You should now have four completed optimisations and frequency analysis:
6-31G(d): square planar and tetrahedral
6-311G(d,p): square planar and tetrahedral
Now we will consider the energy of the two conformers, which has the lower energy, the tetrahedral (triplet) or the square planar (singlet)? Normally singlet states are lower in energy than triplet states. However, a tetrahedral geometry is normally lower in energy than a square planar geometry.
Record the energy of your calculations on [NiCl4]2-, mine are detailed below. Yours should be very similar, the same up to 4th decimal place, if they are not speak to a demonstrator to find out where you have gone wrong!
6-31G(d)
tetrahedral: -3349.14928357 au
square planar: -3349.09694596 au
6-311G(d,p)
tetrahedral: -3349.41619595
square planar: -3349.37765320
The results of a single calculation are reported in atomic units (hartree) so that others can reproduce your work, this is the ONLY reason for reporting total energies. Never compare calculations run using different basis sets, the results are meaningless.
The important results are always energy differences, these must always be evaluated for calculations carried out with exactly the same method and basis set (and ideally exactly the same "additional keywords" when they relate to the evaluation of the energy of the system.)
When reporting the difference in energy between two isomers we always treat the lowest energy conformer as the reference and report the energy of the other conformers as positive energies, ie being higher in energy than the reference. In my case the tetrahedral conformer is lowest in energy and so it is treated as the reference (ie 0.00 kJ/mol).
Energy differences are reported in kJ/mol, or kcal/mol if you are in the USA. Never in atomic units. Convert your ΔE into kJ/mol
It is standard to report qualitative energy differences to 1 kJ/mol and very accurate energy differences up to 2dp (two decimal places) in kJ/mol, ie 0.01 kJ/mol. Thus, we need to know the energies in au to an accuracy that will allow us to report energy differences to the correct level. The important question is how much is 0.01 kJ/mol in au? 0.01 kJ/mol is 0.0000038 au! So when reporting energies in au you must record them up to at least 6dp (recording up to 8dp is better and then drop the last two when reporting the data)
Thus, (for my calcualtions) the square planar structure lies above the tetrahedral structure by
ΔE=261 kJ/mol at the B3LYP/3-21G level
ΔE=137 kJ/mol at the B3LYP/6-31G(d) level*
ΔE=101 kJ/mol at the B3LYP/6-311G(d,p) level
*structure is a transition state
You can also very clearly see the importance of using a reasonable basis set in order to obtain a proper minimum and good energies (actually energy differences). However, the geometry is not as sensitive to the basis set and so we use the low level calculations which are much cheaper to give us a good starting geometry.
All the calculations carried out here predict that the tetrahedral geometry is the most stable one. Energy differences of 100 kJ/mol are large, thus we would expect the tetrahedral geometry to be almost exclusively the dominant conformer in solution at room temperature. If the energy difference between the conformers is smaller we could expect a small proportion of other conformers to be present in solution, this proportion increases as the energy difference between the conformers decreases.
However, in stating this we assume that the barrier to interconversion between the isomers is small, if this barrier is large then interconversion is unlikely to occur. Typically if an energy barrier is less than 80 kJ/mol it is surmountable at room temperature and multiple conformers could exist. important You will need to remember this for one of the questions in the quiz.
The image below represents a number of different scenario's related to the relative stability of the complexes and barrier heights for interconversion. Thermosdynamic effects relate to the energy of the reactant and product, kinetic effects relate to the energy barrier between them.
