Orbital approximation Assign the electrons to an atomic orbital and a spin Construct an antisymmetrized wave function using a Slater determinant evaluate the energy with the Hamiltonian that includes the electron-electron interactions Ψ ( r, r,, r ) = 1 N 1 N! φ ( r) φ ( r ) φ ( r ) Z Z Z 1s 1 1s 1s N Z Z Z 1s r1 1s r 1s rn φ ( ) φ ( ) φ ( ) φ ( r) φ ( r ) φ ( r ) Z Z Z N 1 N N N E = Ψ H ΨΨ Ψ
Lithium The antisymmetric 3 electron wavefunction can be written φ ( r) φ ( r ) φ ( r ) Ψ ( r, r, r ) =,, = ( r) ( r ) ( r ) Li Li Li 1s 1 1s 1s 3 Li Li Li 1 Li Li Li 1 3 φ1s φ1s φs φ1s 1 φ1s φ1s 3 3! Li Li Li φs r1 φs r φs r3 ( ) ( ) ( ) There are two possible configurations Li Li Li φ, φ, φ and φ1s, φ1s, φs Li Li Li 1s 1s s Construct the Hamiltonian matrix to see which has the lowest energy
Berylium The antisymmetric 4 electron wave function can be written, Ψ ( r, r, r, r ) = φ, φ, φ, φ Be Be Be Be 1 3 4 1s 1s s s = 1 4! φ ( r) φ ( r ) φ ( r ) φ ( r ) Be Be Be Be 1s 1 1s 1s 3 1s 4 Be Be Be Be 1s r1 1s r 1s r3 1s r4 Be Be Be Be s r1 s r s r3 s r4 Be Be Be Be s ( r1) φs ( r φs r 3 φs r 4 φ ( ) φ ( ) φ ( ) φ ( ) φ ( ) φ ( ) φ ( ) φ ( ) φ ) ( ) ( )
Gold The antisymmetric 79 electron wave function can be written, Ψ ( r, r,, r ) = φ ( r ), φ ( r ),, φ ( r ) 79 79 79 1 79 1s 1 1s 6s 79 The determinant of an N N matrix has N! terms. 79! = 8.95 10 116 (more than the atoms in the observable universe) Start with the valence electrons. Stop adding electrons when the matrix gets too big.
Pauli exclusion The sign of the wave function must change when two electrons are exchanged Ψ 1 ( r) ( r) 1 ( r, r ) = φ φ = ( r) ( r ) ( r ) ( r) ( φ φ φ φ ) 1 1 1 1 1 1 1 1 φ1( r) φ( r) If two columns are exchanged, the wave function changes sign. If two columns are the same, the determinant is zero. The Pauli exclusion principle only holds in the noninteracting electron approximation when the many electron wave function can be written as a product of single electron wave functions, only one electron can occupy each single electron state.
Atomic physics summary Solutions to the Schrödinger equation accurately describe the observed energy levels in atoms. We know the equation that needs to be solved but it is intractable. A common first approximation is the orbital approximation: Assign the electrons to an atomic orbital and a spin antisymmetrized product of spin orbitals. The energy is then evaluated including the electron-electron interactions. E = Ψ H ΨΨ Sometimes assumptions have to be made about the core wave functions and the wave functions of the valence electrons are determined numerically. Ψ
The full Hamiltonian of a molecule H Ze e = + + ZZe A A B i A i me A ma ia, 4πε 0riA i< j4πε 0rij A< B4πε 0rAB Everything you can know about the molecule is contained in the Hamiltonian. This explains life, the universe, and everything!
Born Oppenheimer approximation Fix the positions of the nuclei and consider the many electron Hamiltonian. H elec Ze e = + + ZZe A A B i i me ia, 4πε 0riA i< j4πε 0rij A< B4πε 0rAB This is still too difficult. Neglect the electron-electron interactions.
Separation of variables (Trennung der Veränderlichen) The Schrödinger equation can be solved by the separation of variables if the total Hamiltonian can be written as a sum of Hamiltonians each depending on only one variable. H t (r 1,r,...r n ) = H 1 (r 1 ) + H (r ) +... H n (r n ) H elec Ze = + A i i me ia, 4πε 0riA i< j4πε 0rij e H Ze A elec _ red = i HMO i m = e A 4πε 0r ia i
Molecular orbital Hamiltonian Fix the positions of the nuclei. Solve the Schrödinger equation for one electron. Find the ground state and the excited states. 1 - r 1C + r r1a 1B r AC C A + r AB + B r BC H MO Ze A = 1 m 4πε r r e A 0 1 A
Molecular orbitals Molecular orbitals of a molecule are like the atomic orbitals of an atom. ψ ψ ψ ψ MO1 MO MO3 MO4 ( r ) ( r ) ( r ) ( r ) φ1s ( r ) φs ( r ) φ p ( r ) x φ ( r ) You can put two electrons, spin up and spin down, in each molecular orbital. p y
Molecular orbitals The first approximation for the many electron wave function of a molecule is an antisymmetrized product of molecular orbitals. The energy of this wave function should be evaluated using the electronic Hamiltonian. Ψ ( r, r ) = 1 N 1 N! ψmo 1 ( r1) ψmo 1 ( r1 ) ψmo, N ( r1) ψ ( r ) ψ MO1 ( r ) ψ ( r ) MO1 N MO, N N
Molecular orbitals Calculate the energy of a molecular orbital using the electronic Hamiltonian. Ψ ( r, r ) = ψ, ψ, ψ, ψ, 1 N MO1 MO1 MO MO H elec Ze e = + + ZZe A A B i i me ia, 4πε 0riA i< j4πε 0rij A< B4πε 0rAB E = Ψ H elec ΨΨ Ψ
Bond potentials E = Ψ H elec ΨΨ Ψ Calculate the energies for different atomic distances. The minimum yields the bond length and bond strength.
XPS
Bond angles Find the angle that minimizes the energy. E = Ψ H elec ΨΨ Ψ
Shape of a molecule H elec Ze e = + + ZZe A A B i i me ia, 4πε 0riA i< j4πε 0rij A< B4πε 0rAB
Shape of a molecule In Jmol, double click to start and stop a measurement. http://lampx.tugraz.at/~hadley/ss1/molecules/moleculeviewer/viewer.php
Chemical reactions Calculate the energy at every stage of the reaction. E = Ψ H elec ΨΨ Ψ To calculate the speed of a chemical reaction, solve the time-dependent Schrödinger equation. http://en.wikipedia.org/wiki/file:walden-inversion-3d-balls.png
Chemical reactions + + C 3 H 8 + 5 O 3 CO + 4 H O + Energy Propane Oxygen Carbon Water dioxide It is possible to calculate if the reaction is endothermic or exothermic. E = Ψ H elec ΨΨ Ψ
The molecular orbital can be found by: Linear Combination of Atomic Orbitals Look for a solution to the molecular orbital Hamiltonian, of the form, Here φ n are atomic orbitals.
Molecular orbitals of H The molecular orbital Hamiltonian for H is, r A and r B are the positions of the protons. ψ = cφ + c φ + cφ + c φ + H H H H mo 1 1, sa 1, sb 3, sa 4, sb What about spin?
Molecular orbitals of H The time independent Schrödinger equation, Hψ mo = Eψ mo Multiply from the left by H φ1, sa ( ) c φ H φ + c φ H φ + = E c φ φ + c φ φ + H H H H H H H H 1 1, sa mo 1, sa 1, sa mo 1, sb 1 1, sa 1, sa 1, sa 1, sb H Multiply from the left by φ1, sb ( ) c φ H φ + c φ H φ + = E c φ φ + c φ φ + H H H H H H H H 1 1, sb mo 1, sa 1, sb mo 1, sb 1 1, sb 1, sa 1, sb 1, sb Two equations with two unknowns: c 1 and c
Molecular orbitals of H H H H H H H H H φ1, sa Hmo φ1, sa φ1, sa Hmo φ 1, sb c1 φ1, sa φ1, sa φ1, sa φ 1, sb c1 H H H H E c = H H H H φ1, sb Hmo φ1, sa φ1, sb Hmo φ1, sb φ c 1, sb φ1, sa φ1, sb φ 1, sb Roothaan equations: HAA HAB c1 SAA SAB c1 E * * HAB H BB c = SAB S BB c Hamiltonian matrix S 1 0 0 1 Overlap matrix H AA = H BB
Molecular orbitals of H H H c c = E AA AB 1 1 * HAB H AA c c The eigenvalues and eigenfunctions are: E H H = ± ± AA AB 1 ψ± = φ ± φ H H ( r) ( r) ( r) ( ) 1sA 1sB Both H AA and H AB are negative E + < E -
Molecular orbitals of H In the ground state, both electrons occupy the lower energy symmetric orbital (bonding orbital). 1 ψ ( r) ψ ( r) 1 Ψ = = ψ ( r ) ψ ( r ) + 1 + 1 ( r ) ( ) 1, r ψ+ ( r1) ψ+ ( r) + + 1 Ψ = + + + H H H H H H H H ( r1, r) ( φ1sa( r1) φ1sa( r) φ1sa( r1) φ1sb ( r) φ1sa( r) φ1sb ( r1) φ1sb ( r1) φ1sb ( r) )( ) Use this wave function including the electron-electron interaction to calculate the bond potential.
Student project: Draw the Bond potential of H E = Ψ H elec ΨΨ Ψ http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/hmol.html
r 0 = 0.869 A D 0 = 3.18 ev Student project Henögl / Pranter 014 exact values r 0 = 0.74 A D 0 = 4.5 ev
r 0 = 0.85 A D 0 = 3.63 ev student project: Lang / Röthel exact values r 0 = 0.74 A D 0 = 4.5 ev
Homonuclear diatomic molecules H, N, O,... All homonuclear diatomic molecules use the molecular orbitals of H. ψ = cφ + c φ + cφ + c φ + cφ + cφ + Z Z Z Z Z Z mo 1 1, sa 1, sb 3, sa 4, sb 5 p, A 6 p, B x x The Hamiltonian matrix is as large as the number of atomic orbitals in the molecular orbital sum.
Homonuclear diatomic molecules All homonuclear diatomic molecules use the molecular orbitals of H. 1σ g < 1σ u < σ g < σ u < 3σ g ~ 1π u < 1π g < 3σ u g inversion symmetry from: Blinder, Introduction to Quantum Mechanics
Separtion of variables number of electron pairs shared from: Blinder, Introduction to Quantum Mechanics