A+B+e−A+B+e−Electrostatic forces in the H2+moleculeantibindingregionbindingregionX Coordinate (Angstroms)-4 -3 -2 -1 0 1 2 3 4Energy (eV)-60-50-40-30-20-10010X coordinate (Angstroms)-4 -3 -2 -1 0 1 2 3 4Energy (eV)-60-50-40-30-20-10010(a)(b)V = -e2/rA - e2/rB + e2/ReqV = -e2/r2.77 eVPotential energy diagrams for H and H2+H atomH2+moleculeQuantum description of H2+The Born-Oppenheimer ApproximationThe electrons move so much faster than the nuclei that theelectronic energy depends only on the relative separationof the nuclei and is unaffected by their motion. Therefore,the electronic energy at a particular internuclear distancewill be the same as that for stationary nuclei.RA+B+e−rAerBe⎟⎟⎠⎞⎜⎜⎝⎛−+−=R1r1r1eVBeAe2When the Born-Oppenheimer approximation is applied, HelbecomesOnce ψeland E(R) have been determined as a function of R, the Nuclear motion may be treated;and the electronic motion can be separated from the nuclear motion;[])R(E)R()(TnucnucrrnucREψψ=+)r(R,E(R))r(R,Hel2elrrelReψψ=⎥⎦⎤⎢⎣⎡+)r,R(E)r,RΨ(ReHT2elnucrrrrΨ=⎟⎠⎞⎜⎝⎛++The Schrödinger equation for H2+⎟⎟⎠⎞⎜⎜⎝⎛+−∇−=BeAeerrerm11)(2H222elrhE(R) as a function of R for a stable bond ReqRE(R)0DeMolecular orbital theoryInstead of solving the Schrödinger equation to obtain the wavefunctions,one “constructs” wavefunctions using linear combinations of orbitals fromeach atom of the molecule. This is called the LCAO method.In the case of H2+, the molecular orbital wavefunctions areσg= Ng[ ϕ1s(A) + ϕ1s(B) ]σu= Nu[ ϕ1s(A) − ϕ1s(B) ]z/a0-8-6-4-202468ΨgABz/a0-8-6-4-202468ϕ1sABFormation of a sigma bonding molecular orbitalψA= φ1s(A)ψB= φ1s(B)ψg(AB) = Ng[φ1s(A) + φ1s(B)]z/a0-8 -6 -4 -2 0 2 4 6 8ΨuABz/a0-8 -6 -4 -2 0 2 4 6 8ϕ1sABFormation of an antibonding sigma molecular orbitalψA= φ1s(A)ψB= - φ1s(B)ψu(AB) = Nu[φ1s(A)−φ1s(B)]Polar plot representation of the H2+molecular orbitals••••••••+−1sA1sA1sB1sB1s σg1s σu*++−The variation principleWhen an approximate wavefunction is used to calculate the energy Ea,,dτψψdτHψψEa*aa*aa∫∫=Eais always greater than Eexact.x012345ψ0.00.10.20.30.40.50.60.7ψaψexactExercise: Determine the ground state energy of a particle ina 1-D box using the approximate wavefunctionΨa= N(xL − x2).Application of the variation method to the H2+molecule (αA− E)c1+ (β − ES)c2= 0secular equations(β − ES)c1+ (αB− E)c2= 0 α − E β − ES = 0 secular determinantβ − ES α − E dτSanddτHβdτHαwhereB*AB*AA*AAϕϕϕϕϕϕ∫=∫=∫= ,S1βαES1βαEgu++=−−=Integrals for calculating the energies of the H2+σgand σu*orbitalsEvaluate these integrals using spheroidal coordinates:⎟⎠⎞⎜⎝⎛++=2RR1eS2R-[]R)(1e1R27.21R121-27.21α2R+−+⎟⎠⎞⎜⎝⎛+=−R)(127.21eR121-27.21SβR+−⎟⎠⎞⎜⎝⎛+=−∫=−+=∫=+−=∫=dτ)/r(eKwhereK/R)eS(Eβdτ)/r(eJwhere/R,eJEαdτSBA2*A21sAB2*A21sB*AϕϕϕϕϕϕR/a0123456E(R) (eV)-18-15-12-9-6σuσgEHS1βαES1βαEgu++=−−= E(R) curves for the H2+σgand σu*orbitalsExercise: Set up the secular determinant for the wavefunctionψ = c1φAA+ + c2φBB+ + c3φCC((ααA A −−E) (E) (ββABAB−−ESESABAB) () (ββACAC−−ESESACAC))((ββBABA−−ESESBABA) () (ααB B −−E) (E) (ββBCBC−−ESESBCBC) = 0) = 0((ββCACA−−ESESCACA) () (ββCBCB−−ESESCBCB) () (ααC C −−E) E) ABCABCPictorial illustration showing the formation of H2+molecular orbitalsφ1sAφ1sBσg= Ng(φ1sA + φ1sB)σu= Nu(φ1sA −φ1sB)++++−Molecular orbitals formed from 2s and 2p atomic orbitals+++++−Relative energies of molecular orbitals for 2ndrow homonuclear diatomic moleculesMolecular orbital energy level diagrams for second row homonuclear diatomic moleculesLi2through N2O2through Ne2Exercise: Draw a molecular orbital energy level diagram forthe O2−molecule, write its molecular orbital configuration,and determine whether this molecule is paramagnetic. Whichmolecule would you expect to have a larger bond energy, O2or O2−?OO2−O−O2−: 2sσg22sσu22pσg22pπu4 2pπg3ParamagnetismLiquid O2Liquid N2NNN2N2: 2sσg22sσu22pπu42pσg2Molecular orbital configurations and properties of the second row homonuclear diatomic moleculesdz2dxz,yzdz2dxz,yzyzdxydxyxσπδδ∗π∗MMσ∗Metal-Metal BondingTerm symbols for diatomic moleculesug /12SΛ+∑=∑==ΛisiiimmSLLMMMlorbital mlvalueσ 0π ±1δ ±2φ ±3value of Λ symbol0 Σ1 Π2 Δ3 ΦExercises: Determine the term symbols for the followingconfigurations of H2:a). 1sσg2(gs)b). 1sσg1sσuc). 1sσg2sσgPotential energy curves for H21sσg21sσg1sσu1sσg2sσg1sσg21sσg1sσu1sσg2sσgExercises: Determine the term symbols for the ground state electron configuration of O2Potential energy curves for O2(2pπg1 2pπg1)πuπgMolecular orbital energy level diagram for HFhydrogen fluorine1s2pz2py2px2sHFσσ*nbσ = c1ϕ1s(H) + c2ϕ2pz(F) σ*= c3ϕ1s(H) − c4ϕ2pz(F)Molecular orbital energy level diagram for COThe valence bond methodVB method – construct wavefunctions for electron pair bondsusing atomic orbitals of the two bonded atoms. The resulting molecular orbitals are localized.MO method – construct molecular wavefunctions using atomicorbitals of allatoms of the molecule. The resultingmolecular orbitals are delocalized.Example: Wavefunctions for the ground state of H2.ψVB= NVB[ φA(1) φB(2) + φA(2) φB(1)] 2−½[ α(1)β(2) − α(2)β(1)]ψMO= σg(1) σg(2) 2−½[ α(1)β(2) − α(2)β(1)]whereσg(1) σg(2) = NMO[φA(1) φA(2) + φB(1) φB(2) + φA(1) φB(2) + φA(2) φB(1)] ionic ionic covalent covalentMolecular orbitals for linear methyleneH C HCH2Molecular orbital energy level diagrams for methyleneC CH2H HWalsh diagram for methyleneC CH2H Hpxpypzspzsp2hybridizationsp2sp2sp2psp2sp2sp2Formation of hybridized atomic orbitals in carbons + px+ pys + px+ py+ pysp3sp3sp3sp3sp3hybridizations + pzspsppxpysp hybridizationsp hybrid orbitalssp2hybrid orbitalssp3hybrid orbitalsdsp3hybrid orbitalsd2sp3hybrid orbitalsHomework problem 5-7. Make a polar plot in the xy plane of the wavefunctionΨsp2 = −(1/3)1/2ϕ2s − (1/6)1/2ϕ2px − (1/2)1/2ϕ2pywhere the ϕ’s are H-atom wavefunctions. (Note: you may set r/a = 1since only the angular dependence is of interest in this problem).ϕ2s= N(2 − r/a)e−2r/a= 2Ne−2− Ne−2=