Non Covalent Inreraction

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UNIT 5 NON-COV ALENT INTERACTIONSStructure

5.1 ' IntroductionObjectives

5.2 Th e Structure of Ionic and Molecular Crystals5.3 Ionic Bonding and Lattice Energies .5.4 Intermolecular Forces5.5 Hydrogen Bonding5.6 Classical Molecular Dynamics5.7 Summary

. 5.6 Glossary5.9 Answers to SAQs

5.1 INTRODUCTIONBy now we have learnt that covalent interactions are a consequence of a si gr 'f'~ lcantsharing of electrons between the atoms that take part in the bonding process to formmolecules. In mo lecules such as Hz or N,, both the participating atoms share theirelectrons evenly. In bonding betw een unlike atoms, we get a wide variety ofinteractions that range from covalent interactions (as in CH,), ionic interactions (as inthe NaCl lattice) and phenomen a such as hydrogen bonding. In addition, thenon-bonding or the we ak interactions between molecules characterised by van derWaals forces are alw ays present and are responsible for the formation of liquids andliquid crystals and molecular solids. In this unit, we begin w ith the study of ionicsolids followed by the study of intermolecular forces, hydrogen bonding and liquidstructure and dynamics.ObjectivesAfter studying this unit, you should be able to* characterize different crystalline lattices and estimate latticeenergies,

* calculate the unit cell parameters from d iffraction data,* calculate the pack ing fractions in cubic lattices,* compute intermo lecular forces from simple model potentials and relate themto molecular motions.* analyse the consequences of hydrogen bonding,* characterize the liquid state , and* relate molecular dynamics to m acroscopic properties of liquids.

5.2 THE STRUCTURE O F IONIC AND MOLECULARCRYSTALSA study of solids usually begins w ith a description of crystalline so lids. We obtainbeautiful patterns in crystals essentially due to the presence of basic repeating units.As the crystal grow s in size due to the attachment of the repeating units oz the unitcells, the growth can be arrested and the crystal can be c ut to a required size andshape. The smallest arrangem ent of atoms or molecules, which, on repeating in one,two or three dim ensions results in the crystalline lattice is referred to as the unit cell.Or, repetitions of the unit cells in one or more dimensions in space result in anarrangement which is the crystalline lattice. Periodicity seems to be a comm on theme


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in nature whether in space or time. Depending on the shape and the symm etry of theunit cell, we can classify and charac terize lattices. The specification s of the unit cellmust include the length of the three sides or edges of the un it cell, the angles betweenthe three sides and the number and the locations of the various atoms in the unit cell.We are mostly concerned with three dimensional crystals in the presen t unit. Thefourteen common types of lattices found in the seven crystal systems are: cubic(simple cubic, body centered cubic and face centered cubic), orthorhomb ic (simple,body cen tered, face centered and end centered), tetragonal (simple and bodycentered), monoclinic (simple and end centered), rhombohedral, triclinic andhexagonal. Some of the unit cells corresponding to the above lattices are shown inFig. 5.1.

Simpk cubic f ~ a n t a e d h e x r g dcubic (fee) -'4sd athomanbic

Figure 5.1 : Different types of unit cell s.

SAQ 1 :Draw the unit cells for the following lattices. (i) body centered cubic (ii) facecentered ortho'rhombic and (iii) body centered tetragonal. W hat is thedifference between a unit cell and a lattice?

Due to the long range ordering and symmerty of crystalline lattices, these structu resare easier to characterize. With a reduction in symm etry and the long range order, awide range of other structures such as glasses and liquid crystals emerge and theseare a little harder to cha racterize. The characterization and syn thesis of materials oflower dimen sionality (i.e., in one and two dimensions) and of a finite extent (asagains t an infinite lattice) has a profound influence in technology since some of thesematerials are known to have unusual properties such as high temperaturesuperconductivity.On e of the earliest metho ds of study ing the structure of crystalline solids is X-raydiffraction. In this method, X-rays of varying wavelengths and angles of in cidenceare made to interact with the surface of a crystal. The diffracted light emerging fromthe surface is analysed by plotting its intensity as a function of the an gle of theincident beam. The results can be used to identify the type of the crys talline face onwhich the X-ray beam is incident and also calculate the distances between the atoms


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or the ions of the adjacent layers, and thereby, the unit cell dimensions. In Fig. 5.2, Non-Covalent ntcradbm~:the arrangement of atoms in a cubic lattice and the diffraction pattern arising from itare shown. Rays incident at two angles are shown. The rays R, andR, are incident onthe surface at an incident angle 8. Each plane reflects only a part of the incident light.The first two layers, layer-1 and layer-2 contribute significantly to the intensity of thediffracted light and the contribution of the inner layers to the intensity of thediffracted light progressively diminishes.

layer1

platform

a) X-raydiffraction from the surfaa of a cubic lattice. b) Schematic diagramof anX-ray diffractamrcs.Figure 5.2

Consider the interference pattern between rays abc and defgh. Upto the points b ande, the two rays are in phase. The phase difference between the two rays at the points cand h respectively is due to the path difference efg. Both ef andfg are equal to d sin 8,where d is the distance between the two adjacent layers. If this difference is equal tonh, where n is an integer and h is the wavelength of the X-ray, i.e.,

2dsin8 = n h (Bragg's Law) [5.la]then, a maximum in the intensity of the diffracted light I( 8 ) appears at the angle 8 .At other angles such as 8 ' for the ray R,, the intensity is not a maximum unless theBragg condition (Equation 5.1) is satisfied. From the maxima in the curves of 1(8 ),the value of d can be ascertained.In the single crystal X-ray diffraction, the crystal is mounted on a platform and theangle 8 is varied by rotating the platform. Crystals can be cut along varius planes anda method is needed for labelling various planes of the crystal. These labels are theMillerindices of a plane.

Figure 5.3 : ) A few planes in a lattia.b) 'lhe coordinate systemC) 'The Miller Indices ota few planes.


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Atom & Molecules Consider the plane labelled 1 in Fig. 5.3(a). It cuts the x , y and z axes at 2a, a , andinfinity respectively. The plane is parallel to the z-axis which is perpendicular to theplane of the paper, and therefore, the intersection of this plane with the z-axis is atinfinity. The reciprocals of these numbers multiplied by the respective edge le~~gths(a , b and c).are 112, 1 and 0. To obtain the smallest integer set, multiply each numberby 2 to obtain by (1,2,0). These are the Miller indices of this plane. The Millerindices of plane 2 are also (1.2.0). The Miller indices of planes 3 and 4 are (1,3,0) and(1,1,0) respectively. In Fig. 5.3 (c), the 111 plane is shown. It is seen that this planeintersects all the three axes at the distances a, b and c respectively from the origin,which are the dimensions of the unit cell. There are useful formulae which relate thedistances between adjacent hkl planes and the unit cell edge-lengths a, b and c of thelattice. From the interplanar distance, the unit cell dimensions and the distancebetween the atoms can be derived. As an example, the distance between the adjacent(hkl) planes, h,,,, of a cubic lattice is given by

5.3 IONIC BONDING AND LATTICE ENER GIESWhen sodium and chlorine react, the most stable product that can be obtained is theionic NaCl lattice. In this lattice, the constituent units are Na' and C1- ions. Thelattice is more stable than a diatomic NaCl. The process of the formation of the latticefrom the elements Na and C1, in their normal states at room temperature may bedecomposed into the following steps.

1Na (s)+ M C12 (g) -+ NO (8)+Cl (g) ;A H = AHvap Nu)+j D (C12) L5.2~1Na(g)+Cl(g)- a+(g)+Ct(g) ; A H = I (Na)-A(Cl) [5.2b]Nu+ (g)+ C t g) -+ NaCl (s) ; A ? I =U [5.2c]

The resultant of the above three steps is

and the enthalpy change for this process is the enthalpy of formation of NaCl from itselements. Since enthalpy is a state function (i.e., depends on the initial and final statesand not on the path), the following relation holds.

The formation of NaCl (s) from the elements is an involved process and thevisualization of this process as a sum of steps brings out the role of the lattice energyU , defined here as the energy change in the formation of the lattice from theconstituents of the lattice. In the present case, the constituents are Na' and C1-.AHvq(Na) is the energy required to vapourise metallic sodium,D is the dissociationenergy of Cl,. I is the ionization energy of Na, i.e.. the energy change for the processNa -+ Na' + e, A is the electron affinity, which is the energy change for theprocess X- -+ X +e. All the quantities may be taken in units of ~cal lmol r kJ/mol.Substituting the respective values in the above equation for the NaCl lattice, theabove equation becomes (in kcallmol)

AHf (NaCl) = 24.14 + 28.56+ 118.4- 83.4 + U . [5.51How do we obtain the lattice energy? To get an estimate of this value, consider theintkraction energy between a positive ion and a negative ion, with Na' and C1- as anexample. At large separations between the ions, the only interaction is the Coulombicattraction (or repulsion between two positive ions) - e2/r where r is the separationbetween the ions. At short distances, interionic repulsion dominaies because theelectronic clouds of the'two ions overlap 'too closely'. When two elerons cemeyery close to one another, there is a very strong repulsion between them in addition tothe Coulombic interaction and this is a quantum mechanical effect. When two


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electrons are very close to each other, it is not possible to distinguish them, because Non-CovalentI n h sctionssuch a distinction implies that their positions are known more precisely thanpermitted by the uncertainty principle. Furtherm ore, if the two electrons occu py thesame region of space, they will have all quantum numbe rs identical and this will be aviolation of the Pauli exclus ion principle. This repulsive interaction is referred to asthe exchang e interaction and at sh ort distances, it is severe1 orders of m agnitudelarger than the Coulomb repulsion . At large separations, the exchange interactiondecays to zero very rapidlyand the main i.~ter;ction at large distances is the Coulombinteraction. Th e attractive, repulsive and the total energy of interaction betwe en twoions is shown in Fig. 5.4.

Figure 5.4 :Interaction energy between two opp ositely charged ions as a function of the interionic separation.At the distance r, , there is a minimum in the potential energy and if only one Na' andone C1- w ere present, they would be found at this separation. This distance is found tobe 2.38 A for gaseou s NaCl. This me ans that the gaseous NaCl will have an energy of-e2/2.38 = -139.3 kcal/mol relative to the separated ions, taking only the Coulom bicenergy into account.SAQ 2 :

a) Why is a factor of 112 used in Eq. [5.2a]?b) Draw the interac tion energy as a function of r between two like chardedions.c) Using Eq. [5.2b], obtain an estim ate for the form ation of a diatom icNaC l from gaseo us Na and C1. In this calculation, assum e that the shortrange interactio ns between Na' and C1- are negligib le.

A question that now n aturally arises is that, when the diatomic NaCl is even mo restable than H,, then, why is it that NaCl occurs in nature as a lattice? How do es theextra stabilization of the lattice come abou t? T o understand this, consider a onedimensionalNaCl lattice with a unit cell of leng th r,.

Figure 5.5 :A one dimensional lattice of NaCl.


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Atom & Molefulea Consider the interaction enzrgy of the central Na' ion with all the other ions using apoint charge model. The two nearest neighbours contribute an energy of -22/ro. Thetwo next nearest neighbours contribute +2e2/2r0.The -ve ions contribute towardsattraction and the +ve ions contribute towards repulsion. Summing up all the terms,we have,

The sum in the square brackets is greater than 112 since the first term is 112 and everyone of the remaining terms is positive. The value of U can be shown to be= -1.4 e2/rowhich is larger in magnitude than the Coulombic attraction in the Na C1-pair at r,. Therefore, the one dimensional crystal is stabler than the diatomic NaCl.The calculations of the sum of the Coulombic interactions can be extended to threedimensions and the sum may be expressed as -M e2/ro.The constant M s differentfor different lattices and is characteristic of a given lattice. This constant is called theMadelung constant. For the FCC lattice of NaCl, the value of M s 1.75. The latticeenergy of NaCl, taking into account the Coulombic interactions alone is

- - 1.75x (4.8 x 10-loesu)'2.8 x 1@ cm= - 1.44x 10-l1 ergs = - 207 kcaVmolTo these Coulomb interactions, we need to add the contributions due to the shortrange repulsions. The repulsion terms, as shown in Fig. 5.4, mqy be expressed either

as exponential terms, Beq". or as power law repulsions as CIS. Here, B, C and n areconstants which are to be determined for each system. The values of n range from 9to 12. The values of B and C may be expressed in terms of M and r, as shown below.At the distance r = r,, when the lattice is r~~os ttable, the total potential energy is

dUminimum and therefore,- 0. Taking CIS as the repulsive term, U becomes,drMey-Iand therefore, C= , nd

n

Since the value of n (commonly referred to as the Born exponent) is approximatelyequal to 10, the lattice energy U differs from the Madelung energy U, by about 10%.We would expect the U of NaCl to be about - 207 + 20.7 = - 186.7 kcallmol and theexperimental energy is - 183 kcallmol.Inmost iolric solids, the ions are packed very close to one another so as to maximizethe lattice energy. What is the fraction of the total available space that is occupied bythe ions in a close packed lattice? To answer this question consider the faces of thethree commonly occurring close packed lattices: (a) a simple cubic lattice (b) a bodycentered cubic lattice and (c) an FCC lattice. The faces of the three unit cells areshown in Fig. 5.6. The calculation of the packing fraction in the FCC lattice is shownbelow.Example 1:

Calculate the packing fraction in a FCC lattice.


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Solution:In the FCC unit cell, there are six atom s at the six faces of the unit cell. At eachcom er of the unit cell, there is an atom.A face is shared between two adajacent unit cells and.therefore each unit cellco nh ins half the volume of the atom from this face. Since there are six faces.they contribute a volume corresponding to three atoms to the unit cell. Eachcom er is shared by eight adjacent unit cells. Thus, the eight comer atomstogether con tribute a volum e equal to that of o ne atom to the unit cell.Therefore, an FCC unit cell contains four atoms.The vo lume of the unit cell is equal to a3, where a is the edge length of the un itcell. The packing fraction of a lattice is defined as the ratio of the volumeoccup ied by the atoms in the unit cell to the volume of the unit cell. T ocalculate the volume of the atoms in the unit cel l, we need to know the relationbetween the radius of the atom s and the unit cell edge length. From F ig. 5.6(c),if r is the radius of the atom , then, using the Pythagoras theorem,

(4r)' = a2 + a2 = 2 2Therefore,

The volume of four atoms

The volume of the unit cell =a3The packing frac tion is therefore= 0.74

(a) (b)Figure 5.6 : Faces of unit d s howing h e losest packing in' a) a simple cubic lattice.@) a BCC lattice and(c) a FCC latticeExample 2:

Atomic A1 (at. wt. = 26.98 glmo l) crystallizes into an FCC structure with adensity of 2698 kg/m3. X-rays of wavelength , 0.1537 nm, when diffracted fromthe (1 11) planes of this lattice, gave a maximum intensity at an angle of 19.2'.Calculate the Avogadro number, using the above information.

Solution:T o solve this problem, we need to calcu late (a) the edge length of the unit cell,(b) the volum e of the unit ce ll, (c) volume of one mole of Al and (d) thenumber of unit cells in the volume occu pied by one mole of Al.From Bragg's law,

From Eq. 15.lbI. d =$;Therefore, o = (3)'" x d = 0.40475 nm


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Atom & Molecules massdensity =- p = 2698 kg/m3volumemolecular weight of A1 = 26.98 x kglmolVolume of I mol of Al =0.02698m3= 0.00001 m3.2698Th e volume of a unit cell = (4.0475 x 10 m3Th e number of unit ce lls in 1 mol of A1= 1.5081 x loz3Each unit cell in an FCC lattice contains 4 atoms.Therefore, the number of atoms of Al in one mol, or the Avo gadro number= 4 x 1.5081 x loz3= 6.0326 x loz3r A Q ~ :

( - Calculate the packing fraction of a simple cubic lattice.

Bef ore golng to the next section, let us br~e fly iscuss the nature of molecular crystalsor van der Waals crystals. In molecular crystals, the individual units in the structureare molecules or atoms; not ions. The forces that bind the mo lecules are not as strongas in metallic solids (c.g., Cu ) or as in covalent solids (e.g., diamond). Comm onexamples of molecular solids are the crystalline phases of the inert gas elements,crystalline iodlne and crystalline benzene. The arrange ment of the lattice isdetermined largely by packing considerations. The attractive as well as the repulsiveforces between the molecules are very weak and the molecules largely retain theiridentity, very much as in the case of a liquid. Since the bonding forces are weak, thetemperatures at which these solids melt or sublim e are low in comparison with thecorresponding values for the metallic or covalent solids.Molecules w hich are nearly spherical, such as HBr, H,S. and CH, form cubic closepacked structures with each molecule having 12 neighbours. Th e arrangem ents arethus largely determined by packing considerations. The rare gas atoms form F CCclose packed structures. In this arrangement, layers of atoms are stacked in thefollowing way. First, a close packed layer of atoms is formed in a plane. E achspherical atom is surrounded by six atoms in a hexagonal arrang ement. The secondlayer is placed on the interstitial spaces of the first layer. The third layer of atoms isplaced at the interstices of the second layer. If the locations of the atoms of the thirdlayer are exactly on top of thc atoms of the first layer, then we have a hexagonal closepacked (HCP) lattice. If the third layer is placed on alternate sites above the secondlayer, not corresponding to the sites vertically above the atoms of the first layer, thenwe get a face centered cubic (FCC ) close packed structure. Both the HC P and FCCclosed packed structures have the sam e packing fraction. Th e packing of the rare gas. atoms is thus similar to the packing of hard spheres. In a simpl e cubic lattice, thesecond layer will be placed directly on top of the atoms (and not at the interstices) ofthe first. The interstitial space 1s not used for accomm odating the atoms, thereby,resulting in a lower packing fraction in the case of a simple cubi'c lattice.Th e crystals of diatomics such as I, and triatomics such as CO, are not closely packedand these fotm varied structures. Among large molecules, rod like molecules likepolyenes tend to lie in parallel row s close to one another, forming sheets. Flatmolecules tend to stack into columns giving rise to needle-like crystals. Large


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Atom &Molecules other fo rces in some detail and then find out how these forces can be used to studythe structure as well as the dynam ics of the liquid state.In an electrically neuual system , the sum of all positive charges is equal to the sum o iall negative charges. If one of the tw o is in excess, then we get an ionic species suchas ca2' orG-.ven when the system is neuual. the centers of the positive andnegative charges need not coincide. This may be readily seen by consideringexamples of mo lecules such as HC1 and H 20 . In HC1, the center of positive charge iscloser to the H atom and the center of the negative charge is close r to the C1 atomsince the later is more elecuonegative. The m olecule is said to be dipolar or simplypolar (i.e., it has two 'poles' correspond ing to positive and negative charges) andpossesses a dipole moment represented by $The magn itude of the dipole mom ent isgiven by the product of the magnitude of the charge and the charge separation. If acharge of +e and -e are sepa rated by 1A, then the magnitude of the dipole momentis e x 1 A = 4.8 x 10-l8 esu.cm. Th e quantity lo-'' esu. cm is called 1 Debye. Thedipole moments of m olecules range from 0 Debye to a few Debyes. The values of thedipole moments and some other molecular properties of a few molecules are given inTable 5.1. The dipole moment is a vector quantity. The direction of the vector istaken from the positive to the negative charge. The vector $is given by

+$= xi q i ri [5.10]where q, is the algebraic charge and c i s he vector from the center of mass of themolecu le to the location of q,. Instead of point charges, if we have a chargedistribution represented by p(x , y, z), whose dimensions are charge per unit volume.the dipole moment is given by

+P r I $1, Y, ) p(x,y,z) h d y dz. [5.11]where ? = r ( ~ , ~ , z ) = x i + ~ j + z kAnalogous to the dipole moment zitwhich is a mom ent of first order in r, highe rmoments such as the quadrupole moment (second order in r), the octupole moment(third order in r) etc., can be defined. The net charge on a molecule or an atom is~ i o t h i n ~ut the zeroth moment of the charge density, i.e., the integral of the chargedensity over the w hole space. Given below is the xy component of the quadrupolemoment Q, Q = N x Y P (~ ,Y J)h d~ dz [5.12]The re are six components of the quadrupole moment, name ly the xx, xy, xz, yy, yzand zz. For a diatomic molecule, with the z axis as the molecular axis, only onecomponent among the six is independent and is referred to as the quadrupole momentand is given by

Q = Zi qi ( 3 2 t ) / e . r5.131The division by e is so as to make Q have the dimensions of an area. The values ofthe qu ad ru pl e mom ents of some molecules are given in Table 5.1Anothe r property of a molecule that plays a crucial role in intermolecular interactionsis the polarizability. In a uniform electric field, a molecule is polarized by the field(i.e.. there is a displacemen t of the centers of the positive and negative charges of themolecule) and the induced dipole moment (to be distinguished from the permanentdipole mo ment) is given by c = 2 [5.14 ]where a is called the polarizability. For sma ll ele cu ic fields, the polarizability may betaken to be independent of the field. The po larizability is anisotropic in general, i.e., itis not the same in all directions. For example, the polarizability along or parallel to abond in a diatomic, a,, . s different from the polarizability in a directionperpendicular to the bond, a,.Th e polarizability is a measure of the extent to whichthe charge cloud of an atom or a m olecule can be distorted by an external field.Tightly bound clouds of systems like rare gases and positively charged ions have lowpol uiu bili ties . Loosely bound or relatively diffuse clouds such as alkali metals or


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negative ions have large values of polarizabilities. Molecular polarizabilities correlatefairly well with molecular volumes. Typical values of a are given in Table 5.1.The three values of the quadrupole moments for water correspond to Q Qyy, nd Q,,respectively, with the molecule placed in the xz-plane with the z-axis being themolecular axis.The various interactions between two molecules will now be considered. We firstconsider the dominant interactions between two charged particles or molecules. If theparticles carry charges q, and q, respectively, the main interaction energy betweenthem is q, q,l ri,. This term has already been discussed in the section on ionic solids.The next interaction to be considered is the ion-dipole interaction. The interaction+energy between an ion of charge q, and a dipole of moment CL/ is given by - p .E,where E is The electric field due to the ion at the dipole$ This energy depends onthe angle 0 that the dipole vector makes with the axis between the ion and the centerof the dipole. The energy is given by + +V( r ,O)iomiiPok = - p .E = - qi cos0/? [5.15]Table 5.1 : Values of dipole momen ts, quadrupole moments and polarizabilitiesfor a few systems-A tom/Molecule Dipole Mom ent Quadrupo le PolarizabilityMoment

LiHHFHz0HCI 1.109 3.7 2.63


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Atom & Molecules . SAQ 5 :Prove the validity of the equation C5.151.

The next interaction to be considered is the ion-induced dipole interaction. If an ion iand speciesj are separated by a distance r, the interaction between the ion i and theinduced dipole moment in species j is given by .

SAQ 6 :Will the quadrupole moment of a molecule multiplied by the molecular size(e.g., the length) give the molecular polarizability? You may want to refer toTable 5.1; .

The next interaction to be considered is the interaction between two dipoles pi and p?In this case, the interaction energy depends on the distance r between the dipoles aswell as the angles defining the orientation of each dipole. For the dipole i, the anglesare 8, and cp,. For the dipolej, the angles are 8, and + The angle 8 is measured with

-3 + +respect to the vector rij= ri- rj. The centers of the two dipoles z a n d $are given by


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ion-dipoIe, dipole-dipole, ion-induced dipole and dipole-induced dipole. The otherinteractions such as the dipole-quadrupole, quadrupo le-qua druple and so on, areeven smaller in m agnitude and a re not considered in this discussion.All the above interactions were electrostatic in nature, in which atleast one of thespecies was charg ed. Now we consider an interaction- eferred to as the dispersioninteraction. The d ispersion interaction is present between all molecules and atomseven when they are not charged and d o not possess an electric moment of any order.The se forces have a quantum m echanical origin and can not be explained by classicalelectrostatics. The dispersion fo rce between atoms/molccules i and j arises from acorrelated motibn of electrons of the species i and j . Although this interaction can beunderstood in terms of the wavefunctions of the systems 1 and j , a classically intuitiveargument may be given as follows. C onsider two molecules 1 and j that do not possessany electric moment. A t any instance of time, although each of the system is neutral,i t will possess a transient dipole mom ent due to the instantaneous positions of itselectrons relative to the nuclei. Of course, the time average ( i.e., an average takenover a long interval of time) of this instantaneo us dipole mom ent is zero. Th isinstantaneous dipole of molecule i can "polarize" the electronic motion of molecule jdue to the field that it creates at the molecule j . Thu s, the electronic clouds of the twomolecules will execute a correlated motion. An uncorrelated or an independentmotion of the two electronic clouds w ill not lead to the dispersion interaction. Thisargument is only pictorial and it should be remembered that a quantum m echanicalresult can not alway s be justified by a classical argument. You may want to drawpictures of instantaneous dipole moments of two nearby molecules to follow theabove arguments. Try with two He atoms.The d etailed calculations of the dispersion energy are quite difficult. Therefore, theresult of a simple model will be given below.

This formula brings out the role of polarizabilities ai nd a, of the moleculesinvolved in the interaction. The interaction is an inverse sixth power law. Thecoefficients of the interaction are system specific. Thbfrcquencies vi and v , are thecharacteristics of n~olccules and j and are nearly equal to the frequencies of thelowcst clectronic transition energies of the molccules i and j. These attractive forcesbetween molccules are known as the van der Waals forces.A quantitative estimation of the intermolecular forces ( ds agalnst the covalent or thechem~ca lorces ) can be made in the studies involving molecular collisions and in thestudles ol the liquid state. The m ag n~ tud e f these interactions may be calculated for afew ty p~ ca t alues of interparticle separations. The parameters taken in the followingcalculations are some of the typical values of parameters that occur in molecules.Thesc calculated values are given in Table 5.2. In this table, the ionic charges q, and9, are taken to be +e and -e, where, e is the electronic charge. The dipole moments F~and p, are taken to be 1.5 D. The polarizabilities are assigned the values 3 A3. Thed~ po le s re all aligned along the vector r,,, so that all the angles 0, to cp, are zero. Forthe Lennard-Jones potentla1 tb be discussed below, the parameters taken are &/k, =120 K and o= 3.4 A. Th e various intermolecular potentials discussed in this sectionare shown as a function of r in Fig. 5.7, and the values at r , = 5 and 15 A are shownin Table 5.2.Tab le 5.2 : Values of typical non-c ovalen t interactions at a few interparticle ,separations. All values in kcallmal.

Type of interaction r , = 5 A r,, = 10 A r , = 15 Aion-ion - 66.4 1 - 22.14ion-dipole - 4.15dipole-dipole - 0.52 - 0.019ion-induced dipo le - 0.79 - 0.0098dipole-induced dipole - 0.01 24 - 0.000017Lem ard Jones potential

-.- 0.0212 - 0.0000032

Non-Covalentlnc;;?rl!ms


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Atom & Mokcules

\ ion-induced ipole

Figure 5. 7 : Various intermolecular interactions as a function of the ntermolecular separation.~ S A Q:1 Fill in the second blank column of Table 5.2

Two argon atoms or two methane molecules will not bind chemically to one another.Precise calculations of the interaction energies between such systems is very difficult;it is certainly harder than the calculation of the energy levels of the constituentindividual molecules! The accurate calculation of these intermolecular energies isbecoming possible only in the last quarter of the twentieth century. However, onemay choose an empirical potential based on the potentials of related systems or byusing typical values of molecular sizes and the van der Waals radii. Using thesepotentials, one may then test whether this potential is capable of predicting certainexperimental features. What are the experimental properties that are related tointermolecular forces? This is indeed a very poor question! Indeed, every property ofa collection of molecules (an isolated molecule does not react with anything) or thereaction between molecules is dependent on and is governed by the intermolecularforces.The question can now be framed better :Which are the most direct measuresor the criteria that can be used in determining intermolecular forces? Consider thecase of an ideal gas first. In this case, as is well known, there are no intermolecularforces. The equation of state is given by PV = nRT.The role of intermolecular forcesis to give rise to deviations from the ideal gas behaviour. The first deviation from theideal gas behaviour is contained in the second virial coefficient which is defined inthe equation


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OD

B = 2 n N, [ 1- a-"(r)'knT ] dr [5.21]0

Thus , the deviation from the ideal gas behav iour, which is obtained from the pressurevolume data, can be used to obtain an estimate of the intermolecular potential u(r).There are seve ral forms of the intermolecular potentials and here, only the simplestforms that are useful in atomic and m olecular systems will be given.The hard sphere potential is defined by

UHS(r) = 00 . if r I a ,= O , i f r > a [ 5 . 2 2 ]

The squ are well potential is given byU,& r)=m if r I a

= - ~ , i f a < r I a + 8=O,if r > a + & [5.23]

The Lemard-Jon es potential is given by12 6 [5.24]

U i i ( r ) = 4 ~ [ [ ~ ) (?)IThis is also called a 1 2 - 6 potential since r-l2 and r4 terms appear in it. In thisequation, a represents an estimate of the contact d~stance etween the interactingparticles and E is the depth of the potential. The attractive term ( a ~ r ) ~ncorporates thecorrect form of the asymptotic, long range, dispersion interaction. The repulsive term( c ~ l r ) ' ~ncorporates the short range steep repulsion. This form is very widely usedand the parameters of this potential fo r a few system s are given in Table 5.3.Table 5.3 : he pa ramete rs for th e Lennard-Jones potentials.

System ElkB.(K ) a lAAr-Ar 145 3.8Kr-Kr 20 1 3.6Xe-Xe 288 3.9Na-Ar 7 2 3.8Li-Xe 175 3.8

CH4-CH4 145 4.2

It is noticed that the contact distances am ongst various systems are very sim ilar.S A Q 8 :

Draw sk etches of the hard sphere and the square-well potentials. In theLennard-Jones po tential, what is the value of r at which the potential isminimum?

Ing ase s and liquids there is a continuous movement of molecules due to moleculartranslations, vibrations and rotations. Rotations cause a significant reduction or

Non-Covalent Interadions


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Atom & Molecules damping of those intermolecular forces which depend on the relative orientations ofthe molecules. For example, if two dipoles A and B are parallel to each other, then theinteraction energy is negative. Let us denote it by -X. If the two dipoles areantiparallel. then the energy is + X. If there was a free rotation of the molecules, thenthe angle averaged dipole-dopole interaction would have been zero! But we do knowthat the lower, or the more negative values of energy are more favoured than thehigher values of energy at all finite temperatures. In fact, the probability of theoccurrence of a given value of energy U, is proportional to exp (-U,/kBT). This factoris commonly referred to as the Boltzmam factor. (The absolute temperature isdenoted by T and kB is the B o l t z m a ~onstant )The angle averaged potentials aretemperature dependent, and for the dipole-dipole interaction, the angle averagedpotential is given by

(UdiP-d,dav= -1.5 P B ) ~ / ~ B T ~ ~ B i [5.25]Here ( ) refers to angle averaging using the B0 l t z m a ~robability distribution.This averaging is also referred to as thermal averaging. In this averaging, each energyis weighted by the Boltzmann factor corresponding to that energy. While averagingover the orientations, the distance between the dipoles rABs not changed. Thisorientationally averaged dipole-dipole interaction may be contrasted with the angleaveraged (not Boltzmann average) dipole-induced dipole interaction which is givenby,

I 2 2 6Udipoltinduceddipo[~ v = - a ~A+ PB)/'AB [5.26]This later average is independent of temperature because the induced dipole momentis always parallel to the inducing field. The angle averaged dipole-dipole interaction(Boltzmam weighted), the averaged dipole-induced dipole interaction and thedispersion interaction possess the same functional form. They all vary as r4.- - - -5.5 HYDROGEN BONDINGThe term hydrogen bonding is invoked to explain certain special properties of somesystems containing (1) a hydrogen atom bonded to a highly electronegative atom and(2 ) other electronegative atoms present in the vicinity of this hydrogen .atom whichcan significantly alter the electron density in the vicinity of the hydrogen atomwithout actually breaking the parent bond. Hydrogen bonding is present in a varietyof systems ranging from water molecules to the DNA molecule and is responsible fora variety of properties of systems ranging from the strengths of fibers to even theprocess of genetic coding and human memory.To get an insight inlo the nature of hydrogen bonding, let us begin by considering theboiling points of a few liquids. The boiling point of a liquid is the temperature atwhich its vapour pressure is equal to the atmospheric pressure. A higher boiling pointimplies that the molecules of the liquid are held together more strongly. This isbecause a higher temperature is required to evaporate a sufficient number ofmolecules for the boiling to start. Molecules with large values of dipole moments andother higher order moments and molecules possessing large values of molecularpolarizabilities are generally known to have relatively large values of boiling points.The polarizabilities generally increase with an increase in the number of electrons inthe system. As we go from He to Ne, Ar, Kr and Xe, the larizabilities increase

%Orespectively as 0.203.0.392. 1.63,2.41 and 4.01 ( in 10- cm3units). The boilingpoints of these systems are 4.22,27.3,87.3, 119.9 and 165.1 K respectively. It is seenthat the boiling points increase with an increase in the polarizability. Next consider afew relatively small molecules such as propane, dimethyl ether and ethylene oxide.Their polarizabilities are 6.4.6.0 and 5.0 (in lo-" cm3) respectively, which are quiteclose to one another. The dipole moments of these molecules are 0, 1.3 and 1.9Dand their boiling points are 23 1,248 and 284 K respectively. We observe that anincrease in the values of the dipole moments of the molecules also leads to anincrease in the boiling points. If the size of the molecules increases considerably, thenthe dispersion interactions become more significant and the role of the dipolemoments and polarizabililies tends to become less important.In compounds containing hydroxyl, amino or imino groups and also in moleculessuch as HF, the boiling points are considerably higher than for the molecules


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possessing sim ilar values of dipole moments an d polarizabilities. A s examples.consider C,H,OH (ethanol) and C 2H 40 ethylene oxide). Their dipole mom ents are1.7 D and 1.9D and they have similar polarizabilities. W e may expect ethylene oxideto boil at a higher temperature. B ut its boiling point T , is 284 K w hile the boilingpoint of ethanol is 35 2 K. H F with a value of a = 0.84 x lo-%cm3 and a p= 1.91 Dhas a boiling point of 291 K. Similarly methanol (a 3.0 x cm3,p= 1.7 D an dT, = 338 K), ammonia (a= 2.34 x lo-%cm3,p= 1.467 D and T, = 24 0 K) and water(a 1.59 x 10-" cm3,p= 1.86 D and T, = 373 K ) show very high boiling points incomparison w ith other molecules w ith similar values of a and p.This additional interaction present in these mo lecules is much weaker than the usualcovalent bonds and is slightly stronger than the van der Waals forces. The strengthsof these interactions range from 3 to 10 kcallmol. This interaction is attributed to theassociation between the molecules by the formation of a weak bond, called thehydrogen bond that results when a hydrogen atom is sandwiched between two veryhighly electronega tive atoms. The lengths of these bonds ar e in the range of 1.5 to 2 A.Som e common examp les are given in Fig. 5.7.

Figure 5.8 :Some examples of hydrogen bonds. .liHydrogen bonding is not restricted to liquids alone. The fact that urea or boric.acid 'are solids at room temperature i s attributed to the formation of extensive hydrGgenbonding. Aqetic acid molecules pair as dimers due to hydrogen bonding. In salicylicacid, an intramolecular hydrogen bond is formed between the hydroxyl H and theadjacent 0 of the COO H group. Hydrogen bonding is not restricted to sma ll 2 .molecules alone. Indeed the utility of nylon as a fiber is due to hydrogeh bondingbetween adjacen t chain s of molecules. Th e dase pairing in the genetic material, D NAis also due to hydrogen bonding.

5.6 CLASSICAL MOLECULAR DYNAMICSHaving discussed the various forms of the potentials, it is worthwhile t o consider howthese forces inf-luencc the physics a nd the chem istry of the condensed m atter. W eshall use classical mechanics o r Newton's laws, and not quantum mechanics in this

Non-Covalent nteractions


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section. In a gaseous system. the average distance between two molecules is quitelarge, typically 50 A at room temperature. This large sep aration enables a nearlyrandom m otion between the mo lecules of the gas. Ifwe consider solids and liquids,the m olecules therein are very close to one another. The intermolecular distances aretypically 3 to 5 A. In a solid, whether it is ionic. covalent, metallic or molecular, thelattice particles do not move away from hei r eq uilibrium sites. Vv'hatever mobilityrema ins, it is due to the presence of vacancies or defects.In liquids, there is a continuous motion of a ll the particles and this motion may bestudied in a very simple manner if the form of the interm olecular forces is known. Inthe last two decades, an emormous effort has been put in developing the forms of thepotentials between severa l systems, particularly those systems relevant to life such asproteins and nucleic acids. Several computer packages are available to study themolecular motions in such systems. In the present unit, the major features of such anapproach will be illustrated by conside ring the system of liquid argon.Consider a collection of N argon atoms in a cubic box of edge length L. The numberN is in the range 50 to 1000.A cross section of the box is shown in Fig. 5.9(a). The 'number N is so chosen that the value NIL^ correspo nds to the actual density of liquidAr at the tempera ture of interest. The initial configuration of the atoms in the system,i.e., the initial values of the coord inates of all the partic les is dete rmined by placingall the atoms randomly in h e box. They are placed in such a way that no two particlesare too close to one another. Typically each particle is at least 2.5 A. away from everyother particle. The random placem ent can be done by using a random numbergenera tor such as those present in most scientific calculators today.

a) A simul~~iorrcel la m o l ~ l a rynamics. b) Demonstn~ion f periodic bgundary con dtionsin two dimensions.Figure 5.9

SAQ 9 :Why a re the initial coordinates so chosen that no pair of a toms is closer than2.5 A? From the k nn ard -Jo ne s potential energy diagram of A r, see what willhappen if two atoms of Ar come as close as, say, 1 A.

Having chosen an initial set of coo rdinates of all atom s, we need the initial velocitiesof all the atoms. The initial velocities are obtained from a M axwell-Boltzmanndistribution of velocities at the temperature of intere st. According to this distribution,the fraction of molecules having velocities between v and v+dv is given by

According 10 this distribution function, velocities which are very small or very largeoccur very rarely. Th is result is from the kinetic theory of gases. In thisdistribu tionformula, m is the mass of the gas molecule and T is the absolute temperature. The


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assignm ent of velocities is similar to the assignm ent of p ositions except that thevelocities are chosen from the plot off lv ) vs v. This distribution function will againbe used in Units 9 and 10.SAQ 10 :

Plot a graph'offlv) vs v for Ar at 150 K.

Non ralent Interactions

Given the initial positions and velocities, we would like to follow the path or thetrajectory of all the atoms in the sy stem. This is very much like following the futurelocations of all the planets in our solar system knowing the present positions andvelocities of all the planets and the fo rces of interaction between the planets. Or evenmore com mon ly, the evolution of a satellite in the vicinity of the earth. The equ ationsof classical mechanics or the Newton 's law s of motion can be used for this purpose.If you begin wond ering how we have come back to the use of Newton's lawssuddenly, your worry is more than justified. But it has been found that Newton's lawscan be used quite effectively to monitor the motion of atom s and molecules in aliquid. They can not be used to describe the motion of sm aller objects such aselectrons and other tinier particles.The m ethod that is being described here is known as the molecular dynam icssimulation. Having obtained the initial velocities and positions of all the atom s, wewant to know the p ositions and velocities of the atom s after a time 6t. The v alue of 6tis in the range of 1 to 10 fs (femtose cond). In the fall of bodies from a height onto theearth's surface, we may recall having used the familiar formula, s = ur + at2/2. In thisformula, s is the distance travelled by the object, u is the initial velocity, a , theconstant value of the acceleration due to gravity and r is the time of travel. On andvery near the surface of the earth, the acceleration due to gravity may be assumed tobe a constant. But in the case of molec ular motions, the acceleration on each of themolecu les is continuou sly changing. This acceleration depends on the relativedistances between the molec ules which are continuously changing in a liquid.Consider the particles i and j. The interaction energy between the particles is given by

12

[5.28]where, ri j s the distance between the particles i and j and ?=xi. yi. zi) is thecoordinaie of the particle i and ?the coordina te of partic le j. The distance r, is given

2 1/2by [(xi- ,)~+ (yi -$)+ (zi- ,) ]The force o n the particle i due to its interaction with particlej is given by the gradientof the potential. Recall that force multiplied by distance or the integ ral of F.ds ispotential energy.

The particle i interacts with all the remain ing particles in the box. The total force onparticle i can be obtained by adding the forces on it due to all the other particles.F, (total) = C a u

j + i a r iThe sum mation excludes the interaction of the particle i with itself; i.e., j is not equalto i in the summation.


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SAQ 11 :What is the formula for the total force on particlej ?

The new position of the particle i after a time &. (i,e., at the time t + 61) can beobtained from its positions at times t and t - & by the Verlet algorithm which is givenbelow.

r,(t+61) = 2 ri(t) - i(t-61) + [Fl/m] 6? [5.31]and the velocity of the particle at time t is given by

vi(t) = [ r1(1+6t)- i(t- 61) ]/2& [5.32]Using this algorithm, positions and velocities of all the particles can be generated forseveral time steps (usually in tens of thousands). The configurations (collection of allthe coordinates in a given time step) at all the time steps is referred to as themolecular dynamics trajectory. Molecular dynamics simulations have become anindustry now and they provide a very powerful method for studying the structure anddynamics of liquids and solutions. Its relevance in the present unit is to bring out thecritical role of intermolecular forces.Example3 :

The Verlet algorithm, as outlined in the above discussion, allows one tocalculate the new (or future) position of the particle if two of its previouspositions (i.e., the present and the immediate past) are known. When one startsthe simulation, only the present positions and velocities are known. How arethe new positions determined in this case?Solution :In the beginning of the simulation = 0, and only r,(O)s and vi(0)s are known.To obtain r,(0+6t) =r,(6t) use the familiar Newton's formula

r,(&)= r, (0)+ vi (0) 61+0.5 Fi(0) 6t2/mExample4 :

For a three particle system of Ar atoms, choose an initial set of configurationsand velocities, take & = 0.05 ps, and calculate the first four steps in themolecular dynamics simulation of these three particles using the Verletalgorithm.

Solution :The three particles are placed in the xy-plane as follows. One particle is at theorigin, another particle is on the x-axis, 4 A from the origin and the thirdparticle is on the y-axis, again 4A from the origin. The (x,y) coordinates of theparticles are (0,0), (4,O) and (0.4) respectively. The initial velocities arearbitrarily chosen to be (0.0). (0.1) and (1.0) respectively. The unit of velocityis Alps. The unit of force is 10 JI(mo1. A). The MKS unit of force is Newton orJlm. The values,of the components of the forces, positions and velocities aregiven in table 5.4. The time step of the simulation is 0.05 ps. Mass is taken indmo! units.Tor Ar, take m = 39.95. The units of mass, length and time whichare taken here as drnol, A and ps respectively are very convenient in thesecalculations. Verify that these units give energy in units of 10 Jlmol.


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Non-Covalmt Interactions

Particle No. 1 Particle No. 2 Particle No. 3ComponentsX Y X Y X Y

Step Number 0Forces 55.47 55.47 - 68.3 12.8 12.8 - 68.3Positions 0.0Velocities 0.0Step Number 1Forces + 55.56Positions 0.00174Velocities 0.0347Step Number 2Forces +54.3Positions 0.00695Velocities 0.0695Step umber 3Forces 51.4Positions 0.0156Velocities 0.1383Step Number 4Forces +46.3Positidns 0.0274Velocities 0.2044SAQ 12 :

Repeat the calculation of the example for a two particle Ar system withstarting velmities of (0,O) and (0, I) and ~nitial ositions of (0,O) and (4.0)respectively~~ositionsre in 8, and velocities in bps. The force on eachparticle at a given time is calculated from the positions of all the particles atthat lime.

-

For the interaction bctween molecules, several empirical potentials have beenconstructed. Only one example of such a potential, namely the water-water potentialis given here. It is very convenient to think of a molecule as a collection of interactingsites. The sites on a molecule do not interact amongst themselves, but interact onlywith the interacting siles of all the other molecules that are present in the system. Themodel presented here for water is known as the SPC (simple point charge) model.Three point charges are placed on each molecule. The charge on hydrogens is + 0.41e and that on oxygen is - 0.82 e. Here, as usual, e is the magnitude of the electroniccharge. In addition to this Coulombic interaction, there is a Lennard-Jones center oneach oxygen. This means that the interaction betwecn oxygens of any two watermolecules is given by

where r is the oxygen-oxygen distance,'^ = 0.3428 nm (k~/mol)"'~ndA = 0.37122nm (k~ /mol) '~ .ere, nm refers to nanometer, i.e., meter. In addition to thisLennard-Jones interaction between the two molecules, all lhc charges on onemolecule interact with all th25harges on all the other molecules. In each water


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molecule, the 0-H distance is taken to be 1 A and the H-O-H bond an gle is taken tobe the tetrahedral angle of 109"28'.

Figure 5.10 :The SPC model of water.This mod el is found to yield the properties of bulk water such as the diffusioncoefficient (which measures the extent of random m otion in the liquid), the liquidstructure and the water bulk density. The dipole mom ent of this model of water is2.351 D. This is much higher than the dipole mom ent of an isolated water moleculewhich is 1.86 D. This indicates that the models that yield good bulk properties (ormac rosco pic pro per ties ) need not correspond to the exact results for the isolatedmolecu les or the isolated pairs. This is an indication that the interactions in the bulkmedium are significantly different from those of isolated molec ules or isolated pairs.This is precisely the reason for stud ying the three -body and other higher n-bodyinteractions. These are discussed a little later in the present unit.SAQ 13 :

Show that the dipole moment of water in the SPC model discusse d in thissection is 2.35 D. How will you change the charge s on H and 0 o get a dipolemoment of 1.86 D?

You may have been wondering by now- "How do es the simulation on a systemcontaining at m ost a few' hundred particles correspon d to the behaviour of a liquid?"Even a drop of a liquid contains about 1019molecu les. It was found that if the centralcubic box containing about 100 to 500 mo lecules is surrounded on all sides byidentical boxes, extending to infinity, then we have a large system represe nting aliquid. If a pa rticle of the cenual box enters a box to its right, then the pa rticle in theleft box, which is the image of the particle that has left the cenu al box, enters thecentral box from the left. Such conditions are called periodic boundary condition s andare applied in all the three directions. Periodic boundary conditions in twodimen sions are demonstrated in Fig. 5.9(b). It was discovered by 1960, thatcalculations performed on a hundred molecule system with periodic bound aryconditions yielded properties that matched those of real liquids qu ite well. For .exam ple, the average potential energy of the system which may b e taken as the sumof all pairwise interactions present in the system, correspon ded to the internal energyof a macro scopic system which will be considered in the next unit. It is thus possibleto make the vital connection between m olecular properties and prop erties of the bulksystems (i.e., thermo dynam ics and chemical kinetics to be studied in Units 5 to 10)containing an extrem ely large number of particles.Th e importan t features of the structure of liquids is illustrated in Fig. 5.11.Fig. 5.1 1(a) describes the structure of fused (liquid) sodiu m (labelled L) andcompares it with the suuc mre of solid sodium (labe lled S). In a liquid, theatoms/m olecules are in continuous motion. One typical arrangemen t in fused sod ium


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is shown in the upper part of Fig. 5.1 l(a ). Let us call the atom R as the reference atom(any other atom can be used a s the reference). We may think of the neigh bours of Ras constituting a set of neares t neighbours, a set of next nearest neighbours and s o on.Starting from the atom R, let us count the number of atom s enclosed in a sph ere of4radius r of volume - ?. This num ber, designated as n(r), is called the running3coordination number. Th e area under the thick curve in the lower part of Fig. 5.1 1 a )(labelled L) gives this n(r ) for liquid sodium. For examp le, n(4.0 A) = 6.6 and this isthe area show n by the dots in the figure. As r increases, n(r), the area under the curve47tr2 pg (r) a lso increases, becau se more pa rticles get included in sp heres of largerradii. Here, p is the bulk density of the liquid in units of particles per A3. The dashedline labelled U s the plot of 4 x 3 ~s r. This is a parabo la, as expected. Deviationsfrom this curve are the deviations from a fully uniform or random distribution ofparticles about the reference atom R. These dev iations, shown by the Line L spec ifythe structure of the liquid. The vertical dashed lines in figure with the heights of 8 , 6 ,12 , 24, 8, 6 and 24 at values of r near 3.7 ,4.4, 6.1,7 .1, 7.4, 8.6 and 9.3 A respectivelyindicate the number of atoms presen t at these distances from a reference atom in thesolid. At the remaining distances, e.g., betwe en 3.7 and 4.4, between 4.4 and 6.1 andso on, no pa rticles are present in the solid. This is due to the rigid, regulararrangem ent of atom s in the solid. In a liquid, although this rigidity is lost due to themov ement of m olecules, a structure is still prevalent and this is shown by the peaksof the curve L. T he qu an tity 4 x 3 p g( r) is sim ilar to 4 x 3 ~tudied in Units 2 and 3.The qua ntity 47tr2 pg dr g ives the num ber of atom s, dn(r) contained in a sph ericalshell of radius r and thickness dr. A c ross-section of such a shell is shown in Fig.5.1 1 b). The c enter of the shell labelled 0 orrespond s to the reference atom R of Fig5.1 l(a ). A reference atom has to be present at 0, ecause the distance r is measuredwith respect to a reference atom. The qua ntity

1 1g(r) =- n (r +dr ) - n (r)]/4xr2dr=- dn (r)/4m2 dr]P Pis called the pair distribution function of the liquid. The bulk density of the liquid isgiven by p =N/V where N is the number of particles in the liquid and V the volume.For very sm all values of r , the spherical shells about the reference atom do notcontain any particles and therefore, dn(r) = 0 and so is g(r) = 0. For large values of r ,the spherical shell is very much like the bulk of the liquid and is not influenced by thereference atom and the 'local d ensity' of the shell, namely

[n (r + dr) - n (r)]/(4x? dr)

Non-Covalent Interaction

becomes the bulk density p and thus g(r) become s unity. The pair distributionfunctions in liquids are extensively studied and they give considerable informationabout the structure of liquids and they can be related to the macro scopic (or bulk)properties of the liquid. In Fig. 5.1 1(c), the oxygen -oxygen , the oxygen-h ydrogen andthe hydroge n-hy drog en pair distrib ution func tions (i.e., g,(r), go,,(r ) and g,,(r)) inliquid water are shown. In g d r) , i refers to the atom i in a reference molecule, and jrefers to the atom j at a distance r in any other molecule. You are asked to give thephysical interpretations of goo(r), gH H( r) ng go&) in analogy to g( r) for liquidsodium w hich could have been explicitly w ritten as g, ,, (r).The results of the mo lecular dynam ics simulations on water are shown inFig. 5.11 (c). The 0- 0. 0- H and H-H radial distribution functions are shown in thefigure. These re sults are obtained by averag ing over tens of tho usands ofconfigurations in a system containing more than 100 water mo lecules. The graphsrepresent the average structure of water. Th e 0-0 running coordination num ber has Jthe following meanin g. If we take the oxygen atom on one of the water molecules asthe center of the coordinate frame and c ount the number of oxyg en atoms in aspherical shell of radius r aroulrd it, the number obtaine d is the running coordinationnumber noo (r). This is obtained for various values of r. Since all the water moleculesin a liq u~ d re of equal im portance, the value of n( r) is averaged over all the watermolecules.


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Figure 5.11 : a) The structure of liquid sod ium.b) A section of a spherical shell of radius ran d thickness dr.c) The 0 - 0 . 0 - H and H-H pair distribution functions in water.

SA Q 14 :What is the meaning of the 0-H nd H-H coordination numbers? Define themin analogy to the 0-0 oordination number. From the discussion above, give asimple definition of the pair distribution function in tern's of the local densityin a spherical shell of radius r. In the graph 5.1 1(a), what are the number ofsodium atoms surrounding a reference atom in a spherical volume around it ofradius 8 A in (a) liquid sodium, (b) solid sodium and (c) sodium of the samedensity as the liquid-but behaving as an ideal gas ? For this part, compute theareas under the dashed curve (U , or ideal gas) and the smooth curve L (for theliquid). For the solid, count the total number of atoms (vertical dashes) atdistances (from the reference atom) less than 8 A.


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SAQ 15 :Develo p the formula for the total energy of an N partiele system . Start with 2,3 and 4 particles and then extend to an arbitrary N.

In this sec tion, as well as in the previous sections, an underlying assum ption that ismade is that the total energy is the sum of all distinct pairwise intera ctions. This iscorrect for point charges alone. In this situation, the force on a particle i can becalculated as the sum of the contributions from all the remain ing particles. Butpairwise addititivity of forces or energies is only an approximatio~or molecules zn6this needs to be corrected by the inclusion of the three body d3d the other highern-body ( n > 3 ) forces. The three body potential for a system of 3 particles is definedby the following equation.

I/rold = u12(r12) + ~ 2 3 ( ~ 2 3 )~ 1 3 ( ~ 1 3 )u123(r12?r23*r13) [5.35]The last term above is the three-body potential. It is that part of the potential that cannot be expresse d in terms of the pairw ise interactions alone. Consider the exam ple ofthree particles i , j and k. If only pa irwise interactions are present, then the total energyis given by

We hav e already seen that molecules have non-zero values of polarizability. Thepolarization energy of the molecule i when only molecule j is present, added to itspolarization energy when only k is present, is not the same as the polarization energyof i when bothj and k are simultaneously present. This is because , the presence ofmolecule k alters the very moment of the mo leculej which is to polarize the moleculei . Similarly, j alters the moment of k which polarizes i . The same argument holds forthe polarization energies of mo lecules j and k. The dispersion energy of threemolecu les or atom s is thus given by k

Non-Covalent Interadions

where,cGk= u ( 3 ~ 0 s i cos 8, cos Ok+ 1)

Here, a is a constant whose values are of the order of molec ular pola rizab il~ties ndc,, has a value similar to the values of c,)s.The formula for U,,, is called theAxilrod-T eller formula. It give s the contribution that can not be expressed in tsrms ofthe sum of pairwise interactions alone. The an gles 8,9, and 8, are the internal angles iformed by the three atoms and are show n in Fig. 5.12. Figure 5.12 :Intemai angles formedby the three atomsThe a nalysis presented above can be extended to higher terms such as the 4-bodyinteractions and so on. But these are progres sively smaller in magn itude. In gene ral,the three body contributions are of the order of 10%of the total energy of the system.The n-body interactions are important especially when the distances between theinteracting molecu les are sma ll. At large distances, the two body apptcxim ation isquite good. The compu tation of three body forces is quite laborious and therefore, ithas been a commo n practice to look for effective two-body potentials which inc lade


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Atom & Molecules the contributions or tne n-uouy potentials in an average way. For very accuratecalculations, we have to include the 3-body forces.

5.7 S U M M A R YIn this unit, our studies began w ith comm on non-covalent interactions such as theionic interactions that are present in crystalline lattices. This lead to a cha racterizationof the comm on types of crystal lattices. The structure of these lattices can be obtainedfrom the diffraction experiments. The dim ensions of the unit cells can be obtained byanalysing the diffraction patterns arising from the va rious faces or the planes of thecrystal.Th e study of the non-ionic interactions lead to the forms of different types ofintermolecular forces that are important in determining the collective behaviour ofmolecular systems. It is well known that if intermolecular forces were absent, thenthe earth would be a collection of ideal gases, and among other things, this unit wouldnot have been there! Different intermolecular interactions have been discussed, theirformulae have been provided and examp les have been given so that you are ab le touse them in calculations. All of technology (other than nuclear technology) isapproaching the molecular level and a good u nderstanding of intermolecularforces is becomin g more and more critical with the passage of time. As a specialcase of intermolecular interactions, hydrogen bonding was discussed andseveral examples of hydrogen bonding were given. They range from liquid water toDNA. Molecular properties, along with inter molecular forces determine themacroscopic properties of bulk systems such as solids and liquids.The use of intermolecular forces in understanding the behaviour of liquids wasillustrated through the technical details of the molecular dynamics simu lations. TheVerlet algorithm for classical molecular dynam ics was given and examples of the useof the algorithm for oversimplified systems were given. For larger systems acomputer is necessary, but often not sufficient!

5.8 GLOSSARYIonic SolidsUnit CellCrystalline Lattice :Lattice Energy :Bragg's Law

Packing Fraction :Intermolecular Forces :

Solid form of m atter wherein the constituent species areions.The smallest arrangement of atoms/molecules, which onrepetition in space yields the lattice.The structurelarrangement that results w hen.the unit cellis repeated in space.The en ergy change for the process of ions leading to.the.formation of the Lattice.2d sin 8 = nh where d is the distance between layers ofa crystal, 8 is the angle of incidence of X-rays of wavelength h and n is the order of reflection which is usually1.The fraction of volume of the unit cell occupied by theatomslmolecules of the unit cell.Th e weak forces between molecules such as thedipole-dipole, ion-induced dipole, an d dispersion forceswhich do not lead to the formation of a chemical bondbetween the molecules.= 2,q, where q, is the charge at the ith site in amolecule and 3 s he distan ce vector of that site from

the center of mass.Th e extent of distortion of a mo lecular chargedistribution in the presence of an electric field defined


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-9through~,,,,,,a E, where $is the induced dipolemoment and ?is the applied field.

Hydrogen Bonding : The interaction of the order of 3 to 10 kcallmol,resulting when a hydrogen atom is bonded to anelectronegative atom (such as 0,N, F) and placed nearanother electronegative atom about 2 A away from it.

Molecular Dynamics : The motion of molecules determined by the nature offorces acting on the molecules

Verlet Algorithm : The algorithm x,,, = 2 x - x,,, +-x (602 where themnew position is determined in terms of the current andold positions, the force at the current position and thetime step 6t

SP C Model of Water : A model of water wherein the charges on 0 and H are-0.8476e and 0.4238e respectively, r,, = 1 A, and theHOH angle is 109" 28'. The oxygen atom of one watermolecule interacts with the oxygen atoms of the otherwater molecules through a 12-6 potential. Theinteraction energy between two SPC water moleculesconsists of one 12-6 potential between the oxygens and9 Coulombic terms.

5.9 ANSWERS TO SAQs

Non-Covalent nteraci;;?c

To draw the BCC unit cell, add an atom at the center of the simple cubic unitcell. To obtain an FCC unit cell, start with a simple cubic (SC) unit cell andadd a particle at the center of each face of the SC unit cell. Start with atetragonal unit cell and add a particle at its center to obtain a body centeredtetragonal unit cell. A lattice is an infinite array of points in space. If at each

7 lattice point of a periodic lattice, a unit cell of atoms/molecules (which is therepeating unit) is placed, then the result is a crystalline lattice.S A Q 2.

a) The factor of 112 is used because we need one mole of chlorine atomsfor which we need only 112 mole of C1, molecules.b) For two like charged ions of charge Z,e and Z,e, the function tc plot is~ , ~ , e ~ / r .his is similar to plotting the function llr vs r.C) Assume a bond length of 2.8 A. The energy change for the processNa (g) +C1 (g) ---9 Nu+ .. .C t is

e2 1394 4I (Na)- A (Cl)-- 494 - 364- kJ/mol =- 368 U/mol2.8 A 2.8To get the energylenthalphy of formation of gaseous NaCl from Na (s) and C1,we have to add the heat of vapourization of Na and the correspondingdissociation energy of Cl,.

S A Q 3.In a simple cubic lattice, there is one atom per unit cell. If the radius-of thesphere is r, and the edge length of the cube is a , then, the volume of the sphere

and the volume of the unit cell = a3 .The packing fractionR-- -- - 0.523..6


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Atom & Molecules S A Q 4.Molecule Critical Critical Pressure Triple Point, KTemperature, K (atm)Water 647.3 218.5 273.16ammonia 405.5 11 1.5 195.5benzene 561.6 47.9 278.5CCl,F, 384.2 39.6 118.3

S A Q 5.

-++The dipole moment = q d.The interaction energy U= -y .E andR = distance between the ion and the center of the dipole.u = - Qq Q( R - 6/2) cos 9 ) + (R + (6 cos 9)'where Q = charge on the ion. Expand the denominatoras a Taylor series.-bSince-


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SAQ lo . Non-CovalentInteractionsChoose the units of (112)rn2 the same as that of kT. ind the value of v forwhichf (v) is maximum and calculatef (v)for values of v in steps of 0.25 v,wheref (v,) =f*,,

+ N a u,F on particle. = - C --=, for pairwise forces.i = 1 a r j

SAQ 12.Particle No. 1 ParticleNo. 2Components

Step 0 ForcesPositions

e VelocitiesStep 1 Forces

PositionsVelocitiesStep 2 ForcesPositionsVelocities

Step 3 . ForcesPositionsVelocities '

SAQ 13.The OH bond moment in SPC water is 0.4238 x 4.8D = 2.034D. he dipolemoment of water = 2 x 2.034 x cos - 109.5~ = 2.35D. To reproduce ther: 1/ 1 86actual dipole moment of water, the charge on H has to be 0.4238 x'- 0.3354 e2.35and that on 0 has to be - 0.6708 e.Repeat this calculation with the charge on .oxygen = -0.82e and the charge on hydrogen 0.41e..

SAQ 14.The pair distribution function g(r) is defined as the ratio of the'lccal density ofparticles at a distance r (from a given particle) and the bulk density. As r -+ 0,g(r)+ and as r + , g(r)+ 1. In fact, the structure of most normal liquids isshort ranged and g(r) + 1 beyond 10 to 15 A. For ideal gases, g(r) + 1 for allvalues of r. The OH running coordination number no, (r ) gives the number of4H atoms in a volume of - ~c3 surrounding an oxygen atom.3

SAQ 15.Label the particles as 1,2, ...N. Assume only pairwise interactions. For a twoparticle system the interaction is u12 (r). For a three particle system, it isu12 (rI2)+ u13 (r13)+ UZ3 (r23).For an N particle system, there are N (N - 1)/2terms and the formula is


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SUGGESTED READINGSAtomic Sp ectra and Atomic Structure, G. Herzberg. Dover Publications,Inc. ,New York (1944).Textb ook of Physical Chem istry (Second Edition), S. Glasstone, MacmillanStuden t Edition, New Delhi (1974).Fundamentals of Physical Chemistry, S.H.Maron and J.B.Lando, MacmillanPublishing Co. , nc . ,New York. (1974).Introduction to M odern Physics. F.K.Richunyer ,E.H.Kennard andJ.N.Cooper, T ata-McGraw-Hill Publishing Co. Ltd. ,New Delhi (1976).University Chemistry (Third Edition), B. H. Mahan, Narosa PublishingHouse, New Delhi (1985).Coulson's Valence (Third Edition). R. McWeeny, ELBS and the OxfordUniversity Press, Oxford (1979).Physical Chem istry (Fourth Edition), P. W. Atkins, ELBS and the OxfordUniversity Press. Oxford (1991).Physical Chemistry (Fifth Edition), G. M. Barro w, Narosa PublishingHouse, New Delhi (1990).Fundamentals of Molecular Spectroscopy (Third Edition), C. N. Banwell.Tata McGraw-Hill Publishing Company. N ew Delhi (1985).Th e Nature of the Chem ical Bond (Third Edition), L. Pauling, Oxford andIBH Pu blishing Comp any, Calcutta (1967)Chemistry of the Elements, N. N. Greenwoo d and A E arnshaw, PergamonPress, New York (1989).Valen ce Theo ry, J. N. Murrell, S. F. A. Kettle and J. M. Tedd er, ELBS andJohn W iley & Sons, London (1977).Quantum Chemistry, D. A. McQ uarrie, University Science Books and theOxfo rd University Press, Oxford (1983).Physical Chemistry. R. S. Berry, S. A. Rice and J. Ross, Academ ic Press,New York (1980).Intermolecular Forces: Their O rigin and Determination, G. C. M aitland, M.Rigby, E. B. Smith and W. A. Wakeham , Clarendon Press, ox fo rd (1981).Physical Chemistry (Third Edition), G. W. Castellan. Narosa PublishingHou se, New Delhi (1990).Principles of the Solid State, H. V. Keer. Wiley E astern, New De lhi (1993).Comp uter Simulation of Liquids, M. P. Allen and J. Tildesley , OxfordUniversity press (1987).

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