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# Introductory Physical Chemistry CH 331

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This 41 page Class Notes was uploaded by Sienna Shields on Thursday October 15, 2015. The Class Notes belongs to CH 331 at North Carolina State University taught by Stefan Franzen in Fall. Since its upload, it has received 14 views. For similar materials see /class/223995/ch-331-north-carolina-state-university in Chemistry at North Carolina State University.

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Date Created: 10/15/15

Chermstry 331 Lemma 3 Ahsmpheh speeha e1 amms The SehTeeThgeTeeuaheh TeT hydmgen The eTeehehTe shueme e1 amms NC State T T The SoTaT Spectrum quotwere an eepshme shah emssmh eeuee Tmehhmnhes The ms ehse m Svecmc ms h he eh rm ism mamaquot The SchmdngYEmzhumm hyumgm e Samrahun m anzmes Raesz and argmzmzns weegehaum waveTuhnTms Emecmuh vdues Spmmseepy a ammo hyumgm unr equalinn inr lnnlrnun The Smmdmevequa nn Thhee Meme 7 2 The upemnmeT Squared T 3 3 3 7 T E The pmmdm ma pheheannTaTmnhmhmemm nmad dXVandzthe mnmamsae e mandv 11 lm nillm nnlnnlill s The Oummb mlemz bawaemhe eTehm 2m thapmun Ts v e2e14hznx Ehmmunmluvbmmhepmnnmddactmn Ts x we v o v H 27m 2m 2 Samrzhunm hmmhe emhe anda es resuns h aheTenthe eeuahehmhe cmlevrmr mass emmhmes H ehiam e ZeiAhznx gt1 mm heTmpmehTem m m hemua nn huh au We mnah e shunahenuey when a pmcedwe Maxm a eparannn m memes Ts used The any are 1 swam h when when 2 Dmde mm was hv leISM 3 Mump vhmh eaeswzw lmlm mewavmmmm summs mmezng zv argubvmmerlum mm mm s m 1h quotMl mum rm whencamanmmchnleyml mun mmuav We shape mm nfnamsemhbxme m Whema hanmn v mm m mm mm mm M PLE mama mmmv we Wm mmmm been mm mamaquot m hneav mmmnm mm anvmmmexnumhm quotMl m mayhem hanmmchaxthe nrmman arm mm mm anguhnmmnnh quotMl rv m pnfEn E EMMAWWHEMHVWNEvamthelnrm mm are m mgmavmmmm mum quotMl w A mmmaanm r 4 a A 3 We are m angmammmnxhvl emmpmmenergyemzm v humM mm curlzmsme cunnmmmtm emrgy mm mama harms Tugmmmmm 0mm Mermsz wemvap emdenergyws vr 7 22mm munM The radial euuation for hydrogen Making the above approximations we have an radial hamiltonian energy operator 2 ves 2 R quotM 2p 47t8ur 2pZ R ER The solutions have the form 1 an p39 e39 z anlr where p ZZr and an 4 rtenftZmeZ Nnvl is the normalization constant anlr is an associated Laguerre polynomial The Billquot radius The quantity aJ 4 rtenh7meZ is known as the Bohr radius The Bohr radius is an 0529 A Since it emerges from the calculation of the wave functions and energies of the hydrogen atom it is a fundamental unit In socalled atomic units the unt of length is the Bohr radius So 1 A is approximately 2 Bohr radii You should do a dimensional unit analysis and verify that an has units of length Solutions of the radial euuation The normalization constant depends on n andl IV quot44 n Each of the radial equation solutions is a polynomial multiplying an exponential The normalization is 2391quot 1 3 quot5 obtained from the integral The associated Laguerre polynomials are m 2 mm RDIRDIr dr 1 n2Ell1 x72l27x D n 2 l 1 Gm 7 at The volume element here is erIwhich is the I n a n gm 7 aka 73x my part of the spherical coordinate volume element n a 1 L3x74l47x rzsinedrd dtb n3 2 imam MAPLE worksheet A MAPLE worksheet attached to this lecture illustrates the normalization of the rst three radial functions The worksheet includes plots of the functions When examining a plot keep in mind that you can plot the wave function orthe square of the wave function We often plot the square of the wave function because the integral of the square of the wave function gives the probability Enerov levels of hydrogen atom The energy levels of the hydrogen atom are speci ed by the principal quantum number n M64 32712 0 n2 r All states with the same quantum number n have the same energy All states of negative energy are bound states states of positive energy are unbound and are part of the continuum The Rydberg constant The energy levels calculated using the Schrodinger equation permit calculation of the Rydberg constant w One major issue is units Spectroscopists often use units of wavenumber or cml At rst this seems odd but hv hcA hcv where i isthe value of the transition in wavenumbers 64 R I1632n2802 2 in cml Shells and subshells a All of the orbitals of a given value of n for a shell r n 1 2 3 4 correspond to shells K L M N Orbitals with the same value of n and different values of Iform subshells 1 0 1 2 correspond to subshells s p d Using the quantum numbers that emerge from solution of the Schrodinger equation the subshells can be described as orbitals Hvdrouenic orbitals s orbitals are spherically symmetrical The is wavefunction decays exponentially from a maximum value of 11903 2 at the nucleus p orbitals have zero amplitude at r0 and the electron possesses an angular momentum of h I 1 The orbital with m 0 has zero angular momentum about the z axis The angular variation is cose which can be written as r leading to the name pZ orbital l39lllllf g 1 S radial wavefunction The is orbital has no nodes and decays exponentially Rls 21a3Ze39 Z n 1 and I 0 are the quantum numbers for this orbital E The Radial Ilistrihution in Hvdrouen 2s and 2H orbitals EXllBGlalllJII values The expectation or average value of an observable ltrgt is given by the general formula lt r gt 11 w lFrll dr El As written the above integral describes the expectation value of he mean va ue oft era ius r integration overthe angular pan gives l because the spherical harmonics are normalized The volume element can be written dr dr The mean value is lt rgt 1 P r PerrI W Wradr I U MAPLE worksheet IJII Enneotation Average values There is a MAPLE worksheet attached to this lecture that illustrates the use of expectation values The example of the position ltrgt is given for the 1s 2s and 3s radial wave functions The expectation value or average value ofr gives the average distance of an electron from the nucleus in a particular orbital Since the 2s orbital has one radial node and the 3s orbtal has two radial nodes the average distance of the electron fromthe nucleus is shown to increase as ltrgt for 3s gt ltrgt for 2s gt ltrgt for 1s sneotrosoonv of atomic Innlrooen Spectra reported inwavenumbers r Rydberg fIt all of the series of hydrogen spectra with a single equation Absorption or emission of a photon of frequency v occurs in resonance with an energy change AE hv Bohr frequency condition n Solutions of Schrodinger equation result in further selection rules SIIBGIIIJSGIJIHG transitions w A transition requires a transfer from one state with its quantum numbers nqu my to another state n2 t mg e Not all transitions are possible there are selection rules Al 1 m 0 it w These rules demand conservation of angular momentum Since a photon carries an intrinsic angular momentum of 1 Illanveleetron atoms The orbital annroximation r The orbital approximation to the total many electron wavefunction I r1 r2 is to rewrite it as a product of oneelectron wave functions Mr l l z sothat I r1r2 wow The configuration of an atom is the list of occupied orbitals r The Pauli exclusion principle states that the spins must be paired if two electrons are to occupy one orbital and no more than two electrons may occupy an orbital Penetration and shielding n In manyelectron atoms the s p d etc orbitals of each shell are not degenerate An electron distance r from the nucleus experiences the nuclear charge Z shielded by all of the other electrons The effective charge is Ze ZEff Z cwhere c is the shielding parameter An s electron has a greater penetration through inner shells than a p electron of the same shell because the p electron has a node at the nucleus The Authau nrinoinle Aufbau means buildingup in German a The configurations of atoms are built up by population of the hydrogenic orbitals a Imagine a bare nucleus with charge Z and then add Z electrons to the orbitals in the following order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s etc Hund s rule an atom in its ground state adopts a configuration with the greatest number of unpaired spins The nrohlem oi mullinle electrons The central difficulty with application of the Schrodinger equation is the presence of electronelectron interaction terms in the potential energy 7 I Zez J 62 V 77 r 4neur 112 47mm electmn 7 nuclear electmn 7 electmn rs No analytical solutions HartreeFock procedure Find solutions that optimize the electron in each orbital in the presence of the eld of all of the other orbitals HartreeFock nrocetlure As an example for He we can write the two electron wave function as a product oforbitals r119 r1 r2 The probability distr bution at r2 r2drZ for electron 2 corresponds to a charge density in classical hysics Therefore we can say that the effective potential felt by electron 1 is Vim I ltr2gt ltr2gtdr2 Tl39le sellconsistent field method The effective or average potential can be used in a one electron hamiltonian operator for electron 1 ef 1 2 2 ff H1 r1TT2V1T7Jr V1201 The Schrodinger equation is solved for electron 1 WW who start with a trial function r2 and solve for Mn Using rpm calculate an effective potential for 2 and solve for r2 Continue until convergence is eached Atomic term SVIIIIIIJIS The letter indicates the total orbital angular momentum quantum number of all electrons in an atom Ll1lz 4421 Hill 421 The left superscript gives the multiplicity 281 where S s1s2 s3 r The right subscript gives the total angular momentum J LS LS 1 L S The term symbol is EHL J I39llIIIII39S rules determine the term SVIIIIIIJI 0f the ground state Each state is designated by a term symbol corresponds to a wave function that is an eigenfunction of L2and with unique energy The state with the larges value ofS is the most stable For states with the same value of S the state with the largest value ofL is the most stable lfthe states have the same value of L and S subshell less than half lled smallestJ is most stable subshell more than half lled the state with the largest value OH is the most stable Chemistry 331 Lecture 3 Absorption spectra of atoms The Schrodinger equation for hydrogen The electronic structure of atoms NC State University The Solar Spectrum There are gaps in the solar emission called Frauenhofer lines 2 The gaps arise from specific atoms in the sun that absorb radiation Electronic SII IIGIIII B 0f quotWII IIQBII The Schrodinger equation for hydrogen Separation of variables Radial and angular parts Hydrogen atom wavefunctions Expectation values Spectroscopy of atomic hydrogen Schriitlinger equation flll hydrogen The lunetlc energy operator The Schrodinger equation in three dimensions is 752 2 V 1 V E 1 2H The operator deIsquared is V 2 62 82 82 8X2 ayz 822 The procedure uses a spherical polar coordinate system Instead of x y and z the coordiantes are 6 I and r SBIII lIiIIEIBI equation for IIVIII IIEIBII l0 f l lll 0f the Il lBllli l The Coulomb potential between the electron and the proton is V Ze247t801 The hamiltonian for both the proton and electron IS 712 2 z 2mN V quot 2me Separation of nuclear and electronic variables results in an electronic equation in the centerof mass coordinates H h22uV2 Z6247t801 1 ulme1mN V V It would be impossible to solve the equation with all three variables simultaneously Instead a procedure known as separation of variables is used The steps are 1 Substitute in LPr6 RrY9 2 Divide both sides by RrY6 3 Multiply both sides by 2er2 The wavefunction solutions of the angular equation are spherical harmonics Ycm 94 2 These functions describe the angular distribution of atomic orbitals and are the wavefunctions for the rigid rotor of polyatomic molecules i The degeneracy of a given orbital is 261and the angular momentum of the electron is lll 1h 1 The form of the spherical harmonics th 64 is quite familiar The shape of the sorbital resembles the first spherical harmonic YOO Attached to this lecture are three MAPLE worksheets that illustrate the s p and d orbitals respectively The idea is to obtain an interactive picture of the mathematical form and the plots of the functions Disclaimer The spherical harmonics have been simplified by formation of linear combinations to remove any complex numbers MAPLE WOI IKSIIEBI 0 snherical harmonics The Y00 spherical harmonic has the form of an s orbital There is only one angular function fort 0 The Y10Y11 and Y11 spherical harmonics have the form of porbitals 3 39 There are three angular functions for 1 1 The Y20 Y21 Y2 1 YZ2 and Y22 spherical harmonics have the form of dorbitals 1 There are five angular functions fort 2 The effective potential result of the S lllli ll 0f the angular part The solutions for the angular part result in a term in potential energy equal to V 2ll12 prZ This term contains the contributions to the energy from angular terms Together with the Coulomb potential the effective potential energy is Veg Ze247T801 2 12ur2 The radial equation for IWIII IIQBII Making the above approximations we have an radial hamiltonian energy operator 2 V2R 292 RmR 2p 47teor 2er2 The solutions have the form RM NH pl eP2 Ln7er Where p 2ZrnaO and a0 4780h2m62 NHe is the normalization constant aneyr is an associated Laguerre polynomial The Bohr radius The quantity a0 4780 2m62 is known as the Bohr radius The Bohr radius is a0 0529 A Since it emerges from the calculation of the wave functions and energies of the hydrogen atom it is a fundamental unit In socalled atomic units the unit of length is the Bohr radius 80 1 A is approximately 2 Bohr radii You should do a dimensional unit analysis and verify that aO has units of length The normalization constant depends on n and l N 2 32 n39l 2nn1l3 n ao 12 The associated Laguerre polynomials are n1 0 L1 x 1 I These are given 2 X for completeness n2 0 Lgx n2 1L x n3 0 L13X n3 1Lix 4 x 2 s s3 3x 128 4 5 n3 2sz Each of the radial equation solutions is a polynomial multiplying an exponential The normalization is obtained from the integral I RRr2dr 1 0 The volume element here is rzdr which is the r part of the spherical coordinate volume element rzsinedrde dd IVIAI39lE worksheet Normalization 0f l0 radial functions A MAPLE worksheet attached to this lecture illustrates the normalization of the first three radial functions The worksheet includes plots of the functions When examining a plot keep in mind that you can plot the wave function or the square of the wave function We often plot the square of the wave function because the integral of the square of the wave function gives the probability Energy levels of IWIII39IIEIBII atom 2 The energy levels of the hydrogen atom are specified by the principal quantum number n rue 1 327t2802 2 n2 All states with the same quantum number n have the same energy All states of negative energy are bound states states of positive energy are unbound and are part of the continuum The Rydberg constant The energy levels calculated using the Schrodinger equation permit calculation of the Rydberg constant One major issue is units Spectroscopists often use units of wavenumber or cm39l At first this seems odd but hv hcA hcv where 392 isthe value of the transition in wavenumbers 4 R1 H9 in cm391 70 327152802712 r All of the orbitals of a given value of n for a shell n 1 2 3 4 correspond to shells K L M N Orbitals with the same value of n and different values of 5 form subshells l O 1 2 correspond to subshells s p d Using the quantum numbers that emerge from solution of the Schrodinger equation the subshells can be described as orbitals HVIII IIQBIIiG orbitals i s orbitals are spherically symmetrical The is wavefunction decays exponentially from a maximum value of 17ta03 2 at the nucleus p orbitals have zero amplitude at rO and the electron possesses an angular momentum of TNW 1 The orbital with m O has zero angular momentum about the z axis The angular variation is cose which can be written as zr leading to the name pZ orbital quotVIII IIQBII 1 S radial wavetunction 20 The 13 orbital has no nodes and decays 15 exponentially g R15 21a032e39P2 10 L1113 n 1 and t O are the quantum numbers for this orbital 05 The Radial Distribution in Hydrogen 28 and 2 orbitals 06 L1 2 P2 p S The expectation or average value of an observable ltrgt is given by the general formula lt rgt I Ifr PdT 0 As written the above integral describes the expectation value of the mean value of the radius r Integration over the angular part gives 1 because the spherical harmonics are normalized The volume element can be written dr r2dr The mean value is ltrgtl lJP kr JPerr lPlPr3dr 0 0 MAPlE worksheet on Expectation Average WIIIIBS There is a MAPLE worksheet attached to this lecture that illustrates the use of expectation values The example of the position ltrgt is given for the 1s 2s and 3s radial wave functions The expectation value or average value of r gives the average distance of an electron from the nucleus in a particular orbital Since the 2s orbital has one radial node and the 3s orbital has two radial nodes the average distance of the electron from the nucleus is shown to increase as ltrgt for 3s gt ltrgt for 2s gt ltrgt for 1s SIIBGII IISGIIIW 0f atomic IIVIII IIEIBII Spectra reported in wavenumbers u Rydberg fit all of the series of hydrogen spectra with a single equation Absorption or emission of a photon of frequency v occurs in resonance with an energy change AE hv Bohr frequency condition Solutions of Schrodinger equation result in further selection rules i r A transition requires a transfer from one state with its quantum numbers n1 6 m to another state n2l5 m2 Not all transitions are possible there are selection rules AK r 1 m O r 1 2 These rules demand conservation of angular momentum Since a photon carries an intrinsic angular momentum of 1 Manyelectron atoms The orbital annroximation The orbital approximation to the total many electron wavefunction Pr1 r2 is to rewrite it as a product of oneelectron wave functions wr1wr2 so that Tm r2 Wr1Wr2 The configuration of an atom is the list of occupied orbitals 3912 The Pauli exclusion principle states that the spins must be paired if two electrons are to occupy one orbital and no more than two electrons may occupy an orbital In manyelectron atoms the s p d etc orbitals of each shell are not degenerate An electron distance r from the nucleus experiences the nuclear charge Z shielded by all of the other electrons The effective charge is Zeff Zeff Z o where o is the shielding parameter An s electron has a greater penetration through inner shells than a p electron of the same shell because the p electron has a node at the nucleus The Aufbau nrincinle Aufbau means buildingup in German The configurations of atoms are built up by population of the hydrogenic orbitals Imagine a bare nucleus with charge Z and then add Z electrons to the orbitals in the following order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s etc 39 Hund s rule an atom in its ground state adopts a configuration with the greatest number of unpaired spins l0 nrohlem 0f multinle BIBGII IIIIS The central difficulty with application of the Schrodinger equation is the presence of electronelectron interaction terms in the potential energy 11 Ze2 e2 V I 4780rl 1321 47580739lj electron nuclear electron electron No analytical solutions 39 HartreeFock procedure Find solutions that optimize the electron in each orbital in the presence of the field of all of the other orbitals As an example for He we can write the two electron wave function as a product of orbitals L11r1 r2 r1 r2 The probability distribution 1r21r2dr2 for electron 2 corresponds to a charge density in classical physics Therefore we can say that the effective potential felt by electron 1 is V r1 r2rin r2dr2 The SBIfG llSiSlBllt field IIIBIIIIIII The effective or average potential can be used in a one electron hamiltonian operator for electron 1 A e 1 2 2 27 H1 V1 V1 7 V16 V1 The Schrodinger equation is solved for electron 1 A eff H1 Mn 81 r1 Start with a trial function r2 and solve for r1 Using r1 calculate an effective potential for 2 and solve for r2 Continue until convergence is reached Atomic IBI III SVIIIIIIIIS l The letter indicates the total orbital angular momentum quantum number of all electrons in an atom L 81 HZJ1 HZ 1 i1 l2 The left superscript gives the multiplicity 281 where S s1 s2 s3 g The right subscript gives the total angular momentumJ LS LS 1 LS 1 The term symbol is ZSHL J I39Illllll39S I IIIBS determine the term symbol 0f the l llllll state Each state is designated by a term symbol corresponds to a wave function that is an eigenfunction of L2 and 82 with unique energy 1 39 The state with the larges value of S is the most stable 139 For states with the same value of S the state with the largest value of L is the most stable If the states have the same value of L and S subshell less than half filled smallest J is most stable subshell more than half filled the state with the largest value OH is the most stable

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