Ch 10. Many-Electron Atoms MS310 Quantum Physical Chemistry - Schrödinger equation cannot be solved analytically - Schrödinger equation cannot be solved analytically when an atom has more than one electron. when an atom has more than one electron. - Approximate numerical methods can be used to obtain the eigenfunctions and eigenvalues of the Schrödinger the eigenfunctions and eigenvalues of the Schrödinger equation for many-electron atoms. equation for many-electron atoms. - New issues such as indistinguishability of e - s, e - spin, interaction between and need to be considered. interaction between and need to be considered. The Aim of the Chapter

MS310 Quantum Physical Chemistry 10.1 He : the smallest many-electron atom We can solve analytically the hydrogen atom, but cannot solve analytically the helium atom. Separate the total motion and internal motion, Schrödinger equation is given by (Laplacian of e 2 is similar to e 1 ) The wavefunction of S.E depends on the all cocordinates of electrons cannot be obtained How one can solve S.E? Orbital Approximation on many-electron eigen functions

MS310 Quantum Physical Chemistry Orbital approximation : wavefunction of many-electron atoms is given by the product of one-electron orbitals φ n (r n ) : similar to hydrogen atom orbital and associated with one-electron orbital energy ε n when using the orbital approximation : big problem is Potential of He atom is given by Orbital : eigenfunction of H atom It doesn’t think about the last term : electron-electron repulsion When dealing with Coulomb potential, electron correlation is neglected.

MS310 Quantum Physical Chemistry What does neglect of the electron correlation in He atom lead to? We know both electrons of He atom are in the 1s orbital → the wavefunctions similar to 1s orbital wavefunction Zeta(ζ) : ‘effective nuclear charge’ : felt by the electron No electron correlation : electron 1 interact with nucleus and feels averaged charge distribution of electron 2. Spatially averaged charge distribution of electron 2 : φ*(r 2 )φ(r 2 ) Negative charge in the volume dτ : -eφ*(r 2 )φ(r 2 )dτ Under this approximation * n-electron Schrödinger E → n·one-electron

MS310 Quantum Physical Chemistry 10.2 Introducing electron spin In Stern-Gerlach experiment, there are 2 deflected beams. Case of l>0 : nonzero magnetic moment → splitting Valence electron of Ag is only 1, 5s → cannot split(l=0) However, splitting is observed! → solved by the ‘spin’, the intrinsic angular momentum s In this case, it splits into 2s+1 beams : s=1/2 z-component element of angular momentum : s z = m s ℏ = ± ℏ /2 Intrinsic : spin independent to environment of electron(r,θ,φ) What about the many-electron case? : ‘doubly occupied’ in the orbitals of many-electron atom one electron has m s = +1/2, the other electron has m s = -1/2 → added to 4 th quantum number to the H atom eigenfunction, ψ nlm l m s (r,θ,φ)

MS310 Quantum Physical Chemistry Define the 2 spin wavefunction as α and β, and spin angular momentum operator and. They satisfy the following properties : σ : spin variable (not a spatial variable). It is orthogonal to spatial variables.

MS310 Quantum Physical Chemistry For example, eigenfunctions of H atom ψ nlm l m s (r,θ,φ) are They also satisfy the orthonormal relation. Hamiltonian doesn’t depend on the spin coordinate. → these two wavefunctions have the same energy.

10.3 Wave functions must reflect the indistinguishability of electrons MS310 Quantum Physical Chemistry Discuss the He : electrons are numbered 1 and 2 Macroscopic object can be distinguished from one another However, no way to distinguish between any 2 electrons! How can indistinguishability be introduced by the orbital approximation? Consider the n-electron wavefunction ψ(1,2,…,n)=ψ(r 1,θ 1,φ 1,σ 1, r 2,θ 2,φ 2,σ 2,…,r n,θ n,φ n,σ n ) See the case of He. Wavefunction is not observable, but square of wavefunction is proportional to electron density and observable. 2 electrons of He are indistinguishable → if we change the electron 1 and 2, there are no change of observable! → ψ 2 (1,2) = ψ 2 (2,1)

MS310 Quantum Physical Chemistry ψ 2 (1,2) = ψ 2 (2,1) means ψ(1,2) = ψ(2,1) or ψ(1,2) = - ψ(2,1) ψ(1,2) = ψ(2,1) : symmetric wave function ψ(1,2) = - ψ(2,1) : antisymmetric wave function For a ground-state He atom, symmetric and antisymmetric wavefunctions are given by Postulate 6 : Wave functions describing a many-electron system must change sign(be antisymmetric) under the exchange of any two electrons. W. Pauli showed only the antisymmetric wave function is allowed for electrons. This postulate is also known as the Pauli exclusion principle.

MS310 Quantum Physical Chemistry Meaning of this postulate → single-product wave function cannot be antisymmetric. Ex) How can solve it? → Slater determinant m=n/2(n:even) or m=(n+1)/2(n:odd) Why determinant can construct antisymmetric wave function? → value of determinant automatically change the sign when 2 rows are interchanged.

MS310 Quantum Physical Chemistry Ground state of He : represented by 2x2 determinant Ground state of He : two electrons have the same quantum number n, l, and m l and only the difference is m s (+1/2 for one e - & -1/2 for the other). Electrons : assigned to orbitals by a configuration Configuration : specifies the values of n and l for each electron ex : ground state of He : 1s 2 F : 1s 2 2s 2 p 5 In a configuration m l and m s are not specified : total energy doesn’t depend on.

MS310 Quantum Physical Chemistry Consider the ground state of Li. If third electron is in the 1s orbital, wavefunction is given by However, third column is same to first column → ψ(1,2,3)=0! Therefore, third electron must in the next level, 2s (L 1 : 1s 2 2s 1 ) and real wavefunction is given by It shows Pauli exclusion principle(at most 2 electron can occupy in the one orbital) Orbitals with the same n value : shell Orbitals with the same n and l values : subshell

MS310 Quantum Physical Chemistry

10.4 Using the variational method to solve the Schrödinger equation MS310 Quantum Physical Chemistry Many-electron atom : only can approximate One of the method : ‘Hartree-Fock self-consistent field method’ combined with the variational method Schrödinger equation for ground state is Ĥψ 0 =Eψ 0 We can rewrite the equation by the integrating : If we cannot know the eigenfunction, how can we approach to energy(also, eigenvalue)? → variational theorem Φ : trial wave function(prove : end-of-chapter problem) Set Φ and optimization parameter α, we can find optimum value

MS310 Quantum Physical Chemistry Example : particle in a box Real ground-state wavefunction is Take the trial function i) α=0 Real minimum energy : 0.125h 2 /ma 2 Trial energy is bigger than real energy and there are small difference between real wavefunction and trial function

MS310 Quantum Physical Chemistry ii) α≠0 Integrate it, we can obtain Calculate the dE/dα=0 and find extreme value of α → α = and α = , minimum : α = In this case, energy of trial function is 0.127h 2 /ma 2 and it is very close to real value.

MS310 Quantum Physical Chemistry See the graph of the optimized trial function Contribution of term : blue line We can see E→E 0 and Φ→ψ 0 Obtain the ‘best’ energy : depends on the choice of trial function We can obtain the lower energy if we choose the trial function Φ(x)=x α (a-x) α This example shows Schrödinger equation can be solved by the variational method.

MS310 Quantum Physical Chemistry 10.5 The Hartree-Fock self-consistent field method Antisymmetric wavefunction is given by the slater determinant It is represented by the modified Hydrogen-like orbitals, φ j (k) Hartree-Fock method : single Slater determinant gives the lowest energy for ground-state, absence of electron correlation We discussed in 10.1, HF method assumed the electrons are uncorrelated and that a particular electron feels the spatially averaged electron charges distribution of remaining n-1 electrons.

MS310 Quantum Physical Chemistry Use HF method : n-electron Schrödinger equation → n 1-electron Schrödinger equations 1-electron Schrödinger equation is given by HF method allows the best one-electron orbitals and the corresponding orbital energies to be calculated. Neglect of electron correlation → effective potential is spherically symmetric → angular part of the wave functions is identical to the solutions for the hydrogen atom

MS310 Quantum Physical Chemistry To optimize the radial part of the determinantal wave function → Variational method in 10.4 Individual entries in the determinant are expressed as a linear combination of suitable basis functions The criterion for a “good” basis set → the number of terms in the sum representing φ j (r) is as small as possible → basis functions enable the HF calculations to be carried out rapidly

MS310 Quantum Physical Chemistry Two examples of basis set expansions for atomic orbitals → Figures 10.3 & p atomic orbital of Ne obtained in a HF calculation

MS310 Quantum Physical Chemistry H 1s AO and the contributions of each member of the m=3 basis set → Figure 10.4

MS310 Quantum Physical Chemistry To solve the Schrodinger equation for electron 1, V 1 eff (r 1 ) must be known → we must know the functional form of all other orbitals For an initial set of φ j (k) → effective potential is calculated → energy and improved orbital functions, φ' j (k), for each of the n electrons are calculated φ' j (k) → φ'' j (k) Until the solutions for the energies and orbitals are self-consistent Coupled with the variational method in optimizing the parameters in the orbitals → effective in giving the best one-electron orbitals and energies available for a many-electron atom in the absence of electron correlation

MS310 Quantum Physical Chemistry Accuracy of a HF calculation depends primarily on the size of the basis set He 1s orbital cannot be accurately represented by a single exponential functions as was the case for the hydrogen atom

MS310 Quantum Physical Chemistry HF radial functions → To obtain the radial probability distribution for many-electron atoms

MS310 Quantum Physical Chemistry Important result of HF calculation : ε i for many-electron atoms depends on the both n and l. ε ns < ε np < ε nd < … : not same as the H atom See the radial distribution of 3s, 3p, and 3d. Probability of finding the electron : 3s>3p>3d Therefore, degree of shielding is different and 3s electrons are more tight binding than 3p and 3d electrons. Similarly, energy of 3p orbital is lower than 3d orbital

MS310 Quantum Physical Chemistry Effective nuclear charge ζ is smaller than real nuclear charge. Why? ‘Shielding’ Electron doesn’t feel only the charge of nucleus, but also feels ‘average charge distribution’ of another electrons! In same period small n : a few electrons effects large n : a lot electrons effects → Z-ζ increases as the across the periodic table

MS310 Quantum Physical Chemistry How can use the orbital energy ε i ? ‘ionization energy’ Highest occupied orbital energy ε i : first ionization energy (first ionization energy : energy of ground state to free) → Koopman’s theorem Lowest unoccupied orbital energy ε i : electron affinity However, accuracy of electron affinity is much less than ionization energy.

MS310 Quantum Physical Chemistry See the case of Scandium(Sc) ε 4s and ε 3d : depends on the configuration of Sc(4s 2 3d 1, 4s 1 3d 2, 4s 0 3d 3 ) because of orbital energy of many-atom electron depends on ‘distribution’ of another electrons. Known distribution of orbital energy is given by this picture It called as the ‘Aufbau principle’

MS310 Quantum Physical Chemistry Electron configuration of 4 th period elements is given by Configuration of Cr and Cu : ‘Hund’s rule’ In K and Ca, ε 4s ε 3d Calculate the ∆E of 4s 2 3d n → 4s 1 3d n+1 is given by ∆E(4s→3d) ≈ (ε 3d - ε 4s ) + [E repulsive (3d,3d) – E repulsive (3d,4s)] (by Vanquickenbourne)

MS310 Quantum Physical Chemistry d electron : more locally distributed than s electron → E repulsive (3d,3d) > E repulsive (3d,4s) > E repulsive (4s,4s) Therefore, even though ε 4s > ε 3d, magnitude of repulsion is greater than ε 4s - ε 3d → promotion 4s 2 3d 1 → 4s 1 3d 2 cannot be occurred. HF calculation : neglecting the electron correlation Therefore, total energy of HF > true energy Difference : correlation energy It increases roughly linearly with the number of electron pairs in the atom. However, it is small amount of total energy. For example, correlation energy of He : 110 kJ mol -1, 1.4% of total energy However the correlation error exist in both reactant and product, therefore this term cancelled out in thermodynamic function

10.6 Understanding trends in the periodic table from HF calculations from HF calculations MS310 Quantum Physical Chemistry Result of HF calculation is 1) orbital energy depends on n and l, and ε ns < ε np < ε nd < … 2) electrons in a many-atom are shielded from the full nuclear charge by other electrons. 3) ground-state configuration results from a balance between orbital energy and electron-electron repulsion HF method : numerical calculation → no simple formula for ζ by the radius r → Ex) VDW radii VDW radii : radius of include the 98% of electron charge Reaction : related to HOMO and LUMO It predict the binding of NaCl i) Na + Cl - : ∆E = E ionization Na - E electron affinity Cl = 5.14eV eV = 1.53eV ii) Na - Cl + : ∆E = E electron affinity Na - E ionization Cl = 12.97eV – 0.55eV = 12.42eV → Na + Cl - is more favorable than Na - Cl +

MS310 Quantum Physical Chemistry

Atomic radius decrease continuously in going across the period. It increases abruptly as n increases. Electronegativity same pattern as the IE.

MS310 Quantum Physical Chemistry - Slater determinant can represent the orbital- type wavefunction satisfying the Pauli exclusion principle. - Result of HF-SCF method for helium atom is approximately same. - Study about the electron configuration, good quantum number. Summary