Sunday, May 27, 2018

Chemistry: Solid state

Classification of solids:

Crystalline and  Amorphous solids:
    S.No.
    Crystalline Solids
    Amorphous solids
    1
    Regular internal arrangement  of particles

    irregular internal arrangement  of particles

    2
    Sharp melting point

    Melt over a rage of temperature

    3
    Regarded as true solids

    Regarded as super cooled liquids or pseudo solids

    4
    Undergo regular cut  
    Undergo irregular cut.

    5
    Anisotropic in nature

    Isotropic in nature

    Based on binding forces:
    Crystal Classification 
    Unit Particles 
    Binding Forces 
    Properties 
    Examples 
    Atomic
    Atoms
    London dispersion forces
    Soft, very low melting, poor thermal and electrical conductors
    Noble gases
    Molecular
    Polar or  non – polar molecules
    Vander Waal’s forces (London dispersion, dipole – dipole forces hydrogen bonds)
    Fairly soft, low to moderately high melting points, poor thermal and electrical conductors
    Dry ice (solid, methane
    Ionic
    Positive and negative ions
    Ionic bonds
    Hard and brittle, high melting points, high heats of fusion, poor thermal and electrical conductors
    NaCl, ZnS
    Covalent
    Atoms that are connected in covalent bond network
    Covalent bonds
    Very hard, very high melting points, poor thermal and electrical conductors
    Diamond, quartz, silicon
    Metallic Solids
    Cations in electron cloud
    Metallic bonds
    Soft to very hard, low to very high melting points, excellent thermal and electrical conductors, malleable and ductile
    All metallic elements, for example, Cu, Fe, Zn

    Bragg Equation:                                          

    nλ = 2dsinθ,
    Where
    • d= distance between the planes
    • n = order of refraction
    • θ= angel of refraction
    • λ = wavelength
    Bragg equation

    Crystal Systems:

    • Total number of crystal systems: 7 
    • Total number of Bravais Lattices: 14
    Crystal Systems
    Bravais Lattices 
    Intercepts 
    Crystal angle 
    Example
    Cubic 
    Primitive, Face Centered, Body Centered 
    a = b = c
    a = b = g = 90o 
    Pb,Hg,Ag,Au Diamond, NaCl, ZnS 
    Orthorhombic
    Primitive, Face Centered, Body Centered, End Centered
    a ≠ b ≠ c
    a = b = g = 90o
    KNO2, K2SO4
    Tetragonal
    Primitive, Body Centered
    a = b ≠ c
    a = b = g = 90o
    TiO2,SnO2
    Monoclinic
    Primitive, End Centered
    a ≠ b ≠ c
    a = g = 90o, b≠ 90o
    CaSO4,2H2O
    Triclinic
    Primitive
    a ≠ b ≠ c
    a≠b≠g≠900
    K2Cr2O7, CaSO45H2O
    Hexagonal
    Primitive
    a = b ≠ c
    a = b = 900, g = 120o
    Mg, SiO2, Zn, Cd
    Rhombohedra
    Primitive
    a = b = c
    a = g = 90o, b≠ 90o
    As, Sb, Bi, CaCO3 

    Number of atoms in unit cells.

    Primitive cubic unit cell:Primitive cubic unit cell
    • Number of atoms at corners = 8×1/8 =1
    • Number of atoms in faces = 0
    • Number of atoms at body-centre: =0
    • Total number of atoms = 1
     Body-centred cubic unit cell:
    • Body-centred cubic unit cellNumber of atoms at corners = 8×1/8 =1
    • Number of atoms in faces = 0
    • Number of atoms at body-centre: =1
    • Total number of atoms = 2
    Face-centred cubic or cubic-close packed unit cell: 
    • Face-centred cubic or cubic-close packed unit cellNumber of atoms at corners = 8×1/8 =1
    • Number of atoms in faces = 6×1/2 = 3
    • Number of atoms at body-centre: = 0
    • Total number of atoms = 4

    Packing Efficiency

    Packing Efficiency = (Volume occupied by all the atoms present in unit cell / Total volume of unit cell)×100
    Close structure
    Number of atoms per unit cell ‘z’.
    Relation between edge length ‘a’ and radius of atom ‘r
    Packing Efficiency
    hcp and ccp or fcc
    4
    r = a/(2√2)
    74%
    bcc
    2
    r = (√3/4)a
    68%
    Simple cubic lattice
    1
    r = a/2
    52.4%

    Density of crystal lattice:

    r = (Number of atoms per unit cell × Mass number)/(Volume of unit cell × NA)
    r = \frac{Number\ of\ unit\ cells\times Mass\ number}{Volume\ of\ Unit\ Cells \times N_A}
    or
    r = \frac{z\times M}{V\times N_A}

    Octahedral and Tetrahedral Voids:                             

    Number of octahedral voids = Number of effective atoms present in unit cell
    Number of tetrahedral voids = 2×Number of effective atoms present in unit cell
    So, Number of tetrahedral voids = 2× Number of octahedral voids.

    Coordination numbers and radius ratio:

    Coordination numbers
    Geometry
    Radius ratio (x)
    Example
    2
    Linear
    x < 0.155
    BeF2
    3
    Planar Triangle
    0.155 ≤ x < 0.225
    AlCl3
    4
    Tetrahedron
    0.225 ≤ x < 0.414
    ZnS
    4
    Square planar
    0.414 ≤ x < 0.732
    PtCl42-
    6
    Octahedron
    0.414 ≤ x < 0.732
    NaCl
    8
    Body centered cubic
    0.732 ≤ x < 0.999
    CsCl

    Classification of Ionic Structures: 

    Structures
    Descriptions
    Examples
    Rock Salt Structure
    Anion(Cl-) forms fcc units and cation(Na+) occupy octahedral voids. Z=4 Coordination number =6
    NaCl, KCl, LiCl, RbCl
    Zinc Blende Structure
    Anion (S2-) forms fcc units and cation (Zn2+) occupy alternate tetrahedral voids Z=4 Coordination number =4
    ZnS , BeO
    Fluorite Structures
    Cation (Ca2+) forms fcc units and anions (F-) occupy tetrahedral voids Z= 4 Coordination number of anion = 4 Coordination number of cation = 8
    CaF2, UO2, and ThO2
    Anti- Fluorite Structures
    Oxide ions are face centered and metal ions occupy  all the tetrahedral voids.
    Na2O, K2O and Rb2O.
    Cesium Halide Structure
    Halide  ions are primitive cubic while the metal ion occupies the center of the unit cell. Z=2 Coordination number of = 8
    All Halides of Cesium.
    Pervoskite Structure
     One of the cation is bivalent and the other is tetravalent. The bivalent ions are present in primitive cubic lattice with oxide ions on the centers of all the six square faces. The tetravalent cation is in the center of the unit cell occupying octahedral void.
    CaTiO3, BaTiO3
    Spinel and Inverse Spinel Structure
    Spinel :M2+M23+O4, where M2+ is present in one-eighth of tetrahedral voids in a FCC lattice of oxide ions and M3+ ions are present in half of the octahedral voids. M2+ is usually Mg, Fe, Co, Ni, Zn and Mn;  M3+is generally Al, Fe, Mn, Cr and Rh.
    MgAl2O4 ,ZnAl2O4, Fe3O4,FeCr2O4etc.

    Defects in crystal:

    Stoichiometric Defects
    1. Schottky Defects
    •  Some of the lattice points in a crystal are unoccupied.
    • Appears in ionic compounds in which anions and cations are of nearly same size.
    • Decreases the density of lattice
    • Examples: NaCl and KCl
    2.  Frenkel Defects
    • Ion dislocate from its position and occupies an interstitial position between the lattice points
    • Appears in crystals in which the negative ions are much larger than the positive ion.
    • Does not affect density of the crystal.
    • Examples: AgBr, ZnS
    Non-Stoichiometric Defects
    1. Metal Excess defect: 
    Metal excess defect occurs due to
    • anionic vacancies or
    • presence of extra cation.
    • F-Centres: hole produced due to absence of anion which is occupied by an electron.
    2. Metal deficiency defect: 
    Metal deficiency defect occurs
    • due to variable valency of metals
    • when one of the positive ions is missing from its lattice site and the extra negative charge is balanced by some nearby metal ion acquiring two charges instead of one

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