Chemistry Concept Notes for JEE: Solid State
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Classification of Solids
Crystalline solids have particles arranged in a regular pattern that repeats over long distances (long range order). They have a sharp melting point and are anisotropic, meaning properties such as refractive index and conductivity depend on direction. Amorphous solids (glass, rubber, plastics) have order only over short distances, melt over a range of temperature and are isotropic; they behave like supercooled liquids.
Crystalline solids are grouped by the nature of the particles and the forces between them:
1. Molecular solids: molecules held by weak dispersion, dipole or hydrogen-bond forces (dry ice, ice, I₂). Soft, low melting, non-conducting.
2. Ionic solids: cations and anions held by electrostatic attraction (NaCl, CsCl). Hard, brittle, high melting; conduct only when molten or dissolved.
3. Covalent (network) solids: atoms joined in a giant network of covalent bonds (diamond, SiC, quartz). Very hard, very high melting, insulators; graphite is an exception because its layers have free electrons.
4. Metallic solids: metal ions in a sea of delocalised electrons. Malleable, ductile and good conductors.
Lattice and Unit Cell
A crystal lattice is the regular three dimensional arrangement of points that represent the particles. The smallest repeating block of the lattice is the unit cell; it is described by three edge lengths a, b, c and three angles α, β, γ. Unit cells are primitive (particles only at corners) or centred: body centred (extra particle at the centre), face centred (extra particle at the centre of each face) or end centred.
The sharing rule decides how much of a particle belongs to one cell: a corner particle is shared by 8 cells (1/8), an edge particle by 4 cells (1/4), a face particle by 2 cells (1/2) and a particle inside the cell belongs wholly to it.
| Crystal system | Edges | Angles | Example |
|---|---|---|---|
| Cubic | a = b = c | α = β = γ = 90° | NaCl, Cu, diamond |
| Tetragonal | a = b ≠ c | α = β = γ = 90° | SnO₂, white tin |
| Orthorhombic | a ≠ b ≠ c | α = β = γ = 90° | rhombic sulphur, KNO₃ |
| Hexagonal | a = b ≠ c | α = β = 90°, γ = 120° | graphite, ZnO |
| Rhombohedral | a = b = c | α = β = γ ≠ 90° | calcite, HgS |
| Monoclinic | a ≠ b ≠ c | α = γ = 90°, β ≠ 90° | monoclinic sulphur |
| Triclinic | a ≠ b ≠ c | α ≠ β ≠ γ ≠ 90° | CuSO₄·5H₂O, K₂Cr₂O₇ |
The seven crystal systems together give 14 Bravais lattices (the cubic system has 3: simple, body centred, face centred).
Cubic Unit Cells
| Property | Simple cubic | Body centred | Face centred |
|---|---|---|---|
| Atoms per cell, Z | 1 | 2 | 4 |
| Atoms touch along | edge | body diagonal | face diagonal |
| Radius r in terms of a | a/2 | √3 a/4 | a/(2√2) |
| Coordination number | 6 | 8 | 12 |
| Packing efficiency | 52.4% | 68% | 74% |
Packing efficiency is the fraction of the cell volume occupied by atoms: \( \frac{Z\times\frac43\pi r^3}{a^3} \). Hexagonal close packing and cubic close packing both fill 74% of space; the rest (26%) is void.
Density of a crystal: \( d=\frac{Z\,M}{a^3N_A} \), with a in cm and M in g mol⁻¹ giving d in g cm⁻³. This relation is also used to find Z, M or N_A when the other quantities are known.
Worked Example 1. Find the density of copper (fcc, r = 128 pm, M = 63.5 g mol⁻¹).
Answer:
a = 2√2 r = 2 × 1.414 × 128 = 362 pm = 3.62 × 10⁻⁸ cm.
Z = 4, so \( d=\frac{4\times63.5}{6.022\times10^{23}\times(3.62\times10^{-8})^3}=8.9\ \text{g cm}^{-3} \).
Close Packing and Voids
In a close packed layer each sphere touches six others. A second layer sits in half of the depressions of the first. The third layer can lie directly above the first (ABAB pattern, hexagonal close packing, hcp) or over the other set of depressions (ABCABC pattern, cubic close packing, ccp, same as fcc). Both have coordination number 12 and packing efficiency 74%.
Close packing leaves two kinds of voids. For N spheres there are N octahedral voids (surrounded by 6 spheres) and 2N tetrahedral voids (surrounded by 4 spheres). In an fcc cell there are 4 octahedral voids (12 edge centres × 1/4 + 1 body centre) and 8 tetrahedral voids (all inside the cell).
A void of radius r⁺ can hold a cation only if the radius ratio r⁺/r⁻ is large enough for the ions to touch: coordination number 3 for 0.155–0.225, 4 for 0.225–0.414, 6 for 0.414–0.732 and 8 for 0.732–1.0. The sphere that fits in an octahedral hole has radius 0.414 r and in a tetrahedral hole 0.225 r.
Structures of Ionic Compounds
| Structure | Arrangement | Coordination number | Formula units per cell |
|---|---|---|---|
| Rock salt (NaCl) | Cl⁻ in fcc, Na⁺ in all octahedral voids | 6 : 6 | 4 |
| Zinc blende (ZnS) | S²⁻ in fcc, Zn²⁺ in half (alternate) tetrahedral voids | 4 : 4 | 4 |
| Caesium chloride (CsCl) | Cl⁻ at corners, Cs⁺ at body centre | 8 : 8 | 1 |
| Fluorite (CaF₂) | Ca²⁺ in fcc, F⁻ in all tetrahedral voids | 8 : 4 | 4 |
| Antifluorite (Na₂O) | O²⁻ in fcc, Na⁺ in all tetrahedral voids | 4 : 8 | 4 |
In NaCl-type crystals the ions touch along the edge, so a = 2(r⁺ + r⁻). In CsCl-type crystals they touch along the body diagonal, so √3 a = 2(r⁺ + r⁻).
Imperfections in Solids
Point defects in ionic crystals are of two kinds. Stoichiometric defects keep the cation : anion ratio unchanged:
1. Schottky defect: equal numbers of cations and anions are missing. It lowers the density and appears in ionic solids with high coordination number and ions of similar size (NaCl, KCl, CsCl).
2. Frenkel defect: a (smaller) cation moves to an interstitial site. The density stays the same and the dielectric constant rises. It occurs when the cation is much smaller than the anion (AgCl, AgBr, ZnS). AgBr shows both defects.
Non-stoichiometric defects change the ratio. Metal excess defects arise from anion vacancies filled by trapped electrons (F-centres), which colour the crystal, or from extra cations in interstitial sites. Metal deficiency defects arise when a cation of variable valency is missing and a neighbouring ion has a higher charge (FeO is Fe₀.₉₅O).
Doping an ionic crystal with an aliovalent impurity creates vacancies: a Sr²⁺ ion replacing two Na⁺ ions in NaCl leaves one cation vacancy per Sr²⁺ added.
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