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Chemistry
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Solid State-I
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Session Objectives Solids Classification of solids
Structure determination by X-ray diffraction Lattices and Unit cells Packing in metallic crystals Voids Packing fraction Density
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Solid state Definite volume and shape.
Regular order of arrangement of particles. Strong intermolecular forces. Low kinetic energy.
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Solids are classified into two types
Solids can be crystalline or amorphous. A crystalline solid is composed of one or more crystals; each crystal has a well-defined, ordered structure in three dimensions. Examples include sodium chloride and sucrose. An amorphous solid has a disordered structure. It lacks the well-defined arrangement of basic units found in a crystal. Glass is an amorphous solid. 2
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Difference Between Amorphous Solids and Crystalline Solids
S.No. Amorphous solids Crystalline solids 1. Don't have definite geometrical shape. Characteristic geometrical shape. 2. melt over a wide range of temperature. They have sharp melting point 3. Physical properties are same in different direction, i.e. isotropic. Physical properties are different in different directions. This phenomenon is known as Anisotropy. 4. Amorphous solids are unsymmetrical. They are symmetrical 5. Don't break at fixed cleavage planes. Cleavage along particular direction at fixed cleavage planes.
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Types of crystalline solids
From this point of view, there are four types of solids. Molecular (Van der Waals forces) Metallic (Metallic bond) Ionic (Ionic bond) Covalent (Covalent bond)
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Molecular solids A molecular solid is a solid that consists of atoms or molecules held together by intermolecular forces. Many solids are of this type. Soft, low melting point, volatile, electrical insulators, poor thermal fusion. Examples include solid neon, solid water (ice), and solid carbon dioxide (dry ice). 2
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Metallic Solids A metallic solid is a solid that consists of positive cores of atoms held together by a surrounding “sea” of electrons (metallic bonding). In this kind of bonding, positively charged atomic cores are surrounded by delocalized electrons. Very soft to very hard, low to high melting point, good conductors of electricity and heat. Metallic lustre, malleable and ductile, moderate heats of fusion Examples include iron, copper, and silver. 2
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Ionic Solids An ionic solid is a solid that consists of cations and anions held together by electrical attraction of opposite charges (ionic bond). Brittle, high melting point, good conductors in the aqueous solution or fused state, high heats of fusion. Salts like NaCl, KNO3, LiF, BaSO4 etc. 2
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Covalent Solids Very hard, high melting point, poor conductors of heat
A covalent network solid is a solid that consists of atoms held together in large networks or chains by covalent bonds. Very hard, high melting point, poor conductors of heat and electricity high heats of fusion. Examples include carbon, in its forms as diamond or graphite, asbestos, and silicon carbide. 2
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Comparison of different type of solids
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Structure Determination
X-ray Diffraction The Bragg equation nλ = 2dsin θ l is the wavelength of the X-rays; n is an integer; d is the distance between crystal planes; θ is the angle of incidence.
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Illustrative Problem The planes of a crystal are nm apart and angle 2q = then for first order diffraction, what is the wavelength of X-ray used? Solution : 2 q = 11.8o or q = 5.9o and n = 1 \ 1. = 2 x Sin 5.9 = pm
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Lattice or space lattice or crystal lattice
Space lattice is an arrangement of points showing how constituent particles of a crystal are arranged at different positions in three dimensional space.
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Crystal Lattices A crystal lattice is the geometric arrangement of lattice points in a crystal. A unit cell is the smallest boxlike unit from which you can construct a crystal by stacking the units in three dimensions. There are seven basic shapes possible for unit cells, which give rise to seven crystal systems used to classify crystals. 2
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Unit-cell shapes of the different crystal systems
Unit Cell = simplest repeating unit of the regular crystalline array Bravais Lattices = 14 possible basic crystal structure unit cell types
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Cubic unit cell CN = 6 CN = 8 CN = 12 Axial length a=b=c Axial angle
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Number of Lattice Points in Cubic Crystal Systems
Atoms per unit cell = 1 Atoms per unit cell = 2 Atoms per unit cell = 4
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Relationship between Atomic Radius and Lattice Parameters
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Illustrative Problem Solution:
A substance AxBy crystallizes in a face centered cubic (fcc) lattice in which atoms ‘A’ occupy each corner of the cube and atoms B occupy the centers of each face of the cube. Identify the correct composition of the substance AxBy. Solution: No. of ‘A’ atoms = No. of ‘B’ atoms = Hence formula is AB3
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Illustrative example Three elements P,Q and R crystallize in a cubic lattice with P atoms at the corners, Q atoms at the cube centre & R atoms at the centre of the faces of the cube then what would be the formula of the compound. Solution Atoms P per unit cell =8x1/8 =1 Atoms Q per unit cell=1 Atoms R per unit cell =6x1/2 =3 Hence the formula of compound is PQR3
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Illustrative Problem Sodium metal crystallites in bcc lattice with the cell edge 4.29 A” .what is the radius of sodium atom ? Solution :
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Closed packed structure
Tetrahedral site Octahedral site
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Closed packed structure
Tetrahedral site Octahedral site
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Closed packed structure
Two efficient ways to pack particles together in a crystal structure are hexagonal closest packed (A, hcp) and cubic closest packed (B, ccp). The difference is in the third layer. In hcp the third layer sits directly above the first layer (ababab…). In ccp the third layer is different from the other two (abcabcabc…). The NaCl crystal structure contains a cubic closest packed arrangement of chloride ions.
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Hexagonal close packed structure
Consists of atoms with ABABAB stacking. Each atom is surrounded by 12 closest neighbors. Packing efficiency is 74%.
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Cubic close packed structure
Consists of atoms with ABCABCABC stacking. Each atom is surrounded by 12 closest neighbours. Packing efficiency is 74%.
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Density Where ‘a’ is the edge length in pm for unit cell,
‘Z’ is the number of particle(atoms) in unit cell. ‘NA’ is Avogadro’s number(6.023x1023). ‘M’ molecular mass.
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Illustrative example Determine the density of BCC iron, which has a lattice parameter of nm. SOLUTION Atoms/cell = 2, a0 = nm = 10-8 cm Atomic mass = g/mol Volume of unit cell = = (2.866 10-8 cm)3 = cm3/cell Avogadro’s number NA = 6.02 1023 atoms/mol
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Illustrative example An element exists in bcc structure with a cell edge of 288 pm. Density of the element is 7.2 g cm–3 what is the atomic mass of the element? Solution: bcc structure means Z = 2 Volume of the cell = (288)3 × 10–30 Density = Or 7.2 = 2 × M / × 1023 × (288)3 × 10–30 M = 52 g mol–1
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Illustrative Example Aluminum has the face-centered cubic structure with a unit cell dimension of 4.041Å. What is density of aluminum? Solution: Density (ρ) = mass of contents of cell/volume of cell Mass of atoms in the cell = 4 x 1 x 26.98/6.022 x 1023 = x g Volume of unit cell = (4.041 x10-8)3 = x cm3 ρ = mass/volume = x g/6.599 x cm3 = g cm-3
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Illustrative example A metal of atomic mass = 75 form a cubic lattice of edge length and density 2 g cm-3. Calculate the radius of the atom. Given Avogadro’s number, NA = 6 x 1023. Solution: We know that It indicates that metal has bcc lattice. For bcc lattice, Or r = x 102 = pm
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