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!)