density decreases.
Frenkel defect: a smaller ion (usually the cation) moves from its lattice site to an interstitial site (AgCl, AgBr, ZnS). Density is unchanged. AgBr shows both defects.
Non-stoichiometric defects: metal excess due to anionic vacancies (F-centres, which give colour) or extra interstitial cations (ZnO turns yellow on heating); metal deficiency when a cation is missing and another cation has a higher charge (FeO is really Fe0.95O).
Impurity defects: doping NaCl with SrCl2 creates one cation vacancy for every Sr2+ added.

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