《无机化学 Chemical Theory》课程教学资源（讲稿）Part II Molecular Spectroscopy Chapter 6 Huckel Molecular Orbitals 休克尔分子轨道

Part IIISymmetry and BondingChapter 6 Hickel Molecular OrbitalsProf.DrXinLu（吕鑫）Email:xinlu@xmu.edu.cnhttp:/ /pcoss.xmu.edu.cn/xlv/index.htmlhttp://pcoss.xmu.edu.cn/xlv/courses/theochem/index.html

Part III Symmetry and Bonding Chapter 6 Hückel Molecular Orbitals Prof. Dr. Xin Lu （吕鑫） Email: xinlu@xmu.edu.cn http://pcoss.xmu.edu.cn/xlv/index.html http://pcoss.xmu.edu.cn/xlv/courses/theochem/index.html

6. Hickel molecular orbitals (HMO)休克尔分子轨道. So far we have just been drawing up qualitative MO diagrams aided by symmetryconsiderations without computing the energies and forms of any molecular orbitals.: Of course, it is now possible to compute the detailed form and energy of the MOsusing a computer program such as Hyperchem, G16, Dmol3, ADF, Molpro etc.: Anyway, it is both useful and instructive to do some MO calculations by hand'. Thistopic will be talked about in this chapter

6. Hückel molecular orbitals (HMO) 休克尔分子轨道 • So far we have just been drawing up qualitative MO diagrams aided by symmetry considerations without computing the energies and forms of any molecular orbitals. • Of course, it is now possible to compute the detailed form and energy of the MOs using a computer program such as Hyperchem, G16 , Dmol3, ADF, Molpro etc. • Anyway, it is both useful and instructive to do some MO calculations ‘by hand’. This topic will be talked about in this chapter

6.1 The LCAO method: The simplest and most intuitive way to construct molecular orbitals is to use the linearcombination of atomic orbitals (LCAO)method, which we have been doing up to nowEach MO is expressed as a linear combination of atomic orbitals, Φi, Φ2, . . ., = CiΦ1 + C2Φ2 + C3Φ3 + ...Φ; ~ ith AO (also known as one of the basis functions) used to construct the MO.c; ~ the coefficient which indicates the relative contribution of an AO Φ, to the MO.: The problem we have to solve is finding the values of the coefficients and thecorresponding energy for each MO.The key principle to solve such a problem is the variation theorem

6.1 The LCAO method • The simplest and most intuitive way to construct molecular orbitals is to use the linear combination of atomic orbitals (LCAO) method, which we have been doing up to now. Each MO 𝝍 is expressed as a linear combination of atomic orbitals, 𝜱𝟏, 𝜱𝟐, . . ., 𝝍 = 𝒄𝟏𝜱𝟏 + 𝒄𝟐𝜱𝟐 + 𝒄𝟑𝜱𝟑 + ⋯ ci ~ the coefficient which indicates the relative contribution of an AO 𝜱𝒊 to the MO. 𝜱𝒊 ~ ith AO (also known as one of the basis functions) used to construct the MO. • The problem we have to solve is finding the values of the coefficients and the corresponding energy for each MO. The key principle to solve such a problem is the variation theorem

6.1.1 Derivation of the secular eguations - Variation Theorem中=c,Φ;? For a MO expressed as an LCAO sum:1the expectation value E of the Hamiltonian is calculated in the usual way:SHdtS(E,c,Φ)H(Z,c,Φ,)dtZijCic, JΦ,H,dtE =(H)J(E,c,Φ) (,c,Φ,)dtEij cic, J Φ,Φ, dtJdtwhich involves computation of the following two types of integrals :end up here with aHi =JΦ,HΦ,dtcertain value of E ifSij =JΦ,Φ,dtwekneweverytermsS,~ the overlap integral between the two basis functions Φ; and Φ,withinthisexpression.Hiy ~ a matrix element of the operator H (the Hamiltonian for the system).: According to the variation principle, we need to minimize E with respect to thecoefficients c, i.e. aE/ ac,=0.: Now we rewrite the equation as,EZij CiCjSij = Zij Cic,Hij

6.1.1 Derivation of the secular equations – Variation Theorem 𝑯𝒊𝒋 = 𝜱𝒊𝑯 𝜱𝒋𝒅𝝉 𝑺𝒊𝒋 = 𝜱𝒊𝜱𝒋𝒅𝝉 which involves computation of the following two types of integrals : Sij ~ the overlap integral between the two basis functions 𝜱𝒊 and 𝜱𝒋 . 𝑯𝒊𝒋 ~ a matrix element of the operator 𝑯 （the Hamiltonian for the system). the expectation value E of the Hamiltonian is calculated in the usual way: 𝑬 = 𝑯 = 𝝍𝑯 𝝍𝒅𝝉 𝝍 𝝍𝒅𝝉 𝝍 = 𝒊=𝟏 𝑵 • For a MO expressed as an LCAO sum: 𝒄𝒊𝜱𝒊 = ( 𝒊 𝒄𝒊𝜱𝒊)𝑯 ( 𝒋 𝒄𝒋𝜱𝒋)𝒅𝝉 ( 𝒊 𝒄𝒊𝜱𝒊) ( 𝒋 𝒄𝒋𝜱𝒋)𝒅𝝉 = 𝒊,𝒋 𝒄𝒊𝒄𝒋 𝜱𝒊𝑯 𝜱𝒋 𝒅𝝉 𝒊,𝒋 𝒄𝒊𝒄𝒋 𝜱𝒊𝜱𝒋 𝒅𝝉 • According to the variation principle, we need to minimize E with respect to the coefficients ci , i.e. E/ ci=0. • Now we rewrite the equation as, 𝑬 𝒊,𝒋 𝒄𝒊𝒄𝒋𝑺𝒊𝒋 = 𝒊,𝒋 𝒄𝒊𝒄𝒋𝑯𝒊𝒋 end up here with a certain value of E if we knew every terms within this expression

6.1.1 Derivation of the secular equations. We then take the (partial) derivative of both sides with respect to the coefficient c,a]-[Ecic,Hjaci(i=1,2,...,N;i.e.,atotal ofNequations!)aEZZgsu=ZCic,Sij + Ec,Hijaciij11.Demanding aE/ ac,=0, then we haveZoSu=-ZoMcjHij(Hij-ESij)cj = 0(i=l,2....N,i.e.,atotalofNequations!)

6.1.1 Derivation of the secular equations • We then take the (partial) derivative of both sides with respect to the coefficient ci . • Demanding E/ ci=0, then we have 𝐸 𝑗 𝑐𝑗𝑆𝑖𝑗 = 𝑗 𝑐𝑗𝐻𝑖𝑗 𝑗 (𝐻𝑖𝑗−𝐸𝑆𝑖𝑗)𝑐𝑗 = 0 𝜕𝐸 𝜕𝑐𝑖 𝑖𝑗 𝑐𝑖𝑐𝑗𝑆𝑖𝑗 + 𝐸 𝑗 𝑐𝑗𝑆𝑖𝑗 = 𝑗 𝑐𝑗𝐻𝑖𝑗 (i = 1,2,.,N; i.e., a total of N equations!) 𝜕 𝜕𝑐𝑖 𝐸 𝑖𝑗 𝑐𝑖𝑐𝑗𝑆𝑖𝑗 = 𝜕 𝜕𝑐𝑖 𝑖𝑗 𝑐𝑖𝑐𝑗𝐻𝑖𝑗 (i = 1,2,.,N; i.e., a total of N equations!)

Derivation of the secular equations. The Nequations can be conveniently expressed in matrix form (Nis the number ofbasis functions):HiNSi1SINS12S13H11H12H13G/0)H21S21H2NS22S23S2NH22H2330H31S31S3NH32H33H3NS32S33-E一00:::...::::::0CNHNISNIHN2HN3HNNSN2SN3SNNN2 H,-type integrals and N2 S,type integrals to be computed!!!These are called the secular equations (久期方程)and in general their solution will leadto N different values of E, each corresponding to a MO: By substituting the corresponding value of the energy E back into the secular equations, thecoefficients ic, corresponding to a particular MO can be found

Derivation of the secular equations • The N equations can be conveniently expressed in matrix form (N is the number of basis functions): • These are called the secular equations (久期方程) and in general their solution will lead to N different values of E , each corresponding to a MO. • By substituting the corresponding value of the energy E back into the secular equations, the coefficients {ci } corresponding to a particular MO can be found. N2 Hij-type integrals and N2 Sij-type integrals to be computed!!!！ 0 0 0 ⋮ 0 = 0 𝑐1 𝑐2 𝑐3 ⋮ 𝑐𝑁 E

6.1.2 The Hiickel approximations.The Hickel approximations1) set S,= J ΦiΦ,dt = 0 (i) or 1 (i-)Then the secular equations look simpler,H12HiNHi1H130001C10H210H22H2N10H23C2H31H320H3NH3300C3= 0-E......::::...:HN1000HN2HN3HNNCNSecularmatrixand can be rewritten as（久期矩阵）H12Hii - EH13HINc1These equations can be solved byH21H23H2NH22 - EC2firstly settingthe determinant of theH31H3NH32H33-EC3=0.secularmatrixnamelythe secular...:.....::determinant(久期行列式),to be zero.HN2HN1HN3HNN-ECN

6.1.2 The Hückel approximations • The Hückel approximations: 1) set Sij= 𝝓𝒊𝝓𝒋𝒅𝝉 = 0 (ij) or 1 (ij) Then the secular equations look simpler, and can be rewritten as Secular matrix (久期矩阵) These equations can be solved by firstly setting the determinant of the secular matrix, namely the secular determinant (久期行列式), to be zero

6.1.2 The Hickel approximations2) Calculating the actual values of the matrix elements H, is itself a formidable task, sowe sidestep this by simply leaving them as parameters,Hi = Φ;HΦ;dt = αi(approx. as the energy of the AO )HickelapproximationsHj = JΦ;Hjdt = βj (resonance integral)β, is zero unless the two orbitals are on adjacent atoms, i.e., directly overlapping!.Some of the β, terms can be: Accordingly, the secular equations becomezero case by case!βiNQ-Eβ13β12C1The values of α, β, can beβ21β23β2Nα2-EC2determined semi-empirically!β31β3203-Eβ3NC3= 0.Quiteeasyfordealingwith元-..............·conjugation systems!βN1βBN2βN3QN-ECNQl:howtodetermineα.&β.?Q2: For an allylic r system, write out the secular equation!

6.1.2 The Hückel approximations 2) Calculating the actual values of the matrix elements Hij is itself a formidable task, so we sidestep this by simply leaving them as parameters, 𝑯𝒊𝒊 = 𝝓𝒊𝑯 𝝓𝒊𝒅𝝉 = 𝜶𝒊 𝑯𝒊𝒋 = 𝝓𝒊𝑯 𝝓𝒋𝒅𝝉 = 𝜷𝒊𝒋 (resonance integral) （approx. as the energy of the AO i ) ij is zero unless the two orbitals are on adjacent atoms, i.e., directly overlapping! • Accordingly, the secular equations become Hückel approximations • Some of the ij terms can be zero case by case! • The values of i , ij can be determined semi-empirically! • Quite easy for dealing with - conjugation systems! • Q1: how to determine c & cc? Q2: For an allylic 𝜋3 𝑥 system, write out the secular equation!

6.1.3 The allyl system. The allyl fragment. the Tt-type MOs formed from these P, orbitals, y = ciΦ, + C2Φ2 + C3Φ3The secular eqs. areβ12β13/α1-E/C1)β21β23α2-EC2=0frameoutofplane2porbitalscofatomβ32α3-EC3β31Can the eqs. be further simplified?!· These are C 2p orbitals. Set α,= α,= α3= α, βi2= β2, = β23 = β32 = β (Hickel approx.),Theseculareqs.thusbecome10(α-E)/βC111(α - E)/β=0C201(C3(α -E)/βNowsetx=(α-E)/B!

6.1.3 The allyl system • The allyl fragment: the π-type MOs formed from these p orbitals, 𝛼1 − 𝐸 𝛽12 0 𝛽21 𝛼2 − 𝐸 𝛽23 0 𝛽32 𝛼3 − 𝐸 𝑐1 𝑐2 𝑐3 = 0 • These are C 2p orbitals. Set 1= 2= 3=  , 12= 21 = 23 = 32 =  (Hückel approx.). The secular eqs. thus become The secular eqs. are 31 13 Can the Can the eqs eqs. . bbe further simplified?! e further simplified?! 1 2 3  = c11 + c22 + c33 Now set x = (-E)/! (𝛼 − 𝐸)/𝛽 1 0 1 (𝛼 − 𝐸)/𝛽 1 0 1 (𝛼 − 𝐸)/𝛽 𝑐1 𝑐2 𝑐3 = 0

6.1.3 The allyl system10)x/C1). Now we have the simplified secular equations as11C2=0x(with x = (α-E)/β)10(C3)1x: As usual, set the corresponding secular determinant to zero:x(x2 -1)-1× (x- 0)+0 ×(1- 0)= 00)1(xx(x2 - 1) - x = 0det=011xx(x2 - 2) = 010*x=021xE1 = α + V2β,E3=α-V2βE2 = α,. Let us start with x = -V2 that gives E, = α + v2β and the secular equations as1/-V20-V2c1 + C2 = 0[A](C1)Three eqs. are-V2C2=0 C1 - V2c2 + C3 = 0[B]11notindependent!(C3/-V2)[C]0C2 - V2c3 = 01

6.1.3 The allyl system • Now we have the simplified secular equations as (with x = (-E)/) 𝑥(𝑥 2 − 1) − 1 × (𝑥 − 0) + 0 × (1 − 0) = 0 𝑥(𝑥 2 − 1) − 𝑥 = 0 𝑥(𝑥 2 − 2) = 0 𝑑𝑒𝑡 𝑥 1 0 1 𝑥 1 0 1 𝑥 = 0 𝑥 1 0 1 𝑥 1 0 1 𝑥 𝑐1 𝑐2 𝑐3 = 0 • As usual, set the corresponding secular determinant to zero: x = 0,  𝟐 𝐸1 = 𝛼 + 2𝛽, 𝐸2 = 𝛼, 𝐸3 = 𝛼 − 2𝛽 • Let us start with 𝑥 = − 2 that gives 𝐸1 = 𝛼 + 2𝛽 and the secular equations as − 2 1 0 1 − 2 1 0 1 − 2 𝑐1 𝑐2 𝑐3 = 0 − 𝟐𝒄𝟏 + 𝒄𝟐 = 𝟎 [𝑨] 𝒄𝟏 − 𝟐𝒄𝟐 + 𝒄𝟑 = 𝟎 [𝑩] 𝒄𝟐 − 𝟐𝒄𝟑 = 𝟎 [𝑪] Three eqs. are not independent!