Reikniefnafræði - Verkefni 1 Haustmisseri 2013 Kennari - Hannes Jónsson

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1 Háskóli Íslands, raunvísindasvið Reikniefnafræði - Verkefni 1 Haustmisseri 2013 Kennari - Hannes Jónsson Guðjón Henning 11. september 2013 A. Molecular orbitals and electron density of H 2 Q1: Present the images of HOMO and the LUMO. Arrange the images of the HOMO and LUMO side by side. Write text underneath to label each molecular orbital as bonding or antibonding. Also, specify whether the molecular orbitals are sigma or pi orbitals. (a) HOMO. Bonding molecular orbital. (b) LUMO. Antibonding molecular orbital. HOMO and LUMO for the H 2 molecule. These molecular orbitals are both sigma orbitals. Q2: What is the optimal bond length in a H 2 molecule predicted by the Hartree-Fock calculations in the 6-31G basis set? Compare with the experimental value for the bond length, 0.74 Å. Hartree-Fock calculations in the 6-31G basis set gives the predicted optimal bond length as: Averaged distance (Å): So compared to the expreimental value of 0.74 Å, the dierence is = Å 1

2 Q3: Record the energy of the H 2 molecule in yor report. What is the zero energy in this case? Using the value of the energy of a H-atom, what is the bond energy? (This is, actually, not a good procedure since the two numbers subtracted are not calculated at the same level of theory. Generally, one relies on cancellation of errors in ab initio calculations and should, therefore, consistently use the same level of approximation for all numbers subtracted or added.) Energy of the H 2 molecule is recorded to be 1 Hartree equals ev, so the energy is Hartree = ev The zero energy is reached when the atoms are displaced so far from each other, that there is no interaction between them. The distance from each nuclei and its electron must also be so far from each other, that the interaction between them is also nonexistent. The bond energy is calculated with the following equation E bond = E H2 2E H and the ground state energy of an H atom is the Rydberg energy, which is This will give for the bond energy E Ryd = ev E bond = ( ) = ev Q4: What is the predicted bond length and what is the energy of the molecule in this basis set? Is the energy higher or lower than for the 6-31G basis set? Is the bond length longer or shorter? Here we did the same calculations in Gabedit as before, but with the 6-31G** basis set. There the calculations give Averaged distance (Å): E H2 = Hartree = ev Which gives lower energy and longer bond length than in the 6-31G basis set, as can be seen in Q2 and Q3. Q5: Explain why the energy is dierent and why the energy must nescessarily change in the direction it did (become lower or higher). Recall, SCF(Self Consistent Field) calculations are variational calculations. Is there any such rule for how the bond length changes with the basis set? In the 6-31G** basis, functions are added to the calculation, where the angular momentum quantum number is raised by 1 (l l + 1). The orbitals p x and p y are orthogonal to the wavefunction, which is approximated by linear combination of functions, and will not have any eect. However, the p z orbital is not orthogonal to this wavefunction, and will therefore give a better approximation of the energy. The variatonal theory tells that the calculated ground state energy is always higher or equal to the real ground state energy, or 2

3 E 0 E calculated Therefore the 6-31G** basis will calculate a lower energy than the 6-31G basis, which is a better result since it is closer to the real value E 0. Because the ground state energy is closer to the real value in the 6-31G** basis, the bond length will also be closer to its real value, because the bond length is directly related to the energy. Q6: Does the bond energy follow the same rule as the total energy, when both the molecule and the atoms are calculated on the same level of theory (why/why not)? No, because the reference energy can change, along with the total energy, in terms of the basis. B. Electron density of HF and LiH molecules HF molecule Q7: Does the electron density near the H atom increase or decrease when it binds to an F atom as compared to H 2? How does the size of the isosurface of the electron density near the H-atoms in the two molecules correlate with the electronegativity of the atoms? (a) Electron density of H 2. (b) Electron density of HF. Electron density of H 2 and HF molecules. One can see that the electron density around the H atom decreases when it binds to an F atom. The electronegativity of a hydrogen atom is 2.20, and is 3.98 for a uorine atom. Therefore it can be seen that the size of the isosurface of the electron density near the hydrogen atom will be smaller for an HF molecule than in an H 2 molecule. 3

4 LiH molecule Q8: Compare the result for LiH with those for the H 2 and HF molecules. Comment on how the electron density at the H-atom in LiH compares with those in H 2 and HF. Explain the trend. (a) Electron density of H 2. (b) Electron density of HF. (c) Electron density of LiH. Electron density of H 2, HF, and LiH molecules. In the HF and LiH molecules the H atom is on the left side of the gure. The electronegativity for H, Li, and F are H = 2.20 Li = 0.98 F = 3.98 That clearly states that the electron density at the H-atom in the LiH molecule is greater than at the Li-atom. Therefore the isosurfaces will be larger at the atom with the higher electronegativity, and equal for atoms with equal electronegativity. This is exactly the trend in these gures. Q9: Include the images of the electron densities in your report. From the electron distribution, assign partial charges (δ, δ+) to the atoms in the molecules. (a) Electron density of H 2. (b) Electron density of HF. (c) Electron density of LiH. Electron density of H 2, HF, and LiH molecules. In the HF and LiH molecules the H atom is on the left side of the gure. In the H 2 molecule, there are no partial charges. In the HF molecule, F has a δ partial charge and H has a δ + partial charge. In the LiH molecule, H has a δ partial charge, while Li has a δ + partial charge. 4

5 Basis set errors #1 Q10: Which spin multiplicity do you have to select for the H-atom? Why? Should you select 'Equilibrium structure search' in this case? The spin multiplicity in the Orca input in Gabedit for the H-atom is 2. The spin multiplicity is 2S+1, and the spin of the electron in the H-atom is 1. So the spin multiplicity is = Here we also choose 'Single point energy' instead of 'Equilibrium structure search'. This is because there is no relaxation to be calculated for this structure. Q11: What is the absolute and relative deviation (in %) of the calculated energy of a hydrogen atom using the two basis sets compard with the exact value? For the 6-31G basis, the calculated energy of the hydrogen atom is And the exact value of the H-atom is So the absolute deviation for this basis is And the relative deviation will be E H = Hartree = ev E H = E Ryd = ev E 6 31G = E H,6 31G E H = ev E 6 31G E H = = % For the 6-31G** basis, the calculated energy of the hydrogen atom is exactly the same E H = Hartree = ev So the absolute and relative deviation will be the same for this basis, compared to the exact value. This result is because of the symmetry of the H-atom. In the 6-31G** basis the function calculated for each p-orbital (p x, p y, p z ) is orthogonal to the s-orbital, as a result of this symmetry. Therefore both bases will yield the same result. Q12: Does the larger basis set yield a stronger og weaker bond? The bond energy for the smaller basis set (6-31G) is E bond = E H2 2 E H = ( ) = ev For the larger basis set (6-31G**) the bond energy is E bond = E H2 2 E H = ( ) = ev 5

6 So the larger basis set (6-31G**) yields a stronger bond. Q13: What is the absolute and relative deviation (in %) of the previously calculated bond energies with respect to these bond energies? Becuase the 6-31G** basis yields a better approximation for the bond energy, it will in this question thought as the real bond energy. The dierence of the bond energies calculated in the 6-31G and 6-31G** bases is And the relative deviation will be Basis set errors #2 E bond = E bond,6-31g E bond,6-31g** = = 0.14 ev E bond E bond,6-31g** = = 3.815% Q14: Do you agree with the following statement? 'Adding basis functions to a basis set will always yield a lower or equal total energy in a variational calculation'. A basis is a linear combination of functions in order to approximate the wavefunction. By adding functions, which must be orthogonal to the pre-existing functions, to the basis, a lower energy will be calculated. However, if the added functions are orthogonal to the wavefunction, they will yield no change, resulting in equal energy calculated. So this statement is true. Q15: Do you agree with the following statement? `A variational calculation using a basis set B which has a larger number of functions than basis set A, will always yield a lower or equal total energy as a calculation using A'. No, because the functions in these bases might be dierent from each other. This does not state that the functions in A are a subset of B, and so the larger basis might not necessarily yield a better solution. C. Optional Orbitals, electron densities and isosurfaces Q16: Given a single hydrogen atom: At what distance from the proton does the probability to nd the electron drop to zero (in Å or bohr radians)? If you guess, give an explanation how you made up the number, if you derived it analytically, please provide the derivation. The wave function for the Hydrogen atom (1s orbital) is ψ 1s = 1 π ( 1 a 0 ) 3/2 e r/a 0 where a 0 = Å is the rst Bohr radius, and r is the distance between the electron and the proton. One can see that the probability distribution, ψ 1s 2, will set the exponential factor to e 2r/a 0. So the probability to nd the electron drops to zero when r. 6