Q1: Display all the MOs for N2 in your report and classify each one of them as bonding, antibonding or non-bonding, and say whether the symmetry of the orbital is σ or π. Sketch a molecular orbital diagram using the values of the energy of the MOs from the output file and compare your diagram with the one in your physical chemistry or freshman chemistry textbook. LUMO + 1 Antibonding π orbitals E8 = 0.1556 Hartree = 4.234 ev LUMO Antibonding π orbitals E7 = 0.1556 Hartree = 4.234 ev 1
HOMO Bonding π orbitals E6 = -0.6259 Hartree = -17.03 ev HOMO-1 Bonding π orbitals E5 = -0.6259 Hartree = -17.03 ev 2
HOMO-2 E4 = -0.6304 Hartree = -17.15 ev HOMO-3 Antibonding σ orbitals E3 = -0.7721 Hartree = -21.01 ev HOMO-4 E2 = -1.5338 Hartree = -41.74 ev 3
HOMO-5 Antibonding σ orbitals E1 = -15.7120 Hartree = -427.5 ev HOMO-6 E0 = -15.7156 Hartree = -427.6 ev MO diagram: LUMO+1: 4.234 ev π* LUMO: 4.234 ev π* HOMO: -17.03 ev π HOMO-2: -17.15 ev σ HOMO-1: -17.03 ev π HOMO-3: -21.01 ev σ* HOMO-4: -41.74 ev σ HOMO-5: -427.5 ev σ* HOMO-6: -427.6 ev σ 4
Q2: Which spin state is lower in energy for the O2 molecule? Explain by referring to the energy of the MOs (recall Hund s rule). Note, that when you do calculations of the triplet state, there is a listing of both alpha and beta MOs. Singlet state MO diagram (only P orbital mixing shown): σ* The singlett state requires electrons in degenerate energy levels (π*) π* to pair up, that is not favorable since the pairing itself is higher in energy than the alternative unpaired triplett state due to repulsion π σ forces. Hund s rule states that for degenerate energy levels the electrons should first occupy both levels before pairing. Triplet state MO diagram (only P orbital mixing shown): _ σ* The repulsion forces are minimized in this case and Hund s rule is π* applied, so this state is energetically favorable. π σ 5
Q3: Display all the MOs for O2 in your report and classify each one of them as bonding, antibonding or non-bonding, and say whether the symmetry of the orbital is σ or π. Sketch a molecular orbital diagram using the values of the energy of the MOs from the output file. If you found the triplet state to be lower in energy, include both the spin-up and spin-down orbital energy in the diagram. Alpha orbitals (spin up): LUMO+1 E1 = 1.0154 Hartree = 27.630 ev LUMO Antibonding σ orbitals E1 = 0.4123 Hartree = 11.21 ev 6
HOMO Antibonding π orbitals E1 = -0.5668 Hartree = -15.42 ev HOMO-1 Antibonding π orbitals E1 = -0.5668 Hartree = -15.42 ev HOMO-2 E1 = -0.7580 Hartree = -20.62 ev 7
HOMO-3 Bonding π orbitals E1 = -0.8607 Hartree = -23.42 ev HOMO-4 Bonding π orbitals E1 = -0.8607 Hartree = -23.42 ev HOMO-5 Antibonding σ orbitals E1 = -1.2003 Hartree = -32.662 ev 8
HOMO-6 E1 = -1.7728 Hartree = -48.240 ev HOMO-7 Antibonding σ orbitals E1 = -20.7695 Hartree = -565.17 ev HOMO-8 E1 = -20.7700 Hartree = -565.18 ev 9
Beta orbitals (spin down): LUMO+1 Antibonding π orbitals E1 = 0.0977 Hartree = 2.66 ev LUMO Antibonding π orbitals E1 = 0.0977 Hartree = 2.66 ev HOMO Bonding π orbitals E1 = -0.5880 Hartree = -16.00 ev 10
HOMO-1 Bonding π orbitals E1 = -0.5880 Hartree = -16.00 ev HOMO-2 E1 = -0.6942 Hartree = -18.89 ev HOMO-3 Antibonding σ orbitals E1 = -0.9925 Hartree = -27.01 ev 11
HOMO-4 E1 = -1.6305 Hartree = -44.368 ev HOMO-5 Antibonding σ orbitals E1 = -20.7136 Hartree = -563.646 ev HOMO-6 E1 = -20.7148 Hartree = -563.678 ev 12
MO diagram: UP LUMO+1: 27.630 ev π* UP LUMO: 11.21 ev σ* DOWN LUMO+1: 2.66 ev π* DOWN LUMO: 2.66 ev π* UP HOMO: -15.42 ev π* UP HOMO-1: -15.42 ev π* DOWN HOMO: -16.00 ev π DOWN HOMO-1: -16.00 ev π UP HOMO-2: -20.62 ev σ DOWN HOMO-2: -18.89 ev σ UP HOMO-3: -23.42 ev π UP HOMO-4: -23.42 ev π UP HOMO-5: -32.66 ev σ* UP HOMO-6: -48.24 ev σ DOWN HOMO-3: -27.01 ev σ* DOWN HOMO-4: -44.37 ev σ UP HOMO-7: -565.17 ev σ* UP HOMO-8: -565.17 ev σ DOWN HOMO-5: -563.65 ev σ* DOWN HOMO-6: -563.68 ev σ 13
Q4: Make a plot of the energy of the H2 molecule versus the distance between the atoms (using for example Excel, Matlab or Gnuplot). 14
Q5: What value should the energy approach when the bond length is increased to infinity? Does the binding curve you calculated indicate this? Energy at various values of distance between H atoms (using RHF): Distance [Å]: Energy [Hartree]: 0.40000000-0.93300420 0.50000000-1.05802482 0.60000000-1.11003090 0.70000000-1.12612316 0.80000000-1.12371294 0.90000000-1.11168637 1.00000000-1.09480796 1.10000000-1.07568569 1.20000000-1.05575928 1.30000000-1.03582611 1.40000000-1.01632567 1.50000000-0.99749729 1.60000000-0.97946955 1.70000000-0.96231013 1.80000000-0.94605220 1.90000000-0.93070722 2.00000000-0.91627125 When the bond distance approaches infinity the bond energy should approach zero, due to there being absolutely no interaction between the two atoms when at an infinite distance from each other, however the remaining potential energy will be for the ground state of two H atoms, that is 2*-13.6 ev = -27.2 ev. So when the bond distance approaches infinity the energy will approach -27.2 ev. The binding curve goes over the -27.2 ev, the final calculated point (2Å) has the potential energy -0.916271 Hartree = -24.9256 ev. 15
Q6: Create one plot with three curves: One for the singlet state in restricted Hartree-Fock (RHF), one for the singlet state in unrestricted Hartree-Fock (UHF) and one for the triplet state. Also, add a horizontal line at the energy of two completely separated hydrogen atoms. Q7: Compare the RHF and the UHF curves. Describe the problem that occurs in the RHF calculation for large bond lengths. What is the reason for it? The reason the triplet state calculation is so different from the others is that the triplet state does not allow bond formation, so the energy plot is based on repulsion forces, eventually resulting in the non-interacting potential energy. RHF requires the atom or molecule in question to remain a closed-shell system, with all orbital levels doubly occupied. If the H atoms are separated they each need to have an electron in their 1s level, resulting in energy levels that are not doubly occupied. RHF therefore forces the 2 H atoms to bond, regardless of distance, resulting in a high energy cost for the bond at long 16
distances. UHF allows the two electrons to have different spatial orbitals and therefore lets the system be open-shell, this allows the electrons to separate and the H atoms to stop interacting eventually. Even though the electrons are each in their own orbital the singlet state is maintained if they still have opposite spin. Q8: Can two H-atoms form a bond in the singlet state? How about the triplet state? Is the formation of the bond(s) out of two isolated atoms exothermic or endothermic? Singlet state H2 has the following MO diagram: σ* This state forms a bond, having a σ bond order of 1. Triplet state H2 has the following MO diagram: _ _ σ* This state cannot form a bond as it has equal occupation in bonding and σ antibonding orbitals, resulting in a bond order of 0. As can be seen from the plot in Q6 (UHF singlet state) the total potential energy of the molecule is lower than if we have two non-interacting H atoms, so the bond formation is exothermic. 17
Q9: How does the binding curve of this calculation differ from the previous UHF calculation? Why is that? Since the orbitals for the hydrogen molecule are symmetric in this calculation, the initial guess for the orbital functions is symmetric as well. Since symmetry is a very powerful tool the next guesses will retain this symmetry. This is however not ideal in this case because it assumes complete symmetry for the spin up and spin down electrons, and thus the unrestricted HF calculation is forced into the restricted version. This results in the exact same plot as the RHF singlet in the plot from Q6. 18