Computational Chemistry

Winter Quarter 1997

( Due 3 / 10 / 97 )


The Problems: (note that there is an appendix to simplify the compilation of some answers-report absolute energies in au to 5 decimal places, relative energies in kcal/mol to 1 decimal place, bond lengths in ångstroms to 3 decimal places, and valence and dihedral angles in degrees to 1 decimal place-written comments should still be provided separately)

1. Last problem set, we (painfully) examined the inversion barriers for ammonia and fluoroammonia. What is the inversion barrier for ammonia as calculated at the CCSD(T)/cc­pVTZ//MP2/cc­pVDZ level? Does the inclusion of triple excitations have much effect? Correct the CCSD(T) barrier for zero-point vibrational energy. Correct for thermal vibrational enthalpy. Finally, correct for all free-energy effects. (To do these last items, a frequency calculation will be required for the optimized geometry). Fill this information in Table 1. How does this barrier compare to the estimate from AM1? Without knowledge of the experimental value, what factors remain that might cause any deviation between the calculated value and experiment.

The best DE barrier is 6.3 kcal/mol. This is reduced to 5.4 kcal/mol by zero-point effects (H0), and this value is unchanged by temperature correction to 298 K. The free energy barrier is 5.9 kcal/mol. The effect of triple excitations (evaluated by comparing the barrier to that calculated at the CCSD level, which is part of the CCSD(T) output) is to raise the barrier by 0.2 kcal/mol, a rather small effect, as is expected for this molecule which has no special correlation effects. The AM1 barrier was 4.2 kcal/mol, which is considerably lower than the CCSD(T) number (if one believes that an AM1 calculation is really a heat of formation and not an electronic energy, one might compare to 5.9 kcal/mol, but this is still a difference of almost 2 kcal/mol out of 6). Other factors that remain are: (1) We have not demonstrated convergence with respect to basis set size. (2) We could consider more refined techniques for electron correlation, but they probably are unimportant given the good agreement between CCSD and CCSD(T), i.e., we seem to be capturing all of the electron correlation effects on the barrier height, at least with our current basis set. (3) Tunneling might lower the phenomenological barrier height for the experimental measurement.

2. FOOF (fluorine peroxide) is an odd molecule, to say the least, but DuPont thinks it is fascinating (oxyTeflon®?) Using the cc-pVDZ basis set, calculate the structure for FOOF at the RHF, MP2, and BPW91 (a density functional) levels and record your results in Table 2. What level of theory appears to be most accurate compared to experiment? Take the total time for your calculation (printed at the bottom of the output file) and divide by the number of geometry optimization steps to get a rough estimate of the time per step for each level of theory. Report this time, and comment on whether this makes any theory seem more attractive than an analysis based purely on agreement with experimental structure. (If one of your jobs runs out of time (after 6 minutes), note the number of steps so you can keep track before you restart).

The RHF level gives a truly terrible geometry for FOOF, because electron correlation is so important for this molecule with lone pairs EVERYWHERE! The MP2 structure is clearly the best, although DFT is very nearly as good. Perhaps surprisingly (at least if you listen to DFT advocates) the time for the MP2 geometry optimization step isn't much slower than for DFT. This is partly because of how Gaussian is coded (it only recently became a DFT code) and partly because you need to get to bigger systems before the difference in time between DFT and MP2 really starts to increase. If we had done frequency calculations, a much larger difference between DFT and MP2 would have been apparent (but we didn't have enough time).

3. Find a transition state where the reaction coordinate is not symmetric. That is, a case where one cannot use symmetry to impose a particular constraint on the transition state structure (like ammonia inversion necessarily being planar-an example of an unconstrained case is the tautomerization of ammonia N-oxide to hydroxylamine from last year's problem set 3 (on server Dionysus and class website), which you can not use). Verify your calculation by reporting the imaginary frequency for the transition state structure. Print out a picture (Chem-3D or your favorite drawing program) of the transition state structure and, using the imaginary vector, describe what the imaginary mode looks like (if you want, you can print out pictures of the structure distorted by the displacements listed in the output, or you can just say something like "the hydrogen is moving to eclipse the sulfur while the oxygen-manganese bond lengthens"). Pick any ab initio level of theory you want.

This is the transition state structure (imaginary frequency 401i at RHF/STO-3G level) for the reversible isomerization of {[(NH3)3Cu]2}O2 between a bis-oxo structure (four membered ring) and a peroxo structure. The imaginary frequency corresponds to simultaneous inward (outward) motion of the oxygen atoms and outward (inward) motion of the copper atoms. The ammonia ligands are pretty much just along for the ride.

{[(NH3)3Cu]2}O2


Table 1. Ammonia inversion barrier (absolute energies in au, barrier height in kcal/mol).

NH3 structure MP2/cc-pVDZCCSD(T)/cc-pVTZ// MP2/cc-pVDZ H0H298 G298
pyramidal-56.382 50-56.472 92 -56.438 11-56.434 31-56.456 16
planar-56.369 38-56.462 84 -56.429 49-56.425 71-56.446 82
barrier height8.26.35.4 5.45.9

a Note that the default for Gaussian94 is 298 K, so all thermochemical information is provided for this temperature without you needing to specify any additional input.



Table 2. Details of FOOF calculations with cc-pVDZ basis set.

LevelrFO, ÅrOO, Å Angle FOO, degwFOOF, degtime per geometry step, sec
RHF1.3681.304106.084.4 12a
MP21.5821.209109.387.4 51a
DFT1.6111.212111.288.7 49a
expt.1.5751.217109.587.5

a These timings may vary considerably depending on individual runs due to overhead, different starting geometries, etc.