LCAO: The basis set consists of atomic orbitals Molecular orbitals (MO‘s) Atomic orbitals (AO‘s), 1s, 2s, 2p Basis sets (VII) In the literature, there is a number of established basis sets. In the following some typical examples are shown. Basis sets of Pople et al. (Gaussian program): STO-3G minimum basis (3 GTOs per STO) 3-21G small “split-valence” basis (double-zeta valence AO) 6-31G* large “split-valence” basis, polarized at X H 6-311+G** polarized triple-zeta basis with diffuse functions Basis sets of Dunning et al.: DZP polarized double-zeta basis TZP polarized triple-zeta basis TZ2P triple-zeta basis with 2 sets of polarization functions TZ2Pf TZ2P + additional f polarization functions Basis sets (VIII) Correlation – consistent basis sets: cc-pVTZ multiply polarized triple-zeta basis cc-pVQZ multiply polarized quadruple-zeta basis aug-cc-pVQZ cc-pVQZ + additional diffuse functions RHF/UHF We have described a HF calculation based on a Slater determinant like this spin orbital Each space orbital is used twice This is a restricted HF (RHF) calculation space orbital RHF/UHF Another possibility: spin orbital space orbital ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) A new space orbital for each spin orbital This is an unrestricted HF (UHF) calculation RHF/UHF In UHF we have twice as many (for closed shells) expansion coefficients to vary than in RHF So, in the HF calculation we willl normally reach a lower energy in UHF than in RHF That is good So why do we not always use UHF? RHF/UHF Because UHF has problems with the spin multiplicity: A = X, Y, Z component of total electron spin: Square of total electron spin: The exact electronic wavefunction satisfies: Possible values of S: 0, ½, 1, 3/2, 2, 5/2, 3, ...... Possible values of MS: -S, -S+1, -S+2, ...., S-1, S 2S+1 possibilities Multiplicity: 2S+1 – singlet, doublet, triplet, quadruplet ..... RHF/UHF RHF wavefunction (Slater determinant) has a defined multiplicity and so it behaves „correctly“: The UHF wavefunction has no defined multiplicity If it is just important to get lowest (most accurate) energy: UHF If the wavefunction is also important (properties = dipole moment, ...): RHF Units Parenthetic remark: Units • SI- and cgs-units • Atomic units Units SI and cgs units Quantity Mass Length Time Force Energy Charge Units Principal difference: Units for charge Q1 Q2 r12 Coulomb energy in SI units: Coulomb energy in cgs units: cgs: Two charges, each of 1 e.s.u., separated by 1 cm, produce a Coulomb energy of 1 erg Units Units for dipole moment: Dipole moment = charge length SI: C m cgs: e.s.u. cm 1 Debye = 1 D = 10-18 e.s.u. cm = 3.33564×10−30 C m Units Atomic units, used by MOLPRO Quantity units Mass Electron mass Angular momentum Planck‘s constant/2 Charge Elementary charge Length Energy Time Frequency Momentum Force Electrical current units Units Atomic units for dipole moment: Dipole moment = charge length SI: C m cgs: e.s.u. cm Atomic units: 1 e a0 = 8.47835281 x 10-30 C m = 2.54175 D Koopmans’ theorem Applications of HF theory • Koopmans‘ theorem • Mulliken population analysis Koopmans’ theorem Molecule M Ion M+ (and electron infinitely far away) How do we get the ionisation energy? Do HF calculation for M, result EHF(n) Do HF calculation for M+, result EHF(n-1) Ionisation energy I = EHF(n-1) EHF(n) Koopmans’ theorem How do we get the ionisation energy in a simpler way? Do HF calculation for M, result EHF(n) Calculate HF energy for M+ with same orbitals as for M, result EHF(M)(n-1) Ionisation energy I0 = EHF(M) (n-1) EHF(n) Some derivation gives Koopmans‘ theorem Ionisation energy I0 = j is j the orbital energy of the orbital from which the „ionised“ electron has disappeared Koopmans’ theorem Ionisation energies of small molecules (in eV) Koopmans HF values from MOLPRO, Basis set 6-31G** Geometry optimized at HF level Experimental values from „CRC Handbook of Chemistry and Physics“, 53rd edition. Deviations of 10-20% 16.3 Exp. Koopmans’ theorem Why is Koopmans relatively successful? Ionisation energy I0 = EHF(M) (n-1) EHF(n) First improvement: Separate HF calculation for M+ lowers EHF(M) (n-1) and gives lower I0 Second improvement: Introduce Configuration Interaction (better theory, see later). Lowers both energies, but energy of M most because M has more electrons. Gives higher I0 So: The two improvements tend to cancel each other and Koopmans‘ theorem is fairly all right. Mulliken population analysis Population analysis??? The molecule has n electrons: • How many electrons „belong“ to particular nuclei? • How many electrons are „shared“ between pairs of nuclei? That is, where are the bonds in the molecule? Mulliken population analysis Population analysis n electrons [R.S. Mulliken, J. Chem. Phys., 23, 1833, 1841, 2338, 2343 (1955)]. Mulliken population analysis One electron in one MO. Differential probability of finding the electron in the volume element dV = dx dy dz Summing over all occupied orbitals gives electron density Mulliken population analysis The space orbitals are normalized and so n electrons Mulliken population analysis In the Roothaan-Hall approximation we have and so and Density matrix element Mulliken population analysis Consequently and so That is n electrons Mulliken population analysis Since S = 1 we can partition the n electrons as follows n electrons and Mulliken interpretation: P is the net population of basis function (atomic orbital) Q is the overlap population of basis functions and electrons is shared between these two basis functions ; this fraction of the Mulliken population analysis P is a net population of basis function (atomic orbital) Q is the overlap population of basis functions P33 P22 and ; Q45 P44 Q12 P11 P66 P55 Mulliken population analysis Now we can divide the electrons completely between basis functions: Gross population of basis function (atomic orbital) Summing over the orbitals gives n electrons Mulliken population analysis We can divide the electrons between nuclei/atoms: Gross atomic population of the atom A on Gross overlap population of the atoms A and B on on Mulliken population analysis qAB large, positive means „bond“ qAB small, negative means „no bonding“ q2 n electrons q12 q23 q1 q13 q3 [R.S. Mulliken, J. Chem. Phys., 23, 1833, 1841, 2338, 2343 (1955)]. Mulliken population analysis H2O Mulliken values from MOLPRO, Basis set 6-31G** Geometry optimized at HF level q2 = 8.68 MOLPRO does not seem to be able to calculate qAB q1 = 0.66 q3 = 0.66 Mulliken population analysis O CH2O+ Mulliken values from GAUSSIAN, B3LYP/STO-3G Geometry optimized at B3LYP/STO-3G level q12 = 0.45 C q23 = -0.03 q24 = -0.03 q2 = 7.76 q1 = 4.80 q12 = 0.36 H q14 = 0.36 q34 = -0.04 q3 = 0.64 H q4 = 0.64 Mulliken population analysis CH2O+ GAUSSIAN calculates = = It follows that + = and we get 7.76+4.80+2 0.64+2 0.36+0.45=15.01

1/--pages