Reikniefnafræði - Verkefni 2 Haustmisseri 2013 Kennari - Hannes Jónsson

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1 Háskóli Íslands, raunvísindasvið Reikniefnafræði - Verkefni 2 Haustmisseri 2013 Kennari - Hannes Jónsson Guðjón Henning 18. september

2 A. Molecular orbitals of N 2 Q1: Display all the MOs for N 2 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 le and compare your diagram with the one in your P-Chem or freshman chemistry textbook. Left: MO diagram for the N 2 molecule. Right: MO diagram as it is in textbook. LUMO+1 LUMO HOMO HOMO-1 HOMO-2 HOMO-3 HOMO-4 HOMO-5 HOMO-6 Molecular orbitals for N 2. 2

3 B. Molecular orbitals of O 2 Q2: Which spin state is lower in energy for the O 2 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 For the singlet state of O 2, the energy is E O2,singlet = Hartree E O2,triplet = Hartree So the triplet state is lower in energy than the singlet state. Hund's rst rule states that(according to Wikipedia): "1. For a given electron conguration, the term with maximum multiplicity has the lowest energy. The multiplicity is equal to 2S + 1, where S is the total spin angular momentum for all electrons. The term with lowest energy is also the term with maximum S." What is needed to do here is to compare the two highest energy occupied electrons of the singlet state and the triplet(spin up), which is in this case the antibonding π orbital. The log le from gabedit gives the following energy and occupation values. State Orbital no. Occupation Energy (ev) Singlet Triplet Here it can be seen that the mean energy value for the triplet state is lower than the energy value of the singlet state. By applying Hund's rst rule one sees that for the triplet state the multiplicity is 2S + 1 = 3, because there are two spin up electrons and for the singlet state the multiplicity is 2S + 1 = 1, because there is one spin up electron and one spin down electron. Therefore one can assume that the triplet state is lower in energy because its multiplicity is higher. 3

4 Q3: Display all the MOs for O 2 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 le. If you found the triplet state to be lower in energy, include both the spin-up and spin-down orbital energy in the diagram. MO diagram for the O 2 molecule. 4

5 LUMO+1 LUMO HOMO HOMO-1 HOMO-2 HOMO-3 HOMO-4 HOMO-5 HOMO-6 HOMO-7 HOMO-8 Molecular orbitals for O 2 in the triplet state with spin up. 5

6 LUMO+1 LUMO HOMO HOMO-1 HOMO-2 HOMO-3 HOMO-4 HOMO-5 HOMO-6 Molecular orbitals for O 2 in the triplet state with spin down. 6

7 C. H 2 energy prole #1 Q4: Make a plot of the energy of the H 2 molecule versus the distance between the atoms (using for example Excel, Matlab or Gnuplot). Energy of the H 2 molecule as a function of atomic distance by manually calculating for seven dierent bond lengths. By doing this calculation again, using the scan function in gabedit, a better graph can be made, as is shown in the next gure. Energy of the H 2 molecule as a function of atomic distance, using 150 values for the bond length. 7

8 D. H 2 energy prole #2 Q5: What value should the energy approach when the bond length is increased to innity? Does the binding curve you calculated indicate this? The energy should approach the value of two independent H atoms when the bond length goes to innity. In the former lab exercise, the energy value of a single Hydrogen atom was found to be E H = Hartree So the graph should approach -1 Hartree as the bond length increases. Instead, the energy rises and seems to approach Hartree. 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. Singlet state in RHF, singlet state in UHF, and triplet state in UHF as a function of bond length. 8

9 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 UHF curves approach -1 Hartree with increasing bond lengths, as one should expect(energy value of two individual H atoms), but the RHF curve rises above that energy value with bond lengths above -1.5 Å. In the RHF method, both atoms can be found together near the same nucleus, because the calculations use the same spatial distribution for the electrons. But because the electrons repel each other, which is not the case here, the computed energy will be too high (because of the proximity of the electrons). Therefore when the bond is supposed to break (at approx. -1 Hartree), this method will be insucient. Q8: Decide, based on the calculated data, whether H 2 can form a bond in the singlet state and in the triplet state. Based on the last gure, one can assume that the singlet state can form a bond because there is a potential well where the molecule can be stable. Such a well is not present for the triplet state. Also, the energy of the triplet state is always above the dotted line which indicates the energy of the two completely separated hydrogen atoms. The triplet state energy seems to approach this energy value, but not reach it. Therefore, based on this calculated data, H 2 can only form a bond in the singlet state. Q9: How does the binding curve of this calculation dier from the previous UHF calculation? Why is that? With no mixing of orbitals the energy rises constantly above the bond length of 0.7 Å. But with mixing of orbitals, the energy lessens its increase, approaching -1 Hartree. There are however two sudden increases in the energy at approximately -1.3 and -1.5 Å with the mixing of orbitals, which are probably connected to the calculation methods/errors in Orca. Without mixing the orbitals, the symmetry in the initial guess is retained throughout the calculation. To break this spin-symmetry, one has to mix the initial orbitals before the calculations begin. The initial guess for the Hydrogen molecule wave function corresponds to a restricted form, for both UHF and RHF. So, Orca is instructed to mix the rst and second orbital of spin up corresponding to a rotation by 45. When the orbitals are not mixed, the initial guess symmetry is retained throughout the calculations, resulting in a "bad"approximation at bond lengths over 1.3 Å (similar to the result in Q7). But, when the orbitals are mixed, this symmetry changes from the initial one, giving a better approximation, where the curve converges at -1 Hartree. 9

10 Singlet state in UHF, with and without mixing orbitals, as a function of bond length. 10