The number of basis functions in a basis set can be thought of as the resolution (eg. pixel count) of your wavefunction. Just like any other image, the resolution can always be improved further by adding more basis functions, and the level that you consider to be sufficiently "accurate" will depend on what you want to do.

That said, larger basis sets generally produce more accurate energy values, so the 6-311+G(2d,p) basis set will almost certainly be significantly more accurate than the 6-31G(d) basis set.

It might be worth reading the manual or surrounding information about these basis sets in particular, so you can get a flavor of what is actually being added by each of them. For example, the first one is a double-zeta basis set with polarization functions on non-hydrogen atoms only, while the second one is a triple-zeta set with diffuse functions on non-hydrogen atoms and polarization functions on all atoms.

It might also be worth noting that if the electronic excitation you are calculating involves electrons moving over a significant distance between two essentially "non-overlapping" orbitals, then you might also want to consider using a range-separated functional like CAM-B3LYP, as standard global hybrid functionals like B3LYP are liable to quantitatively fail to describe these states.

Magnus.

Dear Prof. Magnus,

Thank you very much for the expertise explanation and recommendation. I really appreciate it. Please, I wonder if you would have any spare time to attend to a direct message?

Best wishes,

Jude

B3LYP/6-311G++(2d,p) is more accurate.

More is generally better as mentioned above but the 6-31g* basis sets give extraordinarily good results, as the result of fortuitous cancellation of errors. The triple-zeta 6-311++g** basis set will give marginally better results at the expense of significantly more computational time. Neither provides "chemical accuracy" if you are trying to perform single-point calculations to compare with, say, enthalpies of formation. If you are trying to study a pathway, or find a transition state along the pathway, the diffuse functions and polarization functions contained in the 6-311++g** basis sets will provide a better description of the electron density in these intermediates.

In such calculations, where there are two possible pathways, you hope to find that the barrier height in one is say 30 kJ/mol and the other is 110 kJ/mol. Then, you can have some confidence that the reaction dominantly proceeds along the former path. If the difference between pathways is less than 10-20 kJ/mol, you are really banging into the limits of what the theory can resolve.

Mark

Thank you,@Mark. Actually, I am interested in the emission wavelength/frequency of relatively large molecules. So, the possibility would be geometric optimization of the excited state (obtained from the optimized geometry) and getting the SP energy (Root=1). I am a bit confused on whether it should be difference between the excited state energy and the SP energy obtained from the optimized excited state or the calculation of molecular energy that would give me the emission energy?

I'm not quite sure what you are intending to do. The literature is replete with various uses of "excited state." A DFT frequency calculation will provide vibrational/rotational frequencies that need to be corrected (by a factor) in order to compare with experiment. (https://cccbdb.nist.gov/vibscalejust.asp) These are excited states. I have also seen intermediate states along a pathway called excited states. I believe, instead, that you may be considering chromophores that have electronic transitions that produce visible light. These, I believe, do not arise from conformational shifts in the molecule but from electronic shifts between eigenstates. I do not believe that the excited state should be subjected to geometry optimization in this case.

Thank you everyone for the helpful answers.

Last questions please:

My target is using DFT to predict the emission wavelength of bioluminescent molecules. As suggested, the ground-state (GS) and excited state (ES) geometrics of all the species studied were optimized with the B3LYP/6-311+G(2d,p) model. I am trying to predict emission spectra with the TDDFT method. I know that the difference between the energy of the optimized excited state and the molecular energy would give the emission energy from which the wavelength of the emitted light can be calculated. But it appears that I have four energies for the calculated excitation energy (Root = 1) (see screenshot 1; three are in Hartree, one is in eV).

Which of these should be used with the molecular energy (Screenshot 2) for this emission energy (i.e. the difference)?

PS: Just a newbie in computation chemistry trying hard to learn.

6_31_g (d,p) most occurate