JEE Chemistry The Solid State Notes Set 05

Classification of Solids

Crystalline solids have their particles arranged in a regular, repeating pattern over long distances. They have sharp melting points, are anisotropic (properties depend on direction) and are true solids.
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.
A substance existing in more than one crystalline form is polymorphic; for elements this is called allotropy (diamond and graphite).

Types of Crystalline Solids

Crystal Systems and Unit Cells

A unit cell is the smallest repeating block of a lattice. There are 7 crystal systems (cubic, tetragonal, orthorhombic, monoclinic, hexagonal, rhombohedral, triclinic) and 14 Bravais lattices.
Cubic cells: primitive (simple), body-centred (bcc) and face-centred (fcc). Particles per unit cell: Z = 1 (simple), 2 (bcc), 4 (fcc).
Contribution of a particle: corner 1/8, face centre 1/2, edge centre 1/4, body centre 1.

Packing in Solids

Close packing of spheres in a layer gives each sphere 6 neighbours. Stacking layers as ABAB... gives hexagonal close packing (hcp); ABCABC... gives cubic close packing (ccp = fcc). Both have coordination number 12 and 74% of space filled.
Packing efficiency: simple cubic 52.4% (CN 6), bcc 68% (CN 8), fcc/hcp 74% (CN 12).
Voids: for N spheres there are N octahedral voids (CN 6) and 2N tetrahedral voids (CN 4).

Relations Between Edge Length and Radius

\( \text{Simple cubic: } a = 2r \)
\( \text{bcc: } \sqrt{3}\,a = 4r \)
\( \text{fcc: } \sqrt{2}\,a = 4r \)
Density of the unit cell: \( \rho = \dfrac{Z\,M}{a^{3}\,N_A} \)
Bragg's law for X-ray diffraction: \( n\lambda = 2d\sin\theta \)

Ionic Structures and Radius Ratio

NaCl (rock salt): Cl- in fcc, Na+ in all octahedral holes, CN 6 : 6, 4 formula units per cell, edge = 2(r+ + r-).
CsCl: simple cubic arrangement of Cl- with Cs+ in the body centre, CN 8 : 8, 1 formula unit per cell, r+ + r- = (\( \sqrt{3} \)/2)a.
ZnS (zinc blende): S2- in fcc, Zn2+ in half of the tetrahedral holes, CN 4 : 4, 4 formula units per cell.
CaF2 (fluorite): Ca2+ in fcc, F- in all tetrahedral holes, CN 8 : 4. Na2O has the reverse arrangement (antifluorite), CN 4 : 8.

Imperfections in Solids

Stoichiometric defects: in a Schottky defect equal numbers of cations and anions are missing and the density falls (NaCl, CsCl). In a Frenkel defect a cation moves into an interstitial site and the density is unchanged (AgBr, ZnS).
Non-stoichiometric defects: metal excess (anion vacancies holding electrons, called F-centres, which colour the crystal) and metal deficiency (as in FeO).
Impurity defects: doping NaCl with SrCl2 creates one cation vacancy for every Sr2+ ion added.

Electrical and Magnetic Properties

Conductors have overlapping bands, insulators have a large gap between valence and conduction bands, and semiconductors have a small gap. Doping silicon with a group 15 element gives an n-type semiconductor; doping with a group 13 element gives a p-type semiconductor.
Magnetic behaviour: diamagnetic (all electrons paired), paramagnetic (unpaired electrons), ferromagnetic (domains aligned, e.g. Fe, CrO2), antiferromagnetic (opposed alignment, e.g. MnO), ferrimagnetic (unequal opposed alignment, e.g. Fe3O4).

Worked Examples

Example 1. Copper crystallises in an fcc lattice with edge length 361 pm. Find the radius of a copper atom.
Answer:
\( r = \dfrac{a}{2\sqrt{2}} = \dfrac{361}{2.828} \approx 127.6 \) pm

Example 2. Find the density of a metal (M = 27 g/mol) with fcc structure and a = 405 pm.
Answer:
\( \rho = \dfrac{4\times27}{(4.05\times10^{-8})^3\times6.022\times10^{23}} \approx 2.7 \) g/cm3

Example 3. Find the formula of a compound in which A atoms form ccp and B atoms occupy all tetrahedral holes.
Answer:
ccp gives 4 A atoms and 8 tetrahedral holes, so A : B = 4 : 8 = 1 : 2. Formula = AB2.