Comp. Chem. 2020Univ. of Iceland

Solutions to lab 2ORCA and Chemcraft

Hannes JónssonVilhjálmur Ásgeirsson

Introduction to Hartree-Fock calculations using ORCA andChemcraft

Exemplary report based on the work of Yorick and André

A. Molecular orbitals and electron density of H2

Q1: Present the images of the HOMO and the LUMO in your report. Take care that the windows wereall maximized and the camera settings were identical before you saved the images. Arrange the images ofthe HOMO and LUMO side by side. Write text underneath to label each molecular orbital as bonding orantibonding. Also, specify whether the molecular orbitals are σ or π orbitals.

Figure 1: Molecular orbitals of H2 obtained at the Hartree-Fock/6-31G level of theory. The surfaces areconstructed using an isovalue of 0.1 in atomic units (electrons per Bohr3). Left: Bonding HOMO. Right:Antibonding LUMO. Note that both orbitals are σ-orbitals.

Q2: What is the optimal bond length of an H2 molecule predicted by Hartree-Fock calculations based on the6-31G basis set? How many atomic displacements did ORCA need to find the optimal bond length? Comparewith the experimental value for the bond length, 0.74Å.

The optimal bond length of H2 is predicted to be

rH-H = 0.730Å

using HF/6-31G. The default optimization method in ORCA required four atomic displacements to convergethe atom forces below the given convergence thresholds. The predicted bond length is slightly lower thanthe experimental value of 0.74 Å.

Q3: Record the energy of the H2 molecule in your report. What is the zero of energy in this case? Using theexact value of the energy of an H-atom, what is the bond energy? (This is, actually, not a good procedure sincethe two numbers subtracted are not calculated at the same level of theory. Generally, one relies on cancellationof errors in ab initio calculations and should, therefore, consistently use the same level of approximation forall numbers subtracted or added.)

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Comp. Chem. 2020Univ. of Iceland

Solutions to lab 2ORCA and Chemcraft

Hannes JónssonVilhjálmur Ásgeirsson

HF/6-31G electronic structure calculations of H2 yield a minimum energy of:

EH2 = −1.127 H.

The zero of energy is chosen to be two non-interacting H atoms, or

E0 = −1 H,

so the binding energy of H2 isEH-H = EH2 − E0 = −0.127 H

Q4: What is the predicted bond length and what is the energy of the molecule when using the 6-31G** basisset? Is the energy higher or lower than for the 6-31G basis set? Is the bond length longer or shorter?

The optimal bond-length in H2 is predicted to be

rH-H = 0.733Å.

by using HF/6-31G** and the energy of the molecule is computed to be:

EH2 = −1.131 H.

As expected, the energy is found to be lower when using 6-31G** compared to 6-31G. This decrease inenergy with the larger basis set also yields an increase in bond-length which becomes slightly closer to theexperimental value.

Q5: Explain why the energy is different and why the energy must necessarily change in the direction it did(become lower or higher). Recall, SCF calculations are variational calculations. Is there any such rule forhow the bond length changes with the basis set?

Hartree-Fock calculations are based on the variational principle. In other words, the following inequalityholds for the ground state energy in Hartree-Fock calculations:

〈Ψ0|Ĥ|Ψ0〉〈Ψ0|Ψ0〉

≥ E0.

The equality holds if the exact ground state wave function is used, which requires a complete basis set (andalso a more flexible trial function than a single Slater determinant). As neither 6-31G nor 6-31G** arecomplete, the larger basis set 6-31G** produces a more accurate and therefore lower energy. There is norule for how the bond length changes as the basis set is increased since a lowering of energy can result ineither an increase or a decrease in the bond length.

Q6: Does the bond energy follow the same rule as the total energy, when both the molecule and the atomsare calculated on the same level of theory (why/why not)?

No, since the bond energy is a difference of two values of the energy, namely the energy of the moleculeminus the energy of the atoms, it is not known whether the bond energy increases or decreases when thebasis set is expanded.

B. Electron density of HF and LiH molecules

Q7: Does the electron density near the H atom increase or decrease when it binds to an F atom as comparedto H2? How does the size of the isosurface of the electron density near the H-atoms in the two moleculescorrelate with the electronegativity of the atoms?

The electron density near the H atom decreases if the other H atom is substituted by an F atom. The sizeof the isosurface near the H atom decreases as a result of electron transfer from the H atom to the F atom.

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Comp. Chem. 2020Univ. of Iceland

Solutions to lab 2ORCA and Chemcraft

Hannes JónssonVilhjálmur Ásgeirsson

Q8: Compare the results for LiH with those for the H2 and HF molecules. Comment on how the electrondensity at the H-atom in LiH compares with those in H2 and HF. Explain the trend.

The electron density near the H atom increases when it is bound to a Li atom, consistent with electrontransfer from Li to H. The electron density at the H atom decreases in the order of Li, H and F for thebound atom. This trend can be explained by the increasing electronegativity of the bound atom.

Q9: Include the images of the electron densities in your report. From the electron distribution, assign partialcharges (δ−, δ+) to the atoms in the molecules.

Figure 2: Isocontour surfaces displaying the electron density of HF and LiH from calculations carried outusing HF/6-31G. Isovalues of 0.01 and 0.013458 were used to generate the isosurfaces of HF and LiH,respectively. δ+ and δ- illustrate the partial charges.

Basis set errors

Q10: Which spin multiplicity do you have to select for the H-atom? Why? Why should you not select‘Equilibrium structure search’ or ’OPT’ in this case?

ORCA requires the multiplicity as an input parameter. An H atom has one electron. As electrons arefermions with a spin of 12 , the multiplicity is 2 ·

12 + 1 = 2.

Geometry optimization is used to minimize the forces acting on atoms due to interaction with other atoms.A single H atom is not interacting with any other atom and its energy is, furthermore, invariant to changesin its location in space and therefore geometry optimization of a single H serves no purpose.

Q11: What is the absolute and relative deviation (in %) of the calculated energy of a hydrogen atom usingthe two basis sets compared with the exact value?

The absolute deviation iseabs = (−0.498− (−0.5)) H = 0.002 H.

The relative deviation is

erel =0.002 H

0.5 H= 0.4%.

The energy values are identical to 12 significant digits (default ORCA precision) for the two basis sets.Increasing the basis set from 6-31G to 6-31G** does not change the energy of an atom.

Q12: Does the larger basis set yield a stronger or weaker bond?

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Comp. Chem. 2020Univ. of Iceland

Solutions to lab 2ORCA and Chemcraft

Hannes JónssonVilhjálmur Ásgeirsson

The larger basis set yields a stronger bond since the energy of the molecule is lowered while the energy ofthe isolated atoms is unchanged.

Q13: What is the absolute and relative deviation (in %) of the previously calculated bond energy with respectto these bond energies?

The absolute deviation for both basis sets is twice the deviation in the H atom energy

eabs = (−0.498− (−0.5)) H = 0.004 H.

The relative deviations depend on the obtained total energy. For the smaller basis set the relative deviationis

erel =0.004 H

0.127 H= 3.15%.

For the larger basis set it is

erel =0.004 H

0.131 H= 3.05%.

Q14: Do you agree with the following statement (if so, explain; if not, try to present a counter exam-ple)? ‘Adding basis functions to a basis set will always yield a lower or equal total energy in a variationalcalculation’.

Yes. The total energy in a variational calculation represents an upper bound to the ground-state energy ofthe system. Only in the limit of a complete basis set will the minimum energy be obtained for the given trialfunction approximation (here a single Slater determinant). By expanding an incomplete basis by addingmore basis functions can only increase the variational flexibility, at worst the additional functions do notadd the right flexibility to improve the variation in which case the energy obtained from the calculation willbe the same.

Q15: Do you agree with the following statement (if so, explain; if not, try to present a counter example)?‘A variational calculation using a basis set B which has a larger number of functions than basis set A, willalways yield a lower or equal total energy as a calculation using basis set A’.

No. The results also depend on the quality of the basis set. The issue is not the number of basis functions,but rather the how well they mimic the true wave function. For example, one Gaussian function suffices todescribe the harmonic oscillator exactly while even a set of several Lorentzian functions will always yielda higher energy than the true ground state energy. More complicated systems differ only by the fact thatthe true ground state wave function cannot be described exactly. However, differences in quality betweendifferent basis sets are still observed and cannot always be compensated by the quantity of functions included.

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