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III Sem B.Sc., , UNIT - 3, PHYSICAL CHEMISTRY, , , Second law of thermodynamics, It is a branch of science which deals with the quantitative relationship between heat and other forms of, energy.,  First law of thermodynamics: (Law of conservation of energy), Statement:, “Energy can neither be created nor be destroyed but can be converted from one form to another.”, According to this law the total energy of the universe remains constant.,  Limitations of first law of thermodynamics - Need for the second law of thermodynamics:, First law of thermodynamics suffers from the following draw backs., 1. It does not explain why chemical reaction do not proved to completion and why spontaneous or natural, processes are unidirectional., 2. It does not define the extent of convertibility from one form of energy into another., 3. It does not explain why heat cannot be completely converted into work without leaving some change in, the system or surroundings., 4. It does not explain the direction of natural processes., 5. According to the law, heat energy of 100% efficiency is attainable but our experience says that such, engines are not available in practice,  Spontaneous Process, A process which takes place on its own accord without any external assistance is called spontaneous, Process., Or, A process which can takes place by itself under a given set of conditions once it has been properly initiated, if necessary is said to be a spontaneous process.,  Non-Spontaneous Process, The reverse process which does not proceed on its own is referred to as non-spontaneous process., E.g:, 1., 2., 3., 4., , Flow of heat from hot body to a cold body., Expansion of gas from the region of higher pressure to lower pressure., Flow of electricity from a higher potential point to a lower potential point., Dissolution of ‘Zn’ in dil H2SO4., , Once of system is in equilibrium state, it does not undergo any further spontaneous change.,  Statement of second law of thermodynamics, 1. All spontaneous process are thermodynamically irreversible., 2. The complete conversion of heat into work is impossible without changing the state of any other body., 3. Total energy of the universe remains constant while the total entropy of the universe increases.,  Concept of entropy, Entropy is a thermodynamic state quantity that is a measure of the randomness or disorder of the, molecules of the system., The energy which can be converted into useful work is called available energy, whereas the remaining, portion of the energy which cannot be converted into useful work is called unavailable energy. Entropy is, regarded as measure of unavailable energy., Department of Chemistry, , Page 1

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III Sem B.Sc.,  II law of thermodynamics in terms of entropy:, In a reversible process the entropy of the system surroundings taken together remains constant while in, irreversible process it increases.,  Free energy, Free energy is the maximum amount of energy available to a system during a process that can be converted, into useful work.,  Helmholtz free energy:, Is defined as A= E-TS, ‘E’ is internal energy ‘S’ is the entropy ‘T’ is temperature.,  Gibb’s free energy:, It defined as G= H-TS, ‘H’ is the enthalpy of the system ‘S’ is the entropy.,  Relation between ‘A’ and ‘G’, Since, H = E+PV, G = H -TS, G = E+PV-TS, G = A+PV, E-TS= A, The change in free energy at constant temperature between two states of a system will be, G2- G1, (H2-TS2) – (H1-TS1), (H2-H1) – (TS2-TS1), Where H and S are enthalpy and entropy changes of the system, If G is ‘-ve’ reaction is spontaneous., If G is ‘+ve’ reaction is non-spontaneous., If G is zero reaction is at equilibrium.,  Gibb’s Helmholtz equation at constant pressure, G= H-TS, H= E+PV, G= E+PV-TS, On differentiating we get, dG= dE+Pdv+Vdp-TdS-SdT, from first law of thermodynamics, dq= dE+PdV, dG= dq+Vdp-TdS-SdT, For a reversible process dS=, , or TdS= dq, , dG= TdS+Vdp-TdS-SdT, dG= VdP-SdT, At constant pressure dG= -SdT, Department of Chemistry, , Page 2

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III Sem B.Sc., = 2.303 8.314 300, Jmol-1,  Claussius Clayperon equation, Let us consider two phases A and B of the same component in equilibrium with each other at constant, temperature ‘T’ and pressure ‘P’., A, , P, , Let one mole of the pure substance be transformed from phase ‘A’ to phase ‘B’. the change in free energy is, given by,, = GB-GA, As the system is in equilibrium, GB-GA=0 or GB=GA →, , ①, , If the temperature of the system is raised from ‘T’ to ‘T+dt’ the pressure also has to change from ‘P’ to ‘P+dP’ in, order to maintain equilibrium., In such a situation,, GB+dGB= GA+dGA→, , ②, , Subtracting eq① from ②, dGB= dGA, We know dG= VdP-SdT, For Phase A, , dGA= VAdP-SAdT, , For Phase B, , dGB= VBdP- SBdT, , VBdP-SBdT= VAdP-SAdT, On rearranging,, (VB-VA) dP= (SB-SA) dT, , VA and VB are molar volumes of substances in phase ‘A’ and ‘B’. (VB-VA) represents change in volume when one, mole of substance changes from phase ‘A’ to phase ‘B’ ( ) (SB-SA) is change in entropy ( S) then,, =, If ‘Q’ is the heat exchanged reversibly per mol of a substance during the phase transformation at temperature, ‘T’, then change in entropy is,, S=, =, , or, , =, , This equation is called Clausius – clayperon equation, , Department of Chemistry, , Page 5

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III Sem B.Sc.,  Applications:, 1. For liquid, Vapour system, Q= molar heat of vaporization (ΔHv), VB= Volume of one mole vapour (Vg), VA= Volume of one mole of liquid (Vl), Then ,, , =, , 2. For solid, liquid system, Q= molar heat of fusion(ΔHf), VB= Volume of one mole of liquid (Vl), VA= Volume of one mole of solid(Vs), Then,, , =, ,  Integrated form of claussis clayperon equation, ,  Bond energy, The bond energy of a particular bond is defined as the average amount of energy released when one mole of, bonds are formed from isolated gaseous atoms or the amount of energy required when one mole of bonds are, broken so as to get the separated gaseous atoms., For diatomic molecules (like H2, HCl etc ) the bond energy equal to the dissociation energy of the molecules but, for a polyatomic molecule like CH4 the bond dissociation energies of the four C-H bond are different. Hence an, average value is taken., Enthalpy of reaction can be calculated from bond energy data., H= ∑, , -∑, ,  Problems:, 1. Calculate the bond energy of C-H bond, given the heat of combustion of methane (CH 4), graphite and, hydrogen are 891KJ, 394KJ and 286KJ respectively while the heat of sublimation of graphite is 717KJ, and the heat of dissociation of hydrogen molecules is 436KJ., CH4(g) + 2O2(g) →, CO2(g) + 2H2O(l), → ①, C(s) + O2(g) →, , CO2(g), , H2(g) + O2(g) →, C(s) →, , →, , H2O(g), , →, , C(g), , H2(g) →, , →, (g), , We want, , ②, , →, , for the following reaction, , [CH4(g) →, , C(s) + 4H(g)], , →, , Eq①+Eq④+2 Eq, CH4(g)+2O2(g)+C(s)+2H2(g) →, , CO2(g)+2H2O(g)+C(g)+4H(g), 2EQ+ H=-891+717+2 436=698KJ→, , 2Eq, , +Eq②, , 2H2(g)+O2(g)+C(S)+O2(g) →, , 2H2O(g)+CO2(g), H= 2(-286)+ (-394) = -966 →, , Department of Chemistry, , Page 6

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III Sem B.Sc., Eq⑦-Eq⑧, CH4(g) →, , C(g) +4H(g), , This is the energy required the dissociation of 4moles of C-H bonds., Bond dissociation energy for C-H bond=, = 416KJ,  Resonance energy, The extent of stabilization of a substance per mole as a consequence of resonance is called resonance, energy or resonance stabilization energy., E.g: In the case of benzene the molecule of benzene has three double bonds in the ring at alternate positions., The molecule shows exceptional stability., It is possible to calculate resonance energy from heat of hydrogenation., Heat of hydrogenation of cyclohexene, , = -29KCal/mol, , Which contains one double bond., Heat of hydrogenation of three double bond= 3 (-29), = -87Kcal/mol, In a six membered ring, Heat of hydrogenation of benzene = -50Kcal/mol, (experimental value), we say that resonance energy of benzene= -50-(-87), = 37Kcal/mol,  Qualitative treatment of Nernst heat theorem, This theorem gives the variation of enthalpy change (, decreases of temperature., , ) and free energy change (, , ) of a system with, , According to Gibb’s-Helmholtz equation, (, , )P, , From this equation at the absolute zero i.e T=0, , ., , Nernst observed that as temperature is lowered towards absolute zero,, the value of ( )P decreases and then approaches zero, asymptotically., This means that, and, are not only equal at the absolute zero but, the values approach each other asymptotically in the vicinity of this, temperature., Mathematically it may be expressed as, =, , =0, , Graphically the result maybe represented as, The value of, approaches zero gradually as the temperature is lowered towards absolute zero. This is, known as Nernst heat theorem., , Department of Chemistry, , Page 7

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III Sem B.Sc.,  Third law of thermodynamics, Statement: The entropy of all perfectly crystalline solids may be taken as zero at the absolute zero, temperature., The law also stated as follows., “Every substance has finite positive entropy but at the absolute zero of temperature the entropy may, become zero and infact it does become zero in case of perfectly crystalline solids.”, , Department of Chemistry, , Page 8

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III Sem B.Sc.,  Crystallography, “Solids are characterized by incompressibility, rigidity and mechanical strength.”, This indicates that the molecules, atoms or ions that make up a solid are closely packed, they are held together, by strong force and cannot move at random., Thus, in solids we have well-ordered molecular arrangement., Solids have been classified into crystalline and amorphous solids., , , Distinctive properties of crystalline and amorphous solids;, Crystalline, Crystalline solids have a definite and regular, geometry due to definite and orderly, arrangements of their ultimate particles,, namely, molecules atoms or ions, this is, shown by x-ray diffraction studies., Have a sharp melting point., , Amorphous, Amorphous solids have no pattern of, arrangement of molecules or atoms and, therefore do not have any definite, geometrical shape., , Have no sharp melting point, melts over a, range of temperature., Anisotropic: That is their physical properties Isotropic: That is their physical properties, like thermal conductivity, electrical, are the same in all directions., conductivity etc are different directions., Crystalline solids are considered as really, Amorphous solids are considered as super, true solids., cooled liquids and not as solids., E.g: sodium chloride, sugar, Sulphur etc., E.g: glass, plastic, rubber etc.,  Elements of symmetry, Symmetry is one of the foundation properties of Crystals., There are three types of symmetrics association with a crystal., i., , Plane of symmetry, , A crystal is said to possess a plane of symmetry, when an imaginary plane (passing through the centre of, a crystal) can divide a crystal into two parts such that one is the exact mirror image of the other, ii., , Axis of symmetry, , It is a line about which the crystal may be rotated such that it present the same appearance more than, once during the complete revolution. If the equivalent configuration occurs twice, thrice, four and six times, i.e:, after rotation of 180 , 120 ,90 and 60 , the axes of rotation are shown as two fold (diad), three fold (friad),, four fold (tetrad) and six fold (hexad), axes of symmetry respectively., In general if the same appearance of a crystal is repeated on rotating it through an angle of, imaginary axis, the axis is known as n-fold axis., iii., , around an, , Centre of symmetry, , It is a point in a crystal such that any line passing through the point will meet the surface of the crystal at, equal distance in both directions., A crystal may have any number of planes or axes of symmetry but it has only it has only one center of, symmetry., The total number of planes, axes and centre of symmetries possessed by a crystal is termed as elements of, symmetry of the crystal. To explain this we may consider the elements of symmetry possessed by a cubic, crystal., , , , Total number of elements of symmetry in cubic crystal, , Department of Chemistry, , Page 9

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III Sem B.Sc., A cubic crystal possesses a total of 23 elements of symmetry. These elements of symmetry are:, a. Rectangular planes of symmetry, There are three rectangular planes of symmetry in all., b. Diagonal planes of symmetry, There can be a total of six planes passing diagonally through the cube., c. Axes of four symmetry, There can be a total of three four fold axes at right angle to one another., d. Axes of three fold symmetry, There can be a total of four three-fold axes., e. Axes of two fold symmetry, There are six axes of two-fold symmetry, f. Centre of symmetry, There is only one centre of symmetry. Thus the numbers of symmetry elements of various types in a, crystal are, Planes of symmetry= 3+6=9 elements, Axes of symmetry= 3+4+6=13 elements, Centre of symmetry=1, Total number of symmetry elements= 23, , , Space lattice or crystal lattice and unit cell, , “The position of atoms, molecules, or ions in a crystal, relative to one another in space, are designated, usually by points. Such a representation is called Space lattice.”, A space lattice is an array of points showing how molecules, atoms or ions are arranged at different, sites in three dimensional space., A unit cell is the smallest repeating unit in space lattice which when repeated over and over again results in, a crystal of the given substance., , , , , The unit cell is the smallest sample that represents the picture of the entire crystal., The crystal may be considered to consists of infinite number of unit cells., The geometry or shape of the unit cell is thus same as the geometry of the crystal itself., ,  Bravais lattices and crystal systems, From geometrical considerations it was shown by A. Bravais in 1848 that all possible three-dimensional, space lattices are of fourteen distinct types. These are known as Bravais lattices and are derived from seven, Crystal systems., All crystalline solids can be represented by one of there lattice structures., The seven crystal system differ in the length of the unit cell edges (a, b, and c) and the angles between the, unit cell edges. On the basis of the dimensions of a unit cell along its three axes (a, b, and c) and the sizes of the, angles between the axes (, the crystal systems are differentiated in the table given below.,  CRYSTAL SYSTEM, Crystal System, , Cubic, , Tetragonal, Department of Chemistry, , Possible, variations, Primitive,, Body- centred,, Face- centred, Primitive,, Body- centred, , Axial distances or, edge lengths, , Axial angels, , Examples, , a=b=c, , α = β = γ = 900, , NaCl, Zinc blende, Cu, , a=b≠c, , α = β = γ = 900, , White tin, SnO2, TiO2,, CaSO4, Page 10

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III Sem B.Sc., , Orthorhombic, , Primitive,, Body- centred,, Face- centred,, End- centred, , a≠b≠c, , α = β = γ = 900, , Rhombic, sulphur,, KNO3, BaSO4, , Hexagonal, , Primitive, , a=b≠c, , α = β = 900, γ = 1200, , Graphite, ZnO, CdS, , Primitive, , a=b=c, , α = β = γ ≠ 900, , Monoclinic, , Primitive,, End-centred, , a≠b≠c, , α = γ = 900, β ≠ 900, , Triclinic, , Primitive, , a≠b≠c, , α ≠ β ≠ γ ≠ 900, , Rhombohedral, Trigonal, , or, , Calcite (CaCO3), HgS, (cinnabar), Monoclinic sulphur,, Na2SO4.10H2O, K2Cr2O7,, CuSO4.5H2O. H3BO3, , (a) Body Centered Cubic unit cell (BCC): In this arrangement, the unit cell has nine lattice points. In this eight, lattice points are present at the eight corners of the cube and one lattice points is at the center of the cube., E.g.: Sodium metal, Caesium chloride., (b) Face Centered Unit cell (FCC): In this arrangement, the unit cell has 14 lattice points. In this, 8 lattice points, are present at the eight corners of the cube and 6 lattice points are all the center of the six faces., E.g.: Copper metal, Sodium chloride., (c) End Centered Unit cell (ECC): This unit cell has 10 constituent particles, 8 at the corners and 2 at the center, of any 2 opposite faces., , SCC, , BCC, , FCC, ,  Miller indices, The law of rational indices states that the intercepts of any face of a crystal along three crystallographic, axes are either equal to the unit intercepts (a, b, c) or some simple whole number multiples of them i.e: ma, nb,, Pc etc where m, n ,p etc either integral whole numbers including infinity or fractions of whole numbers., Let ABC be a unit plane having unit intercepts a, b and c along, the crystallographic axes i.e along ox, oy and oz respectively. The, intercept of any other plane such as PQR along the axes will be the, integral multiples of a, b, c i.e: ma, nb, Pc. The co-efficient of a, b and c, (i.e: m, n and P) are known as Weiss indices of a plane. In the study of, crystals, it is more common, to use the Miller indices which are obtained, by taking the reciprocals of Weiss co-efficient and multiplying, throughout by the smallest number that will express all the reciprocals, as integers., For E.g: A plane intersecting the x-axis at unit distance and, parallel to y and z-axis have the intercepts equal to a, and hence, the to Miller indices are, i.e: 1,0,0 usually written as (1 0 0 ), , Department of Chemistry, , Page 11

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III Sem B.Sc., Thus the Miller indices of a face of a crystal are inversely proportional to the intercepts of the face along the, three crystallographic axes., , , Problem:, , Calculate the Miller indices of crystal planes which cut through the crystal axes at, i., , (2a, 3b, c), , ii., , (a,, , b, c), , iii., , (a, b, c), , iv., v., vi., , (a, b, c), (6a, 3b, 3c), (2a, -3b, -3c), , i., , 2a, 3b, c, reciprocals of intercepts, clearing the fraction multiplying by 6., Hence the Miller indices are (3, 2, 6), , ii., , (a,, , b, c), , Miller indices (1,0,2), iii., , (a, b, c), , iv., , Multiplying by 2 (2,2,3), (a, b, c ), (1, 1, 1), (6a, 3b, 3c), , v., , (1, 2, 2), (2a, -3b, -3c), , vi., , Multiplying by 6, (3,, ), In general the Miller Indices of a plane can be expressed as (h k l) where h, k and l refer to the, reciprocals of the intercepts expressed in units the lattice distance., The Miller indices (h k l) of any plane gives the orientation of the plane in the crystal with reference to, its three axes., The distance between the parallel planes in a crystal are designated as dhkl and is known as the, interplanar spacing. For different cubic lattices these interplanar spacing are given by the general formula, dhkl =, , √, , where ‘a’ is the length of the cube, while h, k and l are the Miller indices of the plane., , Department of Chemistry, , Page 12

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III Sem B.Sc., , , Lattice planes in the types of cubic crystal and their relative distance, , In a cube only 3 planes, namely (1 0 0) (1 1 0) (1 1 1) are important because the contain maximum number of, atoms. In a cube, the relative distances between the different planes can be got by using the above formula., If the side of a cube is ‘a’ then the distance between the (1 0 0) (1 1 0) and (1 1 1) planes is found to be a,, √, , √, , Hence the ratio, d100: d110: d111= a:, , √, , √, , = 1:, , √, , √, , in the face centered cubic lattice the important planes are (2 0 0) (2 2 0) and (1 1 1), d200: d220: d111= :, , √, , √, , = :, , √, , √, , in the body centered cubic lattice, the important planes to be considered are (2 0 0) (1 1 0) (2 2 2), d200: d110: d222= :, , √, , , , √, , = :, , √, , √, , Problems, , Determine the interplanar spacing between the (2 2 1) planes of a cubic lattice of length 4.5 A ., In the cube lattice the inter planar spacing d221 is given by, d221=, , , √, , =, , √, , = 1.5 A, , Diffraction of x-rays by crystals, , It was suggested by Lave in 1913 that it might be possible to diffract x-rays by means of crystals. The reason for, this suggestion was that the wave length of x-ray was about the same order (10-8cm) as the inter planar, distances in crystals. This suggestion came out to be the true when Bragg succeeded in diffracting x-ray from, sodium chloride crystal. This observation has proved to be highly useful in determining structured and, dimensions of crystals as well as in the study of a number of properties of x-ray., , , Bragg’s equation, , Bragg point out that scattering of x-ray by crystals could be considered as reflection from successive planes of, atoms in the crystal. However, Unlike reflection of ordinary light, that reflection of x-ray can take place only at, certain angles which are determined by the wave length of x-ray and the distance between the planes in the, crystal. The fundamental equation which gives a simple relation between the wave length of the x-rays, the, inter planar distance in the crystal and the angle of reflection is known as Bragg’s equation. This can be easily, derived as follows., , Consider a set of parallel planes AA, BB etc in the crystal as shown in the figure. These planes characterize the, arrangement of the atoms or ions in the crystal. A parallel beam of monochromatic x-rays of the wave length, Department of Chemistry, Page 13

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III Sem B.Sc., strikes these planes at an angle of incidence . Some of the rays will be reflected by atoms from the upper layer, AA, with an angle of reflection being equal to the angle of incidence. Some of the rays will be absorbed and, some will be reflected from the second layer BB and so on with successive layers. When the rays reflected from, the successive layer are phase, the constructive interference will occur and a bright diffraction spot would be, obtained from these planes. The condition for the constructive interference is that the path difference between, the reflected rays from successive planes must be an integral multiple of wavelength., The wave PQ reflected from the plane BB travel a greater distance than wave LM reflected from the plane AA,, draw MS and MT perpendiculars to PQ and QR. MS and MT make an angle with MQ whose length is equal to, interplanar spacing ‘d’. further,, SQ=QT=d sin, The additional path (or path difference) travelled by the wave reflected from the second is equal to SQ+QT= 2d, sin ., The condition for the constructive interference between the two waves is, SQ+QT=n, Or, 2d sin = n, Where ‘n’ is a integer taking values 1,2,3….. and is known as the order of reflection. is the Wavelength of the xray., The above equation is known as Bragg’s equation., For a fixed value of and d we can have more than one reflection angle 1, 2….. corresponding to n=1,2,…. but, higher order reflections are less intense by using monochromatic x-ray beam and is varied by rotating the, crystal about its axis., , , Problems, , Calculate the distance between two axes planes which gives first order diffraction at on angle of 26.4 with, molybdenum x-ray of wave length 0.710A ., n = 2d sin, here n=1, , 0.710A, , =26.4, , 1 0.710 = 2dsin26.4, = 2d 0.4446, d = 0.7962A, , , Problems, , What will be the wavelength of x-ray which gives a diffraction angle 2 equal to 16.8 for a crystal, if the, interplanar distance in the crystal is 0.400nm and only second order diffraction is observes?, n = 2d sin, here n=2 d=0.400nm =0.4, 2 = 16.8, , or =8.4, , =, Department of Chemistry, , Page 14

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III Sem B.Sc., = 0.4 10-9 sin8.4, =0.4 10-9 0.1461, =0.05845 10-9m, =0.584A, , , Problem, , X-rays of wave length equal to 0.154nm given a first order diffraction from the surface of a crystal when, the value of is 10.5 . calculate the distance between the planes in the crystal parallel to the surface, examined., [Workout the problem], , , Problem, , The interplanar distance in a crystal used for x-ray diffraction is 2A . The angle of incidence ( ) of x-ray is, 9.0 when the first order diffraction is observed. Calculate the wave length of x-ray, [Workout the problem], , , Problem, , Calculate the angle at which (a) first order reflection and (b) second order reflection will occur in an x-ray, spectrometer when x-ray of wavelength 1.54A are diffracted by the atoms of a crystal, given that planar, distance is 4.04A ., a. For first order reflection n=1. Bragg’s equation is, = 2d sin, = sin-1, = sin-1, = sin-1 (0.191), = 10 591, b. For second order reflection n=2, 2 = 2d sin, = sin-1, = sin-1, = sin-1 (0.381), = 22 241,  Determination of crystal structure of rock salt by Bragg’s equation, , x-ray from a target T are confined into a beam by allowing them to pass through lead slits S 1 and S2 and then, made to fall on a crystal c mounted on the spectrometer tube. The crystal is capable of rotation about the, Department of Chemistry, Page 15

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III Sem B.Sc., vertical axis and angle of rotation can be read on the main scale by the Vernier V 1. The beam after reflection, from the crystal is again confined by the slit S3 and is made to enter the ionization chamber A through the, window W. One of the plates of the chamber is charged to a high potential and the other is earthed. On entering, A, the x-ray ionize the gas and intensity of ionization is measured by the electrometer E. The chamber is rotated, co-axially with crystal table. The crystal table and the chamber are so adjusted that when the crystal rotates, through any angle. The chamber rotates at twice the angle so that angle, so that the reflected ray enter the, chamber., The value of incident angle is gradually increased by rotating the crystal. The intensities of the reflected xrays for various angles are determined. Strong reflections are obtained from those planes which contain large, number of atoms and for those values of which satisfy the Bragg’s equation., The process is repeated for each plane of the crystal. The lowest angle at which the maximum reflection occurs, corresponds to n=1. This called first order reflection. The next higher angle at which the maximum reflection, occurs again corresponds n=2. This is second reflection and so on. Generally, the angle for the first order, reflection is taken as so that the value of n is set as 1., The value of for the first order reflection from the three faces of sodium chloride crystal were found to be, 5.9 , 8.4 and 5.2 respectively. Applying Bragg’s equation and knowing that n and are the same in each case., The distance d between successive planes in the three faces will be in the ratio of, , = 9.61: 6.84: 1.04= 1:0.70:1.14, This ratio is very close to that expected to exist between spacings along the three planes of a face centered cube., We conclude therefore that sodium chloride crystal has face centered cubic structure., , , Problem, , A certain solid crystallizes in the body centered cubic lattice. First order x-ray ( =0.154nm) reflection, maximum from a set of (2 0 0) planes was observed at 16 61. Calculate the edge length of the unit cell., For a first order reflection n=1. Therefore, = 2d200sin, d200=, , =, , = 0.28nm, = 280pm, For a cubic lattice, d200 =, , √, , = 0.280pm, a= 2 280, = 560pm,  Determination of Avogadro number, From a knowledge of interplanar distances determined by x-ray analysis of crystal structure, Avogadro number, can be determined, Suppose, we consider a cube lattice of edge=a. then its volume is a3 suppose it contains Z particles in the cube,, Department of Chemistry, , Page 16

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III Sem B.Sc., Then,, Volume of Z particles is a3, or volume of one particle is, Let M be the molecular weight,, , be the density and v be the volume occupied by 1 mole., , Then =, Or V=, Volume, , corresponds to 1 particle, , Volume V corresponds V, =, , particles, particles, , N=, We have n = 2dsin for first order reflection = 2dsin, d=, using x-ray of known wavelength and finding, distance can be determined. Then using, d=, , at which maximum reflection occurs first, d the interplanar, , √, , ‘a’ can be determined,  Liquid crystal, Certain organic compounds with long chain molecules on heating show two phases transformations, one a state, of turbid liquid and the other a clear liquid. The changes are exactly reversed on cooling and at the same, temperatures. The turbid liquid thus obtained are found to be anisotropic, a property possessed by crystalline, substances. Hence these turbid liquids are called liquid crystals.,  Classification of liquid crystals, Liquid crystals are classified as;, i.Smectic liquid crystals, These substances have a layered structure with well defined interlayer spacing. When viewed in polarized light,, smectic phases appear to have fan like appearance. They are always uniaxial and are not affected by magnetic, field. Example: ethyl-p-azoxy benzoate, ethyl-p-azoxy cinnamate, ii.Nematic liquid crystals, In polarised light, they appear to be thread like structures they are affected by magnetic field.Example: p-azoxy, anisole, p-azoxy phenolate, , Department of Chemistry, , Page 17

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III Sem B.Sc., iii.Cholestic liquid crystals, The molecules are said to lie in sheets at angles slightly different between each sheet. Such crystals that have, same smectic and same nematic characters are called cholestic liquid crystals. Example: Cholesteryl benzoate,, cholesteryl formate., , , Uses:, 1. Liquid crystals are used for digital display of numbers and letters in watches, calculators and other, electronic devices., 2. Thin films of cholestic liquid crystals are used to determine temperature and in the rings to study a, person’s mood., 3. Used as solvent for the spectroscopic study of anisotropic molecules., 4. Used in gas liquid chromatography., , Department of Chemistry, , Page 18