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# Physical Chemistry CHEM 3510

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This 52 page Class Notes was uploaded by Mr. Clementine Gottlieb on Wednesday October 21, 2015. The Class Notes belongs to CHEM 3510 at Tennessee Tech University taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/225695/chem-3510-tennessee-tech-university in Chemistry at Tennessee Tech University.

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CHEM 3510 Fall 2006 Unit I Introduction A Introduction to Physical Chemistry 1 Physical Chemistry is the part of chemistry dealing with application of physical methods to investigate chemistry Physical Chemistry main subdivisions are 3 Quantum Mechanics El deals with structure and properties of molecules b Spectroscopy El deals with the interaction between light and matter c Computational Chemistry El deals with modeling chemical properties of reactions using computers d Statistical Mechanics El deals with how knowledge about molecular energy levels or microscopic world transforms into properties of the bulk or macroscopic world e Thermodynamics El deals with properties of systems and their temperature dependence and with energetics of chemical reactions f Electrochemistry El deals with processes in with electrons are either a reactant or a product of a reaction g Chemical Kinetics El deals with the rates of chemical reactions or physical processes CHEM 3510 Fall 2006 B Classical Physics Review 1 2 Classical Physics was introduced in the 17th century by Isaac Newton At the end of 19th century classical physics mechanics thermodynamics kinetic theory electromagnetic theory was fully developed and was divided into a the corpuscular side or particle domain the matter b the undulatory side or wave domain the light Some useful classical physics equations a Total energy E E K V o Kis o Vis b Kinetic energy K K zlmvz 217 2 2 2m 0 m is the mass 0 v is the velocity or speed 0 p is the momentum c Frequency V Greek letter nu c N a V CV l 27 O xlis o cis o 17is O wis Classical mechanics was successful in explaining the motion of everyday objects but fails when applied to very small particles These failures led to the development of Quantum Mechanics CHEM 3510 Fall 2006 C The Classical Wave Equation 1 It is a prelude to Quantum Mechanics because it introduces or reminds you concepts that are similar to the ones in Quantum Mechanics The classical wave equation describes various wave phenomena a a vibrating string b a vibrating drum head c ocean waves d acoustic waves The classical nondispersive wave equation for a ldimensional wave 62uxl i 62uxl 6x2 v2 52 El uxl is El v is El Iis a The classical wave equation is a partial differential equation a linear partial differential equation because uxl and its derivatives appear only to the first power and there are no cross terms b The x and I are c The uxl is Example A ldimensional wave describing the motion of a vibrating string 3 E 3 E a The displacement uxl must satisfy certain physical conditions the amplitude should be zero at the end of the string El El CHEM 3510 Fall 2006 b These conditions are called because c To solve the differential equation we assume that uxt factors into a function of x times a function of t d This technique or method is called e Solving further the equation 7 Substituting ux t in the equation above 2 2 3 Rod X2x d T20 alx alt 7 Dividing by ux t X xTt 1 d2Xx 1 1 d2Tt K X06 dxz v2 T0 altz 7 In order for this equation to be true for every x and t each side should be equal to a 1 X V2 96 constantK called the separation constant 7 The problem of nding ux t transformed into two problems of nding X x and Tt by solving the following linear differential equations with constant coefficient they are ordinary differential equations m mzo abc2 m Kv2Tt0 altz f Solving forXx 3 7 Trivial solution is obtained that is X x 0 if K 20 7 IfK lt 0 set K 2 is real 2 2 d X2 2Xx0 dx 7 The general solution for this equation is X xcle c czeil c 7 Considering Euler equation em cos x i isin x 3 XxAcos xBsin x 7 This solution of X x should verify the boundary conditions X00 3 A 0 Xl0 3 Bsin l 0 3 l nn39 where n 1 2 CHEM 3510 gt gtlt E 4 Fall 2006 Look more closely to the solutions Number of wavelength that ts in 21 gtgt gt By generalization lt o This is called The solutions are a set a functions called Xnx Bn sin x Bn sinz x l xln Also a 27wn 2 m2 m 11010 Where 010 g are called in l CHEM 3510 Fall 2006 h Solving for T I but keeping in mind that 8 2 n7 2 dLZUU 2V2Tt0 dt 7 Similar to above the solution is mz39V l TtD cos wnt Esin wnt where a V i Coming back to uxt uxt XxTt mac 3 uxt Fcos mntGs1na2nts1nT n 1 2 7 There is a ux t function for each n so a better notation would be unxt Fn coswnt Gn sin mntsin n 1 2 7 The sum of all un x t solutions is also a solution of the equation This is called the principle of superposition The general solution is 00 uxt Z Fn cos wnt Gn sin wntsin nl 7 Make the transformation F cos at Gsin at A cosat where Greek letter phi is the phase angle andA is the amplitude of the wave 7 Rewrite the general equation as 00 n 71x 00 gt uxl Z An cosanl n s1n Zun xl n 1 711 El Each un x I is called 0 O O 0 j The time dependence of each mode represents a harmonic motion of frequency vn 3 2 2 i where the angular frequency is 7 m 27zv wnz V27zvn 12 CHEM 3510 Fall 2006 k Solutions El First term is A1 cosl 1s1n 7 quot quot 0 First term is called 0 The frequency is V1 2 V 21 El Second term is A2 coszT VI 2sin 277m 0 Second term is called 0 The frequency is V2 2 Vl O The midpoint has a zero displacement at all times and it is called El Third term is A3 cos3T VI 3sin 377 quot39 Q ll r J I I I u 139 39 i I x 0 Third term is called 0 The frequency is 13 3V 21 o This term has El Fourth term is A4 cos4T VI 4 sin4T7DC v 1 39 l 39 I 1 r x 139 39 CHEM 3510 Fall 2006 1 Let s consider now the case of ux I cosa1l sin c0sazl 3 sin 277 qt 0 NH a 1 4 El This is an example of a sum of standing waves yielding m Thinking backwards any general wave function can be decomposed into n The number of allowed standing waves on a string of length 1 El increases as the wavelength decreases 2 the possible high frequency oscillations outnumber the lowfrequency ones 21 1 7 Consider that I gtgt 2 so we can approximate the set of integers n by a continuous function 712 71 3 dn 2 ld t in 12 7 The negative sign indicates that the number of standing waves decreases as 1 increases 0 The number of standing waves in an enclosure of volume Vuse c not v for the speed 2 dn d t but 13 and v3 dV idl 2 dl Ldv 14 v A 12 c 2 dn l wv vzdv l4 C c3 CHEM 3510 Fall 2006 5 Example A 2dimensional wave equation the equation of a Vibrating membrane y 32M 32M 1 3211 b Where uxyl 6x2 322 v2 52 a Solving this equation x 7 Similar to the onedimensional problem use separation of variables mJ FxyTt 1 dZT 1 62F 62F 2 2 2 2 2 5 v Tt dt F Jay ax 6y 7 Use separation of variables for Fx y Fwy XxYy 1 d2Xx Ldz y X05 abc2 Y0 aly2 7 Solve two equations 1 d2Xx M M Yo dyz 7DiVideby Fxy 20 2 zand 1dYltygt 2 where p2 q2 2 7 Solutions for X x and Y y are XxBsin n 12and YyDsin m 12 2 2 n m Where nm 7239 2 a b 7 Solution for Tt Tnm t Enm cos mnmt an sin mnmt Gnm cosanmt nm N n2 m2 where mnm V nm Wt a2 b2 b The general solution for ux yl xayalj 2020 Eunmbcayal nlml 00 00 mzx m Z Z Anm cosanmt nms1n s1n y nlml a b CHEM 3510 Fall 2006 c Again the general function is a superposition of normal modes unm x yt but in this case one obtains 1 Examples e The case of a square membrane a b the frequencies of the normal modes are given by V7Z39 wnm n2 m2 a El For the cases ofn 1 m 2 and n 2 m 1 one can see that 5V7 0 12 Z 0 21 2 although M12 x yt i L121 x yt f This is an example of CI The frequency 0112 0121 is El This phenomenon appears because of the symmetry 01 b cuiEi43510 Fm12006 D Unit Review L hnponantTennnu ogy equency vvavelength vvavenuaner angukuiiequency independentva ables boundary conditions separation of variables eigenfunctions eigenvalues stationary wave Havehnngave node degeneracy CHEM 3510 Fall 2007 Unit 111 Postulates and Principles of Quantum Mechanics A The Schrodinger Equation 1 Introduction 3 It is the fundamental equation of quantum mechanics b The solutions of the timeindependent Schrodinger equation are called stationarystate wave functions 2 Timeindependent Schrodinger equation a Schrodinger equation is the equation for finding the wave function of a particle and come up based on idea that if the matter possesses wavelike properties there must be a wave equation that governs them b Schrodinger equation cannot be demonstrated it can be seen as a fundamental postulate but it can be understood starting from classical mechanics wave equation 7 Classical wave equation 6211 l 6211 6x2 V2 BIZ 7 The solution is uxt yx cos wt 7 The t dependence appear as cos wt or Tt or eXp2m39Vt 7 The spatial amplitude of the wave 1 is obtained from the equation 2 2 Ww zx 0 where a 272V 272391 abc2 V2 d 2 2 dx 1 7 Rearrange the equation considering 2 EKV V 2 p1l2mE V m g L p 2mE V 2 2 W Z E Vowx 0 dx h 31 CHEM 3510 Fall 2007 2 2 h d x 2 Vxvx Ewx 2m dx c This is the d The solutions wave functions of this equation are called e Schrodinger equation for three dimensions El Rewrite the equation and generalize to three dimensions h2 a2 a2 a2 wltxyz 2m 6x2 332 522 Vx y Zvx y 2 E We y 2 El Rewrite including the notation for 2 2 2 a a 6 2 2 2 15 3x 3y 32 0 An operator is a symbol that tells you to do something a mathematical operation to whatever function number etc follows the symbol h2 2 gt V Vxyz VIEVI 2m El Rewrite including the notation for h2 2 A 0 V VxyzH is 2m gtHlEl This is the simple form of the Schrodinger equation 32 CHEM 3510 Fall 2007 3 Eigenvalueeigenfunction relation in quantum mechanics 3 The eigenvalueeigenfunction relation in quantum mechanics is written in the form 11 11x am Where 121 is x is a is b The wave functions are eigenfunctions of the Hamiltonian operator and the total energy is the eigenvalue 4 Interpretation of the wave function 11 a The 11x11xclx is the probability that the particle to be located between x and x dx the onedimensional case b The function 11 is the of the wavefunction 11 CI The is obtained by replacing i to z39 in the wave function 11 CI The 1111product becomes real 33 CHEM 3510 Fall 2007 B Postulates of Quantum Mechanics l Enunciations a Postulate l The state of a quantummechanical system is completely specified by a function Vx that depends upon the coordinate of the particle All possible information about the system can be derived from lxx This function called the wave function or the state function has the important property that yx Vxclx is the probability that the particle lies in the interval alx located at the position x Note This principle can be stated using the timedependent wave function lI xl instead of the timeindependent wave function Vx Postulate 2 To every observable in classical physics there corresponds a linear Hermitian operator in quantum mechanics Postulate 3 In any measurements of the observable associated with the operator 121 the only values that will ever be observed are the eigenvalues an which satisfy the eigenvalue equation 11 1 2 an 1 Postulate 4 If a system is in a state described by a normalized wave function 11 then the average value of the observable corresponding to Ill12l yalx all space 121 is given by a Postulate 5 The wave function or state function of a system evolves in time according to the timedependent Schrodinger equation me z m w a Postulate 6 All electronic wave functions must be antisymmetric under the exchange of any two electrons Note The principles above are stated for a onedimensional space To generalize to three dimensions one can replace Vx with Vxyz and lI xl with lI xyzl 34 CHEM 3510 Fall 2007 2 First postulate of quantum mechanics a Classical mechanics deals with quantities called dynamical variables position momentum angular momentum and energy A measurable dynamical variable is called The movement of a particle can be described completely in classical mechanics This is not possible in quantum mechanics because the uncertainty principle tells one cannot specify or determine the position and the momentum of a particle simultaneously to any desirable precision This leads to the first postulate of quantum mechanics Postulate l The state of a quantummechanical system is completely specified by a function Vx that depends upon the coordinate of the particle All possible information about the system can be derived from txx This function called the wave function or the state function has the important property that yx Vxalx is the probability that the particle lies in the interval alx located at the position x This principle can also be stated using the timedependent wave function lI xt instead of the timeindependent wave function Vx The onedimensional probability will be replaced by the three dimensional probability 1 x yzwxy zdxdydz The case of two particles 1 x12x2Vx1 2x2dx1dx2 The total probability of finding a particle somewhere must be unity If xvxdx 1 all space CI The wave functions that satisfy this condition are said to be 35 CHEM 3510 Fall 2007 D The functions are called if Iyxyxdx A 1 all space because they can be normalized once they are divided by xZ h Other conditions for the wave function 1x Vfl its first derivative dug06 and its second x 2 derivative d VAX should be h 011x2 0 cl o b O W D If these conditions are met the wave J l functions is said to be 3 Second postulate of quantum mechanics a Postulate 2 To every observable in classical physics there corresponds a linear Hermitian operator in quantum mechanics b An operator is a symbol that tells to do a mathematical operation to whatever follows the symbol D The operators are usually denoted by a capital letter with a little hat over it called a carat like in H D What follows the operator is called c An operator 121 is if CHEM 3510 Fall 2007 d An operator 11 is if it has the property of being linear and if 1 A A lVIAVde lV2AV1dx all space all space A A or I f xAgxdx I gxAfx dx all space all space for any pair of functions 4 and 112 or f and g representing a physical state of a particle El These eXpression are true in derivative form as well e All the quantum operators can be written starting from the operators in the table below using classical physics formulas ClaSSical QM EXpression for operator Variable Operator x X x A h a a P P h x x i 3x 3x I T or I z 2 E 13 lim or h vhnxy 139 a a 2m El Examples p2 0 Kinetic energy K 2m A 2 2 2 1dimensional K p h a 2m 2m 5x2 2 2 2 2 2 3dimensional K h a aa hV2 A 0 Potential energy V V CHEM 3510 Fall 2007 0 Total energy E K V The operator for the total energy is the Hamiltonian Ii 2 2 2 2 2 HI V h a a a V2 h V2 V 21 6x2 322 622 21 o The angular momentum along the x aXis Lx ypz Zpy A 6 6 Lx thE 23 f Commutation El It is a property of operators in which 1213f x is compared to 321 fx 0 A3fx A1 fx O BAfx BAfx El Operators usually do not commute Al f x i BAf x El When 1213f x 3121f x for every compatible x the operators 121 and 3 are El Example 121 i and B x2 dx gt 2xfx x2 diff A A dfx BA 2 2 fx x dx El Rewrite and drop x El Define the as 21 1 211 321 CI The is the zero operator 211 B21 2 3 where 0 is the zero operator 38 CHEM 3510 Fall 2007 g A special property of linear operators for example the operator 121 is that a linear combination of two eigenfunctions of the operator with the same eigenvalue is also an eigenfunction of the operator El Consider that two eigenfunctions have the same eigenvalue 21m a 1 and M am This is a twofold degeneracy El Then any linear combination of 1 and 2 is also an eigenfunction of 121 21c1 1 02 2 c121 1 21ng acl 1 02 2 El Example d2fx 2 m 7 Solvmg 2 m f x gives the eigenfunctions f1x e and dx f2x em 7 A linear combination of f1 and f2 f12x clflx czfx x is also an eigenfunction 2 d clm2elmx 62m2eilmx dx m2clem cze i m2f12x 4 Third postulate of quantum mechanics a Postulate 3 In any measurements of the observable associated with the operator 1 the only values that will ever be observed are the eigenvalues an which satisfy the eigenvalue equation 21 Vln an Vln b For an experiment designed to measure the observable associated to 121 we will find only the values a1 a2an corresponding to the states VlaVZawlln and c Example If 21 if gt flux 2 En lln Schrodinger equation and only the En energies eigenvalues will be experimentally observed 39 CHEM 3510 Fall 2007 5 Fourth postulate of quantum mechanics 3 Postulate 4 If a system is in a state described by a normalized wave function 11 then the average value of the observable corresponding to 11 is given by El 0 J 11l ydx all space a is 2 b Determine the variance 0 of eXperiments El El Variance is a statistical mechanics quantity Assume we have A lln x 2 an lln x 2 ltagt w xvi wnltxgtdx an Also 212 x illi wn x am 00 lta2gt Tw2xawnltxdx a3 00 Variance of the eXperiments become 2 2 2 2 2 0a lta gtltagt an an 0 The standard deviation 0 3 is zero so the only values observed are the an values 40 CHEM 3510 Fall 2007 6 Fifth postulate of quantum mechanics a Postulate 5 The wave function or state function of a system evolves in time according to the timedependent Schrodinger equation A 3 I H Pxt 172 x at b Time dependence of l xt 7 Assume if does not depend on time 7 Separation of variable I xt zxft 1 A m 2 ft 3 W W f0 dr 3 lilyx E zx timeindependent Schrodinger equation WC 2 1 dt h 70 7 The solution is ft eTiEIh ailwt where E hv ha 7139Ent E where E is a constant gt Pxl yxe h assuming a set of solutions En c The probability density and the average values are independent of time the function of time cancels out x 0 max VxeiEquotthVxeiEquotthdx wxwltxazx gt 11 x are called 41 CHEM3510 Fall 2007 C More Properties Propem39es of the Eigen mctions of Quantan Mechanical Operators 1 3 U o Fquot The eigenfunctions of QM operators are The eigen mctions of QM operators are mctions satisfying the equation Elyn any Where an are real although 21 and 11 canbe complex When the operator is Hermitian If the eigenvalues an are real then eigenfunctions have the property no in IllMXWMXW 0 foo D Ifa set ofWave functions satis es this condition then the set is said to be or that the Wave mctions are no D Ifthey are also normalized IV xxIn xdx 1 than the set is foo said to be and Wave functions are Genemlizing no in fully101x41 foo 1ifij Where 5 1s 01fi j Example sin and cos mctions are onhogonal An even function is always orthogonal to an odd function E An even mction is a mction for which E An odd function is a function for Which The eigenfunctions of QM operators form CHEM 3510 Fall 2007 2 Uncertainty Principle written based on operators 3 The uncertainty in the measurements of a and b 0a and 0b are related by l A A 5 IV xABVxdx where 1213 2 1amp3 3121 is the commutator El If 21 and f3 commute then 1213 0 then 7an 2 0 ltgt a and b aaab 2 can be measured simultaneously to any precision El If 21 and 3 do not commute then a and I cannot be simultaneously determined to arbitrary precision b Example El For x the operator is X x El For px the operator is 13x 2 472 CI The commutator of X and 13x is 13xX 13x X13x ihf Where f is the identity operator multiplication to l 2 apex 2l1wxlt miwxxdxl 30p Where 0x ltx2gt ltxgt2 and ap1lltp2gtltpgt2 0x 2 z39h 2 43 CHEM 3510 Fall 2007 D Unit Review 1 Important Terminology Schrodinger equation wavefunctions operator operand Laplacian operator Hamiltonian operator eigenvalueeigenfunction relation compleX conjugate QM postulates wavefunction interpretation Hermitian operator observable average value 44 CHEM 3510 Fall 2007 norm alizednormalizable commutation commutator linear operators variancestandard deviation orthogonal orthonormal even function odd function complete set 45 CHEM 3510 Fall 2006 Unit IX Bonding in Polyatomic Molecules A Hybrid Orbitals 1 Introduction a The concept of hybrid atomic orbitals was introduced to interpret the b The concept of hybrid orbital was also introduced in general chemistry to explain why C is tetravalent El Recall that carbon atom with the configuration 1s2 2s2 214219 may be eXpected to form only 2 bonds El An electron is promoted from the 2s orbital into the 2172 orbital CI The four singly occupied orbitals combines to form four equivalent hybrid orbitals 2 The Sp hybridization a Example BeHZ molecule El two Be H bonds that are equivalent El H Be H angle is 180 b Beryllium electronic configuration 1s22s2 term symbol CI The 2s and 2172 orbitals combine to form two Sp hybrid orbitals atomic orbitals 1 Vsp es i 2172 133 CHEM3510 3 c Fall 2006 The molecular orbitals are formed as a abonding between one 5p hybrid orbital ofBe and the s orbitals of one H atom V BerH Suimas ElliB e2 p2 CsWHas Vim The rpz hybridization a b c d Example BH molecule El three BiH bonds that are equivalent El HiBiH angle is 120 The appropriate hybrid orbitals ie orbitals with proper orientation are constructed by combining the 2 orbital and two 2p orbitals resulting in three rpz hybrid orbitals The amolecular orbital in BH is formed as a linear combination between one hybrid orbital on B and the 15 orbitals of one H atom The orthonormal normalized and orthogonal rpz hybrid orbitals are 1 2 Zr 2 W J 3 p2 22 g2 954 W 43 J3 J5 1 1 iniiZ iiz 3 J3 Jgpz J5 1 lt CHEM 3510 Fall 2006 4 The Sp3 hybridization 3 Example CH4 molecule El four C H bonds that are equivalent El H C H angle is 1095 b Four Sp3 hybrid orbitals are obtained combining the 2s orbital and three 2p orbitals of C The orthonormal Sp3 hybrid orbitals are la 1112s2px 2py 2192 m2s 2px 2py2pz v13 2s2px 2py 2pz W4 2s 2px 2py 2pz 5 The sp3a 2 hybridization a Example SF6 molecule El SiX S F bonds that are equivalent El F S F angles are 90 and 180 CI The geometry is octahedral b The groundstate electron configuration of S is Ne3s23p4 c For S combine 3s orbital with the three 3p orbitals and two 3d orbitals to create hybrid orbitals d An hybridization gives similar results 135 CHEM 3510 Fall 2006 6 Other hybridizations 3 Example H20 molecule El Two O H bonds that are equivalent El H O H angle is 1045 b This geometry is described by the hybrid orbitals that have 1045 between them 111 0452s07l2py 0552pz m 0452S 07l2py 0552pz y H H 2 El These hybrid orbitals are intermediate between the pure p and the sp3 hybrid orbitals El There are two other hybrid orbitals that contain the lone pairs of the oxygen El All these hybrid orbitals are orthonormal 7 The s and p character of hybrid orbitals a For a hybrid orbital whose wavefunction is a combination of s and p atomic orbitals written as 11 cnsns cnpx npx cnpy npy cnpz npz o The s character of the orbital is cns2 O The p character of the orbital is cnpx2 cnpy2 cnpz2 El To understand this result look at the electron density associated with this orbital 2 2 2 2 2 IV dT Cns cnpx cnpy cnpz 1 0 cm2 is the amount of the electron that is in the s atomic orbital giving therefore the s character 136 CHEM 3510 Fall 2006 El Example the hybrid orbitals in BH3 have 2 1 o 033 s character 5 1 2 1 2 2067 character llgl llzl P El Example the hybrid orbitals in H20 have 0 s character 0 p character b A hybrid orbital that is x s character and 100 x p character can be said to be an hybrid orbital El Example the bonding hybrid orbitals in H20 are c An spy hybrid orbital is an orbital that is s character and p character El Example the Sp3 hybrid orbitals in CH4 have 0 s character 0 p character 137 CHEM 3510 Fall 2006 B Molecular Orbital Theory for Triatomic Molecules 1 Molecular orbital energy diagrams a Molecular orbital theory can explain molecular geometry of some triatomic molecules El BeH2 molecule is linear El H20 molecule is bent b The number of valence electrons on the central atom is important El Be has 2 electrons and O has 6 electrons c Obtain the molecular orbitals as a combinations of atomic orbitals on Be or O and H using an LCAOMO procedure d Molecular orbital energy diagram for BeH2 r Be H Be H El H The log MO same as the 1s atomic orbital of Be is not shown CI The 2px and Zpy orbitals of Be do not have the proper symmetry or orientation to interact with H atomic orbitals so they are nonbonding orbitals The configuration is 2 0g21 au2 138 CHEM 3510 Fall 2006 e Molecular orbital energy diagram for H20 O H The Zpy orbital of 0 start interacting with the atomic orbitals of H The configuration is 2a121b223a121b12 El El Only the 2px orbital of O in nonbonding El CI The labels for a bent molecule 239 h ie a1 a2 31 b 2 re ect the quot symmetry of the molecule 1quot i quot I H 2 Walsh correlation diagrams am M a A Walsh correlation diagram is a plot 1 E i 1quot Jr 1 1 of the energy of a molecular orbital 5 as a function of a change in i g Zli l a b The Walsh correlation diagram for an 39 1d AH2 triatomic molecule v fr 5 5D Bond angle 139 CHEM 3510 Fall 2006 The molecular geometry if a molecule is linear or bent depends on which energy is lower Bending destabilizes 20g 10 30g and Zau orbitals Bending stabilizes orbital The case of BeHZ 2 0g21 0 2 configuration is more stable lower in energy than 2010201322 configuration CI The BeH2 molecule is linear The case of H20 2jg210u217zu4 con guration is less stable higher in energy than 201121l223al21bl2 configuration CI The H20 molecule is bent The 17 orbital get stabilized more than the 20g and la gets destabilized at angles close to 180 The exact angle of bending for example 1045 for H20 can be determined using more complicated computational techniques A Walsh correlation diagram for a XY2 triatomic O and Y are second row elements can be used to determine if C02 N02 N02 CF2 2 3 etc are linear or bent 39 i Itquot Molecules with r 3 39 J i 39 u 3 1 valence electrons are predicted to be 5 f1 1 Wt z 1 linear 339 b R 3a L Molecules with valence 3 M 4 electrons are predicted to be bent j A 1 3 a 34 30quot 530 Build angle 140 CHEM 3510 Fall 2006 C Huckel Molecular Orbital Theory 1 Introduction 3 It is a method applied mainly to conjugated and aromatic hydrocarbons b It uses the CI The Sp2 hybrid orbitals of C and s orbitals of H create a a bond framework CI The a bond framework is in xy plane CI The pz orbitals are perpendicular to the plane and form 7rbonds El One can say that 7relectrons are moving in an fixed effective electrostatic potential due to the electrons in the o This is the c The problem of 7relectrons which are delocalized MO occupied by 7 electrons can be treated separately from the problem of aelectrons 2 Application to ethene CH2CH2 a The wave function for 7rorbital in ethene W7 01 2172A 02 2172B b Solving for 01 and 02 using variational method results in the secular determinant H11 E511 H12 E512 H12 E512 H22 E522 El HU are integrals involving an effective Hamiltonian operator that includes interaction of nelectrons with aelectrons El Sij are overlap integrals between 2172 atomic orbitals c The H11 and H22 are El H11 H22 in ethene d The H12 integral is called 141 CHEM 3510 Fall 2006 e Huckel assumptions 1 f i2 El 0 1f 1 i El Hil are assumed to be same and denoted 0c 8 if iis bound to j El 0 if 1 is not bound to j 0 Use the same 3 value for each pair of neighbors f Rewrite the secular determinant 05 E 8 0 gtEai 8 oc E 398 g To get the energy one must have 05 and 8 El Instead of calculating those values one can determine their values from comparison with eXperiment El By doing so gt 8 75 kcalmol h Represent the energies of the 7rorbitals considering 05 as reference a is the energy of an electron into a 2172 orbital a a i The energy of the 7relectrons total 7relectronic energy E 2a82a28 El This is the energy of j Solve for 01 and 02 to get the orthonormal wavefunctions 2 2 2 2 El Conditions are 01 02 and 01 02 l l 1 W1 EZPZA 521723 1 1 W2 2PZA 21723 142 CHEM 3510 Fall 2006 3 Application to butadiene CH2CH CHCH2 a Consider that the molecule is linear although in reality it is not 1 2 3 4 4 b The wavefunctions for 7rorbitals 111 Z 0112 p22 1 j1 c These MOs created as a LCAO leads to the secular deterininant Hii ESii H12 E512 H13 E513 H14 E514 H12 E512 H22 E522 H23 E523 H24 E524 0 H13 E513 H23 E523 H33 ES33 H34 ES34 H14 E514 H24 E524 H34 ES34 H44 ES44 But H ag Hij 39 S6 0 U a E 8 0 0 E 0 3 3 06 3 0 0 8 oc E 8 0 0 8 a E oc E d Rearrange by factoring 8 and writing 2 x x 1 0 0 1 x 1 0 0 0 1 x 1 0 0 1 x e This deterininant leads to a fourthorder equation gt x4 3x2 1 0 f There are 4 solutions x i1618 andxi0618 05 E x gt E a Bx one obtains the four energies g Using associated with the four molecular orbitals 143 CHEM 3510 Fall 2006 a 06186 a 1618 The energy of the n39electrons also called the total n39electronic energy E The delocalization energy D This is the difference in energy between the total n39electronic energy and the energy of localized double bonds or radicals D The delocalization energy gives the relative stabilization of the molecule Edeloc Solve for the coefficients of the atomic orbitals in the molecular orbitals 07 to get the orthonormal wavefunctions 111 037172p21060152p22 060152pz3 03717 2pz4 112 06015 2p21 03717 2p22 03717 2pz3 06015 2pz4 13 06015 2p21 03717 2p22 03717 2pz3 06015 2pz4 114 037172p21 060152p22 06015 2pz3 037172pz4 J The n39MOs are delocalized over the whole molecule D 144 CHEM 3510 4 Application to benzene C5H6 a Fall 2006 The O39bonds between carbon atoms in the ring and between the carbon atoms and the hydrogen atoms form the aframework of benzene Write the secular determinant using thenotationxL x10001 lxlOOO 1 01x1000 6 2 001x10 5 3 0001xl 4 lOOle This determinant leads to a sixthorder equation x6 4x49x2 40 There are 6 solutions x i1 i1 i2 The MO energy diagram is given below The energy of the n39electrons E7 The delocalization energy Edeloc The wavefunctions for benzene top View 1 Wm J 0PM A a7 v J a u 2 23 a J v a2 an 145 CHEM 3510 Fall 2006 D Photoelectron spectroscopy 1 It is a method ofprobing the molecular energy ie the energy of molecular orbitals 2 Molecular orbital energy diagram for CH4 a The small circles represent the number of atomic or molecular orbitals Wm M u is x mm W MLmolquot 3 Molecular orbital energy diagram for C2H4 a This diagram explains the UV absorption of58500 cmquot 4 x 2 2 a Imnm rnsrgs Mimiquot CHEM 3510 Fall 2006 E Unit Review 1 Important Terminology hybrid orbitals Sp spz sp3 clasp3 hybridizations s and p character of hybrid orbitals Walsh correlation diagrams Huckel theory 7relectron approximation a bond framework Hi39ickel assumptions total 7relectronic energy delocalization energy localized energy 147 CHEM 3510 Fall 2007 Unit I Introduction A Introduction to Physical Chemistry 1 Physical Chemistry is the part of chemistry dealing with application of physical methods to investigate chemistry Physical Chemistry main subdivisions are 3 Quantum Mechanics El deals with structure and properties of molecules b Spectroscopy El deals with the interaction between light and matter c Computational Chemistry El deals with modeling chemical properties of reactions using computers d Statistical Mechanics El deals with how knowledge about molecular energy levels or microscopic world transforms into properties of the bulk or macroscopic world e Thermodynamics El deals with properties of systems and their temperature dependence and with energetics of chemical reactions f Electrochemistry El deals with processes in with electrons are either a reactant or a product of a reaction g Chemical Kinetics El deals with the rates of chemical reactions or physical processes CHEM 3510 Fall 2007 B Classical Physics Review 1 2 Classical Physics was introduced in the 17th century by Isaac Newton At the end of 19th century classical physics mechanics thermodynamics kinetic theory electromagnetic theory was fully developed and was divided into a the corpuscular side or particle domain the matter b the undulatory side or wave domain the light Some useful classical physics equations a Total energy E E K V o Kis o Vis b Kinetic energy K K zlmvz 217 2 2 2m 0 m is the mass 0 v is the velocity or speed 0 p is the momentum c Frequency V Greek letter nu c N a V CV l 27 O xlis o cis o 17is O wis Classical mechanics was successful in explaining the motion of everyday objects but fails when applied to very small particles These failures led to the development of Quantum Mechanics CHEM 3510 Fall 2007 C The Classical Wave Equation 1 It is a prelude to Quantum Mechanics because it introduces or reminds you concepts that are similar to the ones in Quantum Mechanics The classical wave equation describes various wave phenomena a a vibrating string b a vibrating drum head c ocean waves d acoustic waves The classical nondispersive wave equation for a ldimensional wave 62uxl i 62uxl 6x2 v2 52 El uxl is El v is El Iis a The classical wave equation is a partial differential equation a linear partial differential equation because uxl and its derivatives appear only to the first power and there are no cross terms b The x and I are c The uxl is Example A ldimensional wave describing the motion of a vibrating string 3 E 3 E a The displacement uxl must satisfy certain physical conditions the amplitude should be zero at the end of the string El El CHEM 3510 Fall 2007 b These conditions are called because c To solve the differential equation we assume that uxt factors into a function of x times a function of t d This technique or method is called e Solving further the equation 7 Substituting ux t in the equation above 2 2 3 Rod X2x d T20 alx alt 7 Dividing by ux t X xTt 1 d2Xx 1 1 d2Tt K X06 dxz v2 T0 altz 7 In order for this equation to be true for every x and t each side should be equal to a 1 X V2 96 constantK called the separation constant 7 The problem of nding ux t transformed into two problems of nding X x and Tt by solving the following linear differential equations with constant coefficient they are ordinary differential equations m mzo abc2 m Kv2Tt0 altz f Solving forXx 3 7 Trivial solution is obtained that is X x 0 if K 20 7 IfK lt 0 set K 2 is real 2 2 d X2 2Xx0 dx 7 The general solution for this equation is X xcle c czeil c 7 Considering Euler equation em cos x i isin x 3 XxAcos xBsin x 7 This solution of X x should verify the boundary conditions X00 3 A 0 Xl0 3 Bsin l 0 3 l nn39 where n 1 2 CHEM 3510 Fall 2007 Look more closely to the solutions Number of wavelength that ts in 21 H A I V El By generalization o This is called CI The solutions are a set a functions called Xnx 2 En sin x 2 En sinz x l xln 27zv mzv El Also angular frequenc1es a 27w 2 T 11010 Where 010 g are called CHEM 3510 Fall 2007 h Solving for T I but keeping in mind that 8 2 n7 2 dLZUU 2V2Tt0 dt 7 Similar to above the solution is mz39V l TtD cos wnt Esin wnt where a V i Coming back to uxt uxt XxTt mac 3 uxt Fcos mntGs1na2nts1nT n 1 2 7 There is a uxt function for each n so a better notation would be unxt Fn coswnt Gn sin mntsin n 1 2 7 The sum of all un xt solutions is also a solution of the equation This is called the principle of superposition The general solution is 00 uxt Z Fn cos wnt Gn sin wntsin nl 7 Make the transformation F cos at Gsin at A cosat where Greek letter phi is the phase angle andA is the amplitude of the wave 7 Rewrite the general equation as 00 n 71x 00 gt uxl Z An cosanl n s1n Zun xl n 1 711 El Each un x I is called 0 O O 0 j The time dependence of each mode represents a harmonic motion of frequency vn 3 2 2 i where the angular frequency is 7 m 27zv wnz V27zvn 12 CHEM 3510 Fall 2007 k Solutions El First term is A1 cosl 1s1n T 0 First term is called 0 The frequency is V1 2 V 21 El Second term is A2 coszT VI 2sin 27706 0 Second term is called 0 The frequency is V2 2 Vl O The midpoint has a zero displacement at all times and it is called El Third term is A3 cos3T VI 3sin 37706 J x 39 I L l I k I s x 39 J 39 0 Third term is called 0 The frequency is 13 3V 21 o This term has 4 4 El Fourth term is A4 cosT VI 4 sinT x v 39 l 39 v i 39 39 I i I i I Q l l J l 39 CHEM 3510 Fall 2007 1 Let s consider now the case of 71x 1 7r 271x uxl cosa1l sm 7 Ecosa2I Es1nT 7r 1 0 ml 2 4 ale El This is an example of a sum of standing waves yielding m Thinking backwards any general wave function can be decomposed into n The number of allowed standing waves on a string of length 1 El increases as the wavelength decreases 2 the possible high frequency oscillations outnumber the lowfrequency ones n 7 Consider that I gtgt 2 so we can approximate the set of integers n by a continuous function 712 21 n 3 dn d t in 12 7 The negative sign indicates that the number of standing waves decreases as 1 increases 0 The number of standing waves in an enclosure of volume Vuse c not v for the speed 47239V 2 dn dt but 13 and v3 dV Ldl 2 dl Ldv 14 v A 12 c 2 dn imp vzdv l C c CHEM 3510 Fall 2007 5 Example A 2dimensional wave equation the equation of a Vibrating membrane b y 32M 32M 1 32M 2 2 Where uxyl ax ay V a a Solving this equation x 7 Similar to the onedimensional problem use separation of variables uxyt FxyTt 1 dZT 1 62F 62F 2 V2Tt 5112 F Jay 6x2 6y2 7 Use separation of variables for Fx y Fxy XXYy 1 d2Xx L 51210 X05 abc2 Y0 aly2 7 Solve two equations 1 d2Xx Xx dxz Yo dyz where p2 q2 2 7 Solutions for X x and Y y are 7DiVideby Fxy 20 2 zand 1dYltygt 2 Xx Bsin n 12 and YyDsin m 12 2 2 n m Where nm 7239 2 a b 7 Solution for Tt Tnm t Enm cos mnmt an sin mnmt Gnm cosanmt nm N n2 m2 where wnm V nm Wt a2 b2 b The general solution for ux yl ux3yal 2 Eunmx3yal nlml 0 0 mzx m Z Z Anm cosanmt nms1n s1n y nlml a b CHEM 3510 Fall 2007 c Again the general function is a superposition of normal modes unm x yt but in this case one obtains instead of d Examples e The case of a square membrane a b the frequencies of the normal modes are given by V7Z39 wnm n2 m2 a El For the cases ofn 1 m 2 and n 2 m 1 one can see that 5V7 12 21 2 although M12 x y I i M21xyr f This is an example of CI The frequency 0112 0121 is El This phenomenon appears because of the symmetry 01 b cuiEi43510 Fm12007 D Unit Review L hnponantTennnu ogy equency vvavelength vvavenuaner angukuiiequency independentva ables boundary conditions separation of variables eigenfunctions eigenvalues stationary wave Havehnngave node degeneracy

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