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# Physical Chemistry CHEM 3510

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This 65 page Class Notes was uploaded by Rebeka Zemlak MD on Wednesday October 21, 2015. The Class Notes belongs to CHEM 3510 at Tennessee Tech University taught by Scott Northrup in Fall. Since its upload, it has received 13 views. For similar materials see /class/225693/chem-3510-tennessee-tech-university in Chemistry at Tennessee Tech University.

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Chem3510 Lecture Notes Introduction 8 Chapter 8 S Northrup Preliminaries A Ovewiew of Physical Chemistry P Chem Quantum Statistical Thermodynamics Chemical Mechanics Mechanics Kinetics microscopic bridge conneciing macroscopic Alomicslr mleO amp macro st aw at cirxn Malecules 2nd Law Maenallranspon sumosmpy Phase transformations Chemical Eq B Dimensions and Uniis 1 See Appendix 1 in Atkins a 51 units base Table A11 b Derived SI units Table A12 c Prefixes Table A13 kilo mega giga tera milli micro nano pico femto 2 Common nonSI units a 1 1 10 E cm 10quot m 57 1 cal 4184J c 1 eV 16 x 10 19J F 1 atm 1013 kPa 760 Torr m 1 bar 105 kPa so 1 atm 1 bar C Mathematical Review See Appendix 2 1 Logs a if y 10quot Iogloy x b Ify ex In y x e271828 natural c log ab b log a d log 1 0 e log ab log a log b f log ab log a log b g 10ab 10310b Derivatives a xt position at time t v velocity a accel F dv dZX ma m7 m7 dt dtz force b 1x nxr 1 1 constant 0 dx dx c ifxgx fx gx gx fx chain rule dx dx dx 1 1 x d d aIn1x dx e ie x efx fx dx dx II 3 Integrals b b a fdxxl b a a a b b 2 2 2 x b a b fxdxi i i 2 2 2 a b a b c faxndx ixn n1 a a b b d fexdxexleb ea a a Review of Classical Mechanics Appendix 3 A Specifying state of a particle xt position at time t pt momentum mvt mass x velocity dx m7 dt 2 EK kinetic energy lmvz L 2 2m Vx potential energy fct of x Total E EKV 2 ELv 2m Newton s Law 2 F ma mass X acceleration mg or d p dt dt mamp dir2 Rotational motion Recall for linear motion p mv gt Analogous for angular J I o J angular momentum I moment of inertia mr2 00 angular velocity rate of change of angular position quot of Recall for linear F ma Angular T where T is torque Harmonic motion in 1 D Force F kx where k is force constant spring constantquot 2 Recall F md X dt2 2 SO kX md X dt2 A solution xt A Sinoot a standing wavequot where p 5 m plot 111291 oscill 4 lquot 41 x S quotemit frequency v 3 271 total energy of oscillator E 12 k A2 square of the amplitude Note since A can be continuously increased by knocking the oscillator harder this implies it s energy can assume any desired value Summary of Classical Physics 1 Classical particles follow precise well defined trajectories governed by Newton s 2nd Law 2 Their translation rotational or vibrational motions can be excited to any energy we so desire 3 Classical physics works beautifully in ordinary everyday experience I CHAPTER 8 Quantum Theory Introduction and Principles The Failures of Classical Physics A Particlelike Properties of Light Classical view of light Newton s optics was strictly a wave theory Wave characterized by wavelength A and frequency v which are related to speed c as for light c 30 x 108 m s391 c V 7 A Photoelectric Effect see below showed that light comes in particles we call photons photon discrete particle of light energy 01 Photon Ephoton hv where v is freq h Planck s constant 6626 x 103934J s Jouleseconds Breakdown of the Equipartition Theorem ET ET says at temp T average value of each quadratic term in the expression for total energy is kBTZ Example 2 E L ikx2 lD harmonic oscillator 2 quad terms E2gltBT kBT k3 Boltzmann s constant 138 x 103923 J K39l ET is saying that total energy is evenly distributed over all the degrees of motional freedom modes of motion available to molecules Will only be true if every mode can be excited to any value Turns out not to be true Breakdown occurs at low T or for molecular vibrations Spectral Distribution of Radiation from a Black Body Blackbody radiator can be imagined as follows 7 Detected radiation Pinhole Container at a temperature T Fi ure 84 Atkins Physkal Ch emislry Eighth Edition 7 2006 Peter Atkins and Julio de Paula Classical prediction of the output of the radiator Rayleigh Jeans Law fails catastrophically at low wavelength the ultraviolet catastrophe Rayleigh Jeans law Experimental Energy density p Wavelength A re 6 Atkins Physical Chemistry Eighth Edition 7 2006 Peter Atkins and Julio 19 Paula Planck 1900 successfully predicted experimental output spectrum ONLY by making the drastic assumption that the vibrational energy of the oscillators making up the walls of the cavity is limited to discrete values and cannot vary arbitrarily This limitation is called the quantization of enerqv The allowed amounts of energy of each oscillator having a fundamental vibrational frequency v is then Enhv n012 integer multiples of hv mm M if i JKY M7 ub cam L1 L9 Cd sQ M3 I avSE cm 048934 cilantro 777 M gt H A39 quot we gt Pkctmx M5 in JEL 0 Line Spectra of Atoms Observed that gaseous atoms only emit and absorb certain frequencies of light and thus only certain energies 53 Therefore atom will only interact with a light photon of certain hvEg EEZ energy to put the atom in the next W available energy level and it will 5 gt 2 v only emIt photons of the same g energy 5 M hv E2 E by E3 E1 lWlWWWWlWlWVlVWV 51 Fig ure 81 2 Atkins Physical Chemistry Eighth Edmun 39 a Photon with any other v won t absorbed or emitted Atom can t absorb part of a photon Bohr Frequency Rule for light absorption and emission Energy difference between two levels AE hv H atom Rydberg observed a pattern in the emission wavelengths f n2 Ri i where k is observed wavelengths in H spectrum n1 2 1 2300 n2gtn1 R 10967758 cm391 Rydberg constant Now know that n1 quantum number of the lower energy level n2 upper level n1 2 corresponds to Balmer series Note gt I l v c a v remember v C A V is the wavenumber cm391 are common units E The Photoelectric Effect Case a no electrons ejected Since particles of light have Insufficient energy Case b electrons are ejected With kinetic energy excess Energy Kinetic energy of ejected electron Energy needed to remove electron from metal hv Fi ure 84 4 Atkins Physical Chemisiry Eighth Edition 57 2006 Peter Atkll39lS and Julio de Paula 1 No e39 ejected unless hv gt I regardless of intensity of light source 2 KE of e39 ejected proportional to freq v and indep of intensity of light mv2 hv CI KE of e39 3 e39 ejected even at low intensities if v above threshold ie The light comes in little packets of energy photons or corpuscles and e39 reacts with them on one to one basis So light behaves like particles Diffraction of Electrons Electron beam Diffracted electrons Diffraction is characteristic of wave like behavior Here s the really astounding thing Beam of e39 behave this way Matter has wavelike proberties The Wave Particle Duality 1923 de Broglie matter has wave properties as well as particle properties 4 inversely proportional to its momentum p mv wavelength of a particle 1 e39 moving at 1100th light speed 3 xi Ogms 1 91 103931k p mv x gx 100 662 x103934JS 24 10391 m 242 A z s39 e of atom 273x10 24kgms X 392 Al p Therefore on atomic scale matter has wavelike properties Mack truck moving at 10 m s39l m 10tons4x104kg A 16 x 103939 m Extremely small No perceptible wavelike character II Quantum Mechanical Picture A Specifying the state of the system 1 No longer use x and v position and velocity as the fundamental variables that characterize a microscopic system Heisenberg Uncertainty Principle impossible to measure simultaneously x and v of a particle to any desired degree of precision In 1D AxAp 2 E rial 2n hba r uncertainty uncertainty in in position momentum HUP an outcome of more fundamental underlying reality To observe something you must act upon it What we are measuring becomes perturbed in process and we are simply seeing how it responds to our perturbation so we never see it as it exists in our absence In QM all information about system is contained in a wavefunction 11 instead of trajectory ll x t 11 is not a physical observable but stores info To get info out of 11 must mathematically operate on it ie To find energy E use the energy operator Fl called the hamiltonian operator For every physical observable there is a corresponding operator The fundamental equation of QM to solve is Schrodinger Eqn instead of Newton s 2nd Law In 1D for single particle in a potential energy V A h2 d2 V W Now let s operate Flzp 7E1p Schrodinger Equation energy 72 dzw d72vwEw x Take 1D single particle free space V0 2 2 Ld7wVwEw 2mdx2 Our job is to find 11 satisfying this equation eg 11 cos kx would work d2 7cos kx llt2 cos kx dx2 2 2 22 So h d cos kxk hcos kx 2m i ii kzhz 2m E Most general solution 1p eiquotX cos kx i sin kx 8 11 has very specific properties which narrow down the possible solutions a continuous b continuous slope E c single valued d never infinite over a finite region d e can be complex eg 1 oc eikX i cos kx i sin kx real imaginary part part But a real function P can be constructed from P 1p1p means complex conjugate eikX 1PgtIlt eikX 9 Meaning Ofll the Born interpretation 1p1p probability distribution for finding the particle at some point in space a lt dX Probability W wdx Wavefunction Probability Figure 8 21 Atkins Physical Chemistry Eighth Edition 6 2006 Peter Atkins and Julio de Paula While 11 is not observable 1p1p is observable eg an atomic orbital is a spatial prob distrib for the e39 10 Normalization condition Probability distribution integrated over all space ought to 1 Examples of normalization Consider particle in a 1D box of length L O S x S L After solving Schrod eqn obtain plausible wave function 1psinE L But it is not normalized yet until we re sure that jzpXlpXdX1 So we say w a sin where a constant to be determined by normalization Now since 11 is a real function 1p 11 so need to solve L fasin a sin3 dx1 L L 0 L azfsin2 de1 0 L 2 4 L fsin2mltdx L 0 Use the following calculus result out of integral table TI fsinzocx dx E 0 2 Do change of variable to get our integral into the form of this one TEX TE Let 7 d idx y L y L XL Y7E As f gt x0 y0 L L It 2 a i 2 snnLjdx nfsm ydy So I IN a El I INN Therefore normalized 1p E sin Do a second example Wave fct for lowest energy state of H atom has the form 112 c e rao where r distance of e39 from proton a0 constant called Bohr radius c constant to be determined by proper normalization Normalization condition 000000 fffxp1pdxdydz1 00 00 00 volume integral of xyz dimension 3D Z Fig ure 8720 Atkins Physiml Chemistry Eighth Edition However r a radial coordinate Therefore use spherical polar coordinates for convenience xrsin0cos yrsin0sin zrcos6 dxdydz volume element r2 sint clr dt dd Fig ure 93 5 Atkins Physiml Chemistry Eighth Edition Our integral then becomes with 1p q cze Zrao 002nm fffcze Zrao r2 clr sinE d0 d4 1 r i OK r l E int int int Which splits apart to give 00 21 31 czfrze zrao drfdcpfsinede 1 o o o r integration 1 int 0int 27 2 4nc2fr2e 2ra0 clr 1 0 Look up this form of integral in table fxze o X o leN Our integral looks like this one if at 2ao 4m2 2 1 Zao 3 4m2 10 1 4 c2 i ma 1 c 7 nag 12 w i e rao nag B Quantum Mechanical Principles 1 Operators for every observable there is a mathematical operator Let s work in 1D Most important hamilitonian operator energy operator potential energy operator just a normal function of x kinetic energy operator linear momentum operator position operator simply multiply by x 2 Operator equations eigenvalue equations General form operator x function numerical factor x same function or Qf wf if operator eqn fits this form we say that f eigenvalue of Q operator 0 eigenvalue of 2 operator Simple example Let 52 i dx2 Let f sin ax Does operator eqn fit the form Qf mf 2 isin ax a2 sin ax dxz T T T 52 f 0 f two derivatives gives you back the sin ax function 2 Therefore sin axis an eigenfunction of L2 dx and a2 is the eigenvalue n d Now what If 2 7 dx d isinaxacosax dx noIthe same fct Therefore sin ax is not an eigenfct of d1 operator x a is not an eigenvalue then Importance of an eigenvalue equation operator value corresp of to x w gtlt1p physncal physncal observable observable T upon e ery observation if that is then this is the H op the energy ll E14 Schrod Eqn is the most fundamental T eigenvalue eqn energy of system is eigenvalue of energy operator or H operator wavefunction of system is an eigenfunction of the H operator Operators for some properties are such that 11 is not an eigenfunction of that operator Examples position momentum Meaning when 11 is not an eigenfct of a particular operator a different value is obtained for every measurement of that property associated with the operator eg repeated measurements of position or momentum of H atom e39 give different results every time Repeated measurements of energy gives same value 5 When measurement produces indefinite values for a physical observable we are interested in the average value called the expectation value observable Q corresp operator Q expectation value of Q lt 2gt a fq de 1p must be properly normalized first Example 1 calculate the average value of position of particle in 1D box of length L 2 Given 1Z sin L X f2pX dX 0 llimi ldx Use sin2 e 12 1 cos 2 e L 2 2nX ltxgt x 1 cos Tjdx Change of variable y ltXgtlililalvgtdlil y0 cosine lawquot 2n L g y cos ydy L yz 27E 2T 7 7 ycosy dy 472 2 0 0 272 472 x 2 ie middle of the box Could have guessed this Although every position measurement gives different value the average value XE Example 2 Calculate average value of KE of particle in 1D box of length L 2 2 A KE operator ii Ek 2m dX2 Ekgt fl Ekwdx First calculate the result of the KE operator operating on the wave function CHAPTER 9 Quantum Mechanics Applications translational vibrational rotational motion I Translational motion A 1 B The Single particle in free space 1D Schrodinger eqn Ipr Exp h2 d2 1p Exp no boundary conditions 2m dX2 General solution 1 Energy eigenvalues kzhz 2m E Note energy of a free particle is not quantized value all values of k are permitted not just integers Therefore the E can take on any Note 1 can also be written as 1p A sin kx B cos kx Since 00 OO A A e396 cos 6 i sin 6 gt e399cosE Ism6 gt CD E particle in a onedimensional box 3 62 The potential function Vx g D VXOforOltXltL O 0 L X Vx 00 elsewhere 2 Schrodinger equation ilw Ew where Fl kinetic energy operator Vx 4 12 d2 77 VX 2m dX2 So we can write 4 12 d2 EEWX VX1lJX Ewx VX 0 in box 00 outside box Above is equivalent to equation d2wx 2m d 2 E1lJX X plus boundary condition Boundary condition x restricted to interval from 0 to L in which case VX 0 and 11 is required to vanish at x0 and xL Boundary condition allows us to drop VX term Now need to find 1px which is a solution of this differential eqn In other words find a fct1p such that if we differentiate with kinetic energy operator we obtain E11 that is a constant times 11 Solution mm Bsinnij n1 2 3 whole series normTaliz quantum number of solutions coefficient state number ml wZI 1P3amp associated energy eigenvalues n2h2 En72 8mL Problem Let s check to see if this is a solution 422 d2 h2 d2 mix 2m dx2 2m dx2 L 2 h d BCOSltn7 xj mt 2ma h2 mt max mt s1n 2m L 2 2 h 3a Sinltn zxj 2m Enipn u or HWn Ean Interpretation Lowest energy quantum state et n1 First excited state et n2 Fig ure 94 Atkins Physical Chemistry Eighth Edition lt7 2006 Peter Atkins and Julio de Paula Energy level diagram m 9 quotA 64 8 5 E E 7 s 49 iv Lu 36 6 25 5 Wavefunctlons for n1 thru 5 16 4 9 3 Limiting conditions 4 2 0 1 1 Figure 92 Atkins thsical Chemistrv Eiahth Edition wwwm uniform probability just like classical result Correspondence principle limit Quantum Results Classical results n a 00 Orthogonality Condition If wn is a correct well constructed solution to the Schrodinger equannthen L fdxwfwj1 ifij O and L fdxwrwfo ifi j 0 mi is said to be orthogonal to IM Illustration of orthogonality Here multiply n1 and n3 wave functions together and then integrate product from 0 to L Should get 0 3 1 I L Practical A lications article in 1 D box Energy levels of delocalized rc electrons in long conjugated molecules Delocalized nbond system 0 O O O Approximately 1 D C C C C C C 6 electrons lt L gt Length nzh2 quot 8mL2 where En is the energy levels for It electrons m is electron mass L z 6 X carbon carbon bond length Spectroscopic transitions in the hexatriene system n4 xx ground 3 state XX n2 XX n1 6 electrons fill 3 levels remember Pauli Exclusion principle excludes gt 2 e39 from occupying a given quantum state x n4 AE hy where y is the x frequency of lst excited 3 light promoting state XX transition n2 XX n1 So can use Particleinbox to predict vspectrum Pauli Excl princ AEhy C Particle in a 2D Box Le a particle confined to a surface 1 Picture a length in xdirection g E 0 a a b length in ydirection E E 1 L2 y Particle 39 LL con ned X to surface H11E1p A 4 12 92 92 H 72 9 2 a 2 partial differential equation now In x y Solved by separation of variables technique see text 2 2 2 3 E1 quot 8hnxz 122 nyny 123oo y m a 11quot n 1 4Sinn nxsinny yj ab 0 b 4 Special case Energy level diagram when abL square box Energy 31 13 10h28mL2 22 8h28mL2 21 12 5h28mL2 nxny 11 2h28mL2 5 Attempt to 1p2draw pictures a b c 11 12 21 22 eg The above is first example of degeneracy two or more different wave functions quantum states having the same energy states 12 and 21 states 13 and 31 D Pamde m 3D BOX 1 a b 8 1173 wry quotm abc a b c 8m a b c Energv 1eve1 dwagram for abc CUBE 222 Energy 1 113 131 311 w n m 7 2 122 212 221 112 121 211 lriply degenerate 111 II Vibrational Motion A The 1D Harmonic Oscillator 1 An important model for vibration of a diatomic Potential energy V O DiSplacement X 1 2 V k X 2 X where k bond force constant stiffness x displacement of bond length b from equilibrium length bo X b b0 x 0 represents bond at its equilibrium distance Fkx Force 2 Classical trajectory xt A sin23wt if vibration starts at XO moving in direction A amplitude of vibration can be any value v frequency of oscillation v i E let s also define ouEZmE 2 M M u reduced mass m1mz m1 m2 3 Quantum mechanical Schrodinger Eqn SE is then lt1 gtlt Potential energy V h2 d2 k 2 X X X E mdxzm 2 w w Displacement X 4 Stationary Solutions 1px vibrational quantum number v O123oo I ULUI ILIGI Energies 1 energy V v i EV V le LLIgt m EV V aha Equot c hw3hw3hm g evenly spaced levels 0 2 where lt 1 k V 27 M OR Displacement X Wavefunctions 2 11w NHvye y22Where y E i and 0 51 3 uk N V normalization constant Hvy the Hermite polynomials 10 Table 91 The Hermite polynomials HM V H10 0 1 1 2y 2 4y2 2 3 8y3 12y 4 16y4 48y2 12 5 32y5 160y3 120y 6 64y6 480y4 720y2 120 Table 9 1 Atkins Physical Chemistry Eighth Edition iv 2006 Peter Atkins and Julio de Pauia Ground state 111026 N e xzZO 2 Gaussian function symmetric function around XO said to be an even function Wavefu nction LP Fig ure 92 3 Atkins Physkal Chemistry Eighth Edition 2006 Peter Atkins and Julio de Paula 1p0x 1p0 x property of even function 11 1st excited state 2x a e quot22 2 x times Gaussnan a 5 9 5 an m function 639 s C a E a W w1X G gt E g i 4 2 O 2 4 V Figure 924 Atkins Ph sicul Chemistry Eighth Edition 0 2006 Peter Atkins and Julio de Pauia ml is odd because 2 2 e x 20 x even function odd function x odd function 2 UN N24xz 2 e39XzZ oc even power even power even of x of x even function x 1 1P2 is even function Here are the first 5 vibrational state wavefunctions 10 05 Wavefunction ir 05 1o 4 2 o V Figure 925 Atkins Physical Chemistry Eighth Edition t 2006 Peter Atkins and Julio de Paula Here are the probability density functions 1010 ure 926 Atkins Physical Chemistry Eighth Edition Q 2006 Peter Atkins and Julio de Paula 5 Notes on Parity symmetry of functions wo has even parity 1P1 has odd parity 1P2 has even parity 113 has odd parity etc even function x even function even function even function X odd function odd function odd function X odd function even function Integrals CD f dx fx 2de fx if fx is even 0 jdx fx o if fx is odd Very useful in solving certain integrals Some functions however are neither even or odd parity Examples Knowing the wavefunctions we can calculate a variety of properties of the harmonic oscillator in any given quantum state v such as average position ltxgt average ltx2gt the average potential energy ltkx22gt Here lt gt denotes expectation value See text for examples III Rotational Motion and Angular Momentum A Rotation in Two Dimensions particle on a ring 1 Physical picture particle of mass m constrained to move on a circular path of radius r in the xy plane Figure 9 27 Atkins Physical Chemistry Eighth Edition O 2006 Peter Atkins and Julio de Paula 2 Treat classically first Energy is all kinetic so classically E p22m Classical angular momentum around the zaxis is Jz or pr So we can write J2 J2 2 where I moment of inertia 2mr 21 3 Apply quantization condition simply using de Broglie equation s1nce Jz i pr and h A oppOSIte Signs denote oppOSIte directions of travel 15 However not all wavelengths it are allowed but only those that exactly repeat themselves around the ring satisfy periodic bc Thus only wavelengths obeying the following condition will work First second circuit circuit 7 HT 1 Second Fjrst circuit CerUIt AJE ml where ml 012 here is the angular momentum quantum and so JZ mlh Wavefunctiona 9 So we have the quantization of angular momentum 4 Now what about the energies E JZ2 miquot2 21 21 Wavefunction 1 O e N 5 The wavefunctions b eimi 10m 5 W B The 2Particle Rigid Rotor rotation in 3D 1 Physical Picture m2 1 33 b quot 2 cantar pr ma 55 4ch quot 1 masses m1 and m2 moment of inertia I ud2 3dimensional problem central force problem that is the 2 particles are held at fixed distance by inwardly directed forces which counteract centrifugal forces Importance good quantum mechanical model for rotation of molecules 2 Full Coordinate System 3 Simplifvind Features Since both particles rotate about a fixed center of mass CM knowledge of the position of 1 particle automatically implies position of 2nd particle Really a 1particle problem with an effective mass called the reduced mass m1mz m1 m2 17 So translational motion of system 2 particle separates as whole motion of the CM problem into internal rotational motion lt characterized by a single effective particle of mass M confined to W motion on sphere of radius d k this is the part of interest The rigid rotor problem is mathematically homologous with a single particle of mass mu on a sphere of radius rd Still 3D problem 4 SE H1pE1p 2 H h V2i7 2M 62 62 62 V2 is Laplacian operator in Cart Coord 6X2 ay2 622 V 0 since particle is constrained to a spherical surface 5 For convenience in central force problems convert to spherical polar coordinates r fixed d 18 2 2 V2w 3 i 1 LI 1 85inew arZ Zar r2 sin293 2 smeae as K I a radial portion angular portion kinetic energy 1 2 of radial motion A 1quot 0 since rd rigid r hA z orbital angular momentum operator A2 the legendrian operator 2 H1pE1p L A214 Ew 2W2 where I pd2 moment of inertia of rotor Solution 1p6 p function of the 2 polar angles 1p6 p Yylmze where Y is special family of functions called the spherical harmonics Quantum state can be specified by 2 quantum numbers E m I 0 1 2 orbital angular momentum quantum number ml l 0Z magnetic quantum number Now refer to Table 93 in Atkins Note H atom wave functions have these very same functions times a radial function Energies of the rotor 2 E 1szI depend only on 2 not my mg 4 L 1 2 2 E43 1 hz 12 0 m10 9 Degeneracy 2 1 10 Angular momentum J and J2 JWh 32 z mlh magnitude of the angular momentum angular momentum about the zaxis Table 93 The spherical harmonics l m B lamaw 3 12 cos 9 41 12 3 sin 9 at 87 1 3 COSZQ 1 12 1 cos 05in 98 J sinzae m 15 327 1 7 3 5cos 973c056 167 H 21 12 5 5052971sin Bet 647 12 105 sinZG cos 9 e w 327 64 12 35 1T sin36 e 39l39 n 11 Appearance of spherical harmonics using m0 l0m0 J l1 m0 Ul2 m0 Lj id l3 m0 0 U V l4m0 Fig ure 936 Atkins Physical Chemistry Eighth Edition 2006 Pete Atkins and Julio de Paula lmll 0 1 2 3 Fig ure 937 Atkins thsiml Chemistry Eighth Edition 2006 Peter Atkins and Julio de Paula Conversion of notation to the rigid diatomic rotor h2 Use EJ JJ 1E where J rotational energy level of rotating diatomic M J2 M O 1 J1 JO IV Intrinsic Spin of Microscopic Particles A Properties 1 Spin emerges when special relativity is applied to quantum mechanics 2 Microscopic particles possess an intrinsic angular momentum about their axis called spin 3 This is treated differently than orbital angular momentum because it has different boundary conditions 4 The magnitude of the spin angular momentum of a particle is determined by the soin quantum number s and is given by ss112h 5 The value of s for a given type of particle ms 2 13 is fixed It cannot change For an electron or d39 proton s12 6 The component of spin angular momentum along an axis 2 defined by an external magnetic field is dictated by the soin magnetic quantum number ms ms can take on values from s s1 s1 s For an electron or proton with s12 ms can take on values 12 and 12 I ms 2 MI 23 Table 94 Properties of angular momentum Quantum number Symbol Valuesquot Speci es Orbital angular momentum l 0 1 2 Magnitude ll 1 Z Magnetic mZ l l 1 l Component on z axis ml Spin 5 13 Magnitude 55 1 2 Spin magnetic m5 i Component on z axis ms Total j lsls 1l s Magnitude jj1 2 Total magnetic m j 1 j Component on z axis mj To combine two angular momenta use the Clebsch Gordan series i1 1341 15 1aJ391 J392 For manyelectron systems the quantum numbers are designated by uppercase letters L ML 8 M5 etc y39Note that the quantum numbers for magnitude 1 5j etc are never negative Table 9 4 Atkins Physical Chemistry Eighth Edition 2006 Peter Atkins and Julio de Paula V Tunneling A Definition passage of a particle through an energy barrier that exceeds the energy of the particle ie passage through a classically forbidden zone B Example v5 barrier barrier E Pad le Winn vtx fquot V E 1 WC 39FY er 5f 09 X 37 Classically the particle of energy E trapped in the well will remain trapped forever In QM it has finite probability of escape 24 C D In fact 11 of the particle has a small contribution through and outside barrier li lv quhde 4 L95 217x 1 1 l 3 GWMJ 9 KY l 1 1 7 A fine casinos l l f are m We bar cr Phi sando 2 12 K spatial decay parameter Transmission probability 0c e39KL L 2 LamMr wr H JL L P Note that the larger K is the more 11 is damped as it passes through barrier As K 1 Transmission Probability 1 K depends on Vb barrier wt Vb 1 K 1 Prob 1 E energy of particle E 1 K 1 Prob 1 m mass of particle m 1 K 1 Prob 1 Exact calculation of transmission probability through square barriers can be obtained by properly piecing together solutions to SE in each region 31 mloLlrzlwm W3 6 w k K s r 2 WM 11 4 V 3 8 x 2L5 Waive H 2 7 ugt 25 Matching boundary conditions provide us simultaneous eqns to solve to obtain weighting coefficient for each region E Result KL KL 2 71 Transmission Probability T1 6 5 where E is the energy of incoming particle V is barrier height F Approx result in limit of high and long barrier KL gtgt1 KL e gtgt e KL Simplest results T16s1 se m T depends on L m E V VI Approximation Techniques A Most interesting systems cannot be solved exactly in terms of known analytical functions Two main approximation approaches are used 1 Variation theory will introduce with molecular orbital theory 2 Perturbation theory here B Timeindependent perturbation theory 1 Hamiltonian split into easy part Oth order part with known eigenfunctions and eigenvalues and the complicating part the perturbation z rm rh 2 Energy eigenvalues are then represented by a series of corrections with the oth term being the eigenvalue of the nonperturbed Hamiltonian EE EDE2 3 Wavefunctions work the same way with the oth order term being the eigenfunctions of the oth order Hamiltonian 0 1 2 1p1p zp zp 26 H The full wave function of manyelectron atoms 1 Full wave function must include spin state designation eg Helium a 2 e system wlt12gtwlt12gtlt gt T l spatial spin part part spin wave functions are combinations of oci or 5i which are one electron spin functions W12 wlw2 l l orbital 4 tentative possibilities approx for 2 spins oc1oc2 51l52 oc1l52 oc2l51 In last two combinations 0L152 or 0L251 due to the indistinguishability condition of two electrons you have to express these as linear combinations and there are two ways to do that 1 77 L 2 0l 2 1l52 510i2 1l52 510i2 0712 So four possible spin arrangements of 2 electrons OC10c2 51l52 0412 0112 Now comes Pauli principle when the labels eg 1 and 2 of any two identical fermions are exchanged the total wavefunction lIJ must change sign antisymmetric wavefunction requirement 1P12 11I21 If just look at spatial wavefunctions alone amp if both e in same orbital 1p11p2 1p21p1 not antisymmetric Therefore the spin portion of total lIJ must be the antisymmetric part 26 Of the four possible arrangements only o12 is antisymmetric so m2 1p11p2012 7 IND31 IND32 Slater determinant Therefo re if 2 e in same spatial atomic orbital only one spin state available o12 which is sins paired Ll But when 2 e are in different orbitals all 4 of the spin combinations are possible a1a2 B1B2 ltL12 112 The first three are parallel triplet the last is paired singlet l i all three combos possess spin no spin angular angular momentum momentum lower in energy ML 1O1 ms 13 M 1 1 s 2 Figure 10 24 Atkins Physical Chemistry Eighth Edition D 2006 Peter Atkins and Julio de Paula 27 III Spectra of Complex Atoms A General More complicated than hydrogenic spectra Still hy E AE differences between atomic energy levels photon A The problem is AE l difference between orbital energies Eg He gt He electronically excited Helium 2s T J F gt ZS amp ZS is T A T1s T1s triplet singlet paired spins parallel spins zero total spin n zero total spin Difference in energy is two effects 1Coulombic e39 e39 2Spin correlation Triplet state is generally lower in energy than the singlet B Selection Rules for allowed transitions in manyelectron atoms 1 No change of overall spin Implies singlet singlet is allowed But singlet 9 triplet is not allowed 28 2 3 Here are the allowed transitions in Helium no singlet to triplet O 1S EnergyeV 20 245 Figure 1025 Atkins Physical Chemistry Eighth Editiuh n 2006 Peter Atkins and Julio de Paula 1083 Sodium s bright doublet emission 17 cm 1 2P 16 973 32 17 cm 1 2P 16 956 2 N D1 D2 E E C C h 3 m on 00 oo 0 Lo Lo ZS12 29 02 Bond order 2 02 By Hund s Rule the two electrons in the lug antibonding level go into separate MO s giving rise to two unpaired electrons Molecular oxygen is paramagnetic F2 has two more electrons into 1mg antibonding level and so has bond order 1 and no unpaired electrons Nez has two more electrons going into antibonding MO and has bond order 0 Unstable Effectiveness of overlap Ato m Atom Molecule v Figure 1131 Atkins Physical Chemistry Eighth Edition Q 2006 Peter Atkins and Julio de Paula The overlap integral S determines the extent of overlap of two atomic orbitals which determines whether they will form good MO s f 1pALpBdt Eg If two ls orbitals approach one another one on A and one on B their overlap integral increases to 1 Overlap of s and p increases up to a point This s and p gives ZERO Then begins to decrease net overlap iquot l i g Figure 1 1 30 At ms in Chemistry Eighth Edition 2006 Peter Atkins and Julio de Paula b Figure 1 128 Atkins Physical Chemistry Eighth Edition 2006 Peter Atkins and Julio de Paula Bond strength The greater the bond order the greater the bond strength The greater the bond order the shorter the bond Synoptic table 113 Bond Synoptic table 112 Bond lengths dlssoc1atlon energles Bond Order Repm Bond Order D0kI mol l HH 1 7414 NN 3 10976 HH 1 4321 HCl 1 12745 NN 3 9417 CH 1 114 HCl 1 4277 CC 1 154 CH 1 435 CC 2 134 cc 1 368 cc 3 120 CC 2 720 CC 3 962 More values will be found in the Data section Numbers in italics are mean values for polyatomic molecules More values will be found in the Data section Table 1 1 2 Aihihs ihysicai chemishy Eighth Edition Numbers 1n 1ta11cs are mean values for 6 2006 Peter Atkins and Julio de Paula polyatomlc molecules Table 1 13 Atkins Physical Chemistry Eighth Edition 2006 Peter Atkins and Julio de Paula 16 C Heteronuclear Diatomics difference between atoms increase z4L L2p MO s are somewhat like homonuclear diatomics but less so as EN I I IL I h f 9 I eg NO I z I BO 25 Now what about HF ENF gtgt ENH no overlap x 2 H H 2 15 lt I I L 25 no A ls ove ap 2p ls 17 1pm O 1911SH 0981sz different weighting contribution MO produced has much greater contrib from 2pZ on F high Ea than from is of H ea polar covalent bond 2e in the MO spend more time near F than near H Shared unequally i F H General Rule atomic orbitals most effectively combine overlap to form MO s when 1 their energies are similar in the separated atoms 2 their shapes are conducive to good overlap Eg these give no overlap due to their shape and sign of wave function 18 The construction of energy level diagrams for arbitrary heteronuclear diatomics is not straightforward and requires calculation The Variation Principle a systematic way of taking linear combinations of atomic orbitals in correct proportions to produce the most optimum approximation to the MO s VP is based on the fact that approximate MO s will always have a higher energy than the true exact MO s energy Eg in HF ifI had constructed the MO as mm 101p15H 101p2pz F with equal coefficients the energy of this 1pm gt Etrue Consequence of VP gt vary the weighting coefficients until you get the lowest energy you can find This will be the best approx you can make to the true wave function Wtrial CAwA CBwB unknown coeff which are to be varied Ema fIPZT IPm f Vary CA and CB until Emal attains its lowest value Suppose wtrial is real Denom ftpim r fCA1pA CH B2dr C1 f1pA2dI C f1pB2dI ZCACB fszzder 1 1 overlap using normalized integral S wave fcts from the separated atoms A B Denominator CZA C ZCACBS Numeramr fwtriallfllptrial C1 fzpA IzpAdr C fzpB IzpBdr ZCACB fzpAI npBdr PV g OLA 0LB OLA 03 Coulomb integrals 5 Resonance integral Numerator c aA cgaB chcB c aA c aB chcBis cg cg chch Etrial Must evaluate integrals OLA 0L3 5 and S to obtain s Then vary CA CB until Emal achieves a minimum ie set E 0 this gives equation acA A EtriaCA l5 Etrial SCB 0 aEtrial Then set T 0 this gives equation 3 B 5 T Etrial SCA 0 T EtriaCB 0 Two equations two unknowns CA and CB This set of eqns has a solution when the determinant is zero OLA T Etrial T Etrial S 0 T Etrial S 0 B T Etrial this is called the secular determinant Defin of determinant A B AD BC C D 20 2 2 OLA E 5 ES o Solve for E a quadratic equation Has 2 roots corresponding to bonding and antibonding orbital energies Simple case A B homonuclear diatomic aAocBoc 12 1 Two solutions 1 bondIn E 0Hf5 C 7 C g 1s A 21s B 12 1 2 anti EL C c A zll sll B Even better approximate wave functions for the molecule can be constructed by taking linear combinations of lots of atomic orbitals not just two This is called expanding your basis set In principle by using a complete basis set you should obtain an exact solution upon using the variation principle In practice a complete set would need to have all atomic orbitals full and empty on all the atoms in your molecule and that is an infinite set 21 HF MNDO CALCULATION RESULTS MDRAC VERSION 600 CALC39D 21 Oct94 XYZ GEOOK LARGE VECTORS FOCK MULLIK BONDS DENOUT lELECTRON INPUT CARTESIAN COORDINATES NO ATOM X Y Z 1 H 00000 00000 00000 2 F 09200 00000 00000 FINAL HEAT OF FORMATION 5973416 KCAL TOTAL ENERGY 49425877 EV ETEIHRONIC ENERGY 57687098 EV CORECORE REPULSION 8261221 EV IONIZATION POTENTIAL 1489120 NO OF FILLED LEVELS 4 PIEECULAR WEIGHT 20006 SCF CALCULATIONS 7 COMPUTATION TIME 0180 SECONDS ATOM CHEMICAL BOND LENGTH NUMBER SYMBOL ANGSTRCWB l H 2 F 095633 EIGENVECTORS ROOT NO 1 2 3 4 5 4377134 1793881 1489120 1489120 580377 18 H 037307 046609 000000 000000 080223 ZS F 091567 032434 000000 000000 023738 2PX F 014956 082314 000000 000000 054779 2PY F 000000 000000 092376 038298 000000 2PZ F 000000 000000 038298 092376 000000 NET ATOMIC CHARGES AND DIPOLE CONTRIBUTIONS ATOM NO TYPE CHARGE ATOM ELECTRON DENSITY l H 02872 07128 2 F 02872 72872 DIPOLE X Y Z TOTAL POINTCH6 1319 0000 0000 1319 HYBRID 0670 0000 0000 0670 SUM 1989 0000 0000 1989 FOCK MATRIX IS 18 H 1 25 F 2 2PX F 2 2PY F 2 2PZ F 2 18 H 1 6329743 28 F 2 12932297 38008471 2PX F 2 11390509 0183063 11327472 2PY F 2 0000000 0000000 0000000 14824289 2PZ F 2 0000000 0000000 0000000 0000000 14824289 BONDING CONTRIBUTION OF EACH MO 07166 11185 00000 00000 18351 TOTAL CPU TIME 023 SECONDS 22 MNDU QUANTUM MECHANICAL CALCULATION HF ENERBWeV 13 4 1L 1 2 AU 3 1L 30 Eg The 2po MD of homonuclear diatomics can be improved by adding in 25 AO s IPtrial wa2pz A 8pr 81 EATPZSA CEIPZSE V Y usual combination extra The Variation procedure will generate 4 eqns and 4 unknowns CA CB C A C B Secular detm will be 4 X 4 It will turn out that CA CB gtgt C A C B Therefore minor contrib from 25 AO s Using computers basis sets can include dozens of AO s to produce and solve huge matrices of numbers and extremely accurate solutions Still must evaluate on 3 and S integrals These are done two ways 1 semi empirical estimate them from spectroscopic data 2 ab initio evaluate them numerically on the computer Simpson s rule 23 E Hybridization can be used in MO theory also Introduced to try to preserve the idea that every MO is constructed by linear combination of only AO s Eg lSON1A In reality in advanced MO cacs every MO is produced by a LC ofaH the atomic orbitals in the basis set Turns out that usually only a single A0 from each atom contributes significantly to a given MO esp in homonucl diatomics Other MO s LiH zp U 25 0MO the bonding one is H is me 0414w25u 033m2px Li 0848w15H From Variation treatment 3 AO s end up contributing to 0 bond in LiH To preserve concept of overlap of 2 AO s to get MO hybridization idea is ntroduced is H is seen as overlapping with a hybrid orbital call it Li Li mixture of Li atomic orbitals prior to interaction with H LLi Li 0715 wzpm Now 110 0323 Li 0685 wls H S ee that hybrid concept is cumbersome here 24 III Polyatomic Molecules MO s are still constructed by LCAO but using more atomic orbitals Ordinary procedure is to use all the AO s from the valence shell of every atom as our basis set A Treat H20 by LCAOMO theory 1 Position nuclei at a trial starting position 0 o o 0 HA H8 2 Choose appropriate basis set is HA 6 AO s in basis set is HB 2s 0 2pX O 2py O sz O 3 Use Variation Principle Will solve 6 x 6 secular determinant Will find 6 roots which are the 6 energies of 6 MO s produced 6 MO s produced 6 energies 4 Walsh diagram plot of energies of the different MO s as we bend the molecule Procedure vary angles until lowest energy is obtained always putting the valence e39 in the lowest energy orbitals possible 25 MOPAC VERSION 600 Water linear 180 degrees FINAL HEAT OF FORMATION EIGENVECTORS ROOT NO 1 2 3836620 2043460 H 1 029829 047730 0 2 090667 000000 0 2 000000 073782 0 2 000000 000000 0 2 000000 000000 H 3 029829 047730 3 1111085 000000 000000 000000 099961 002782 000000 CALC39D 26Oct94 1601362 KCAL 4 1111085 omm 000000 000000 00Z782 099961 000000 5 366569 064111 042185 000000 000000 000000 064111 6 972546 052171 000000 067500 000000 000000 052171 NET ATOMIC CHARGES AND DIPOLE CONTRIBUTIONS ATOM NO TYPE CHARGE ATOM ELECTRON DENSITY 1 H 03664 06336 2 0 07328 67328 3 H 03664 06336 DIPOLE X Y 2 TOTAL POINTCHG 0000 0000 0000 0000 HYBRID 0000 0000 0000 0000 SUM 0000 0000 0000 0000 Water optimum FINAL HEAT OF FORMATION 6092461 KCAL ATOM CHEMICAL BOND LENGTH BOND ANGLE TWIST ANGLE NUMBER SYMBOL ANGSTROMS DEGREES DEGREES I NAI NBNAI NCNBNAI NA NB NC 1 H 2 0 094000 1 3 H 10692268 Z 1 EIGENVECTORS ROOT N0 1 2 3 4 5 6 4047670 1816014 1545354 1Z31455 585841 597957 5 H 1 033152 043899 034044 000000 049342 058139 5 O 2 087433 001750 037543 000000 001728 030659 PX O 2 010787 054937 054118 000000 051874 035296 PY O 2 012305 050143 061997 000000 042241 041308 P2 0 2 000000 000000 000000 100000 000000 000000 S H 3 031444 050373 025673 000000 055562 052228 NET ATOMIC CHARGES AND DIPOLE CONTRIBUTIONS ATOM NO TYPE CHARGE ATOM ELECTRON DENSITY 1 H 01630 08370 2 0 03259 63259 3 H 01630 08370 26

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