Brainstorming ! : for example, If you start with a closed shell singlet system, maybe as atoms get apart it could evolves to an open shell singlet or a triplet in the most simple case so as Matus recommended you have to optimize resulting structures in several multiplicities to explore the most energetically stable spin state of the system.

There is no such thing as "spin multiplication", the correct term is spin multiplicity (http://en.wikipedia.org/wiki/Multiplicity_(chemistry)). Multiplication is the process when you multiply two numbers (http://en.wikipedia.org/wiki/Multiplication). If n is the number of unpaired electrons and M is spin multiplicity and S is the angular spin momentum then S = n / 2, M = n + 1, M = 2S + 1.

Drawing a valance structure is not enough to decide multiplicity. A good example is O2 (http://en.wikipedia.org/wiki/Oxygen#Characteristics). Unless you know the multiplicity beforehand, there is no sure way of figuring it out before doing the actual computation with Gaussian or whatever software you use.

You have to define multiplicity before the computation because it cannot change during the computation even if it is not the lowest energy spin state. So, you have to do the computation with the lowest possible multiplicity and then increase it by two, repeat the computation and then compare energies. The one with the lowest energy is the ground state multiplicity of your species.

Bottom line:

Compute each possible multiplicity and then chose the lowest energy.

Just to add to what others have said ... I think your intuition that it is a singlet is probably correct. However, you can easily see what the energy penalty would be to unpair two of the electrons by calculating the triplet state as well ... you may be surprised and find that it actually lowers the energy (although I rather doubt it). Just be aware that the equilibrium geometries for the two different spin states are likely to be different, so you will need to do separate geometry optimizations (and frequency calculations) for each case.

As far as multiplicities higher than the triplet are concerned, it gets less and less likely that states will be close in energy to the ground state as you increase the number of unpaired electrons. However, for very highly symmetric geometries, you may have a high degree of degeneracy, and thus Hund's rules might lead you to conclude that such a "high-spin" state is likely to be the ground state. On the other hand, such highly symmetric geometries are subject to symmetry-breaking vibronic couplings (e.g. Jahn-Teller distortions) that split degeneracies and lower the energy, in which case the diabatic state picture that you would get from a single-reference electronic structure calculation (like the one you are talking about in your post) is unlikely to be correct.

When you do these calculations, it's highly recommended that you make a Stable Test of the Wavefunctions you used with the "Stable" key word in Gaussian. The case in determining the ground state of O2 is a good example.

Old discussion, but very well fits in my considerations. I would like to calculate (and then understand and calibrate my input file for other cases) the rotational energy barrier for ethene. So I calculate the energy for C2H4 molecule having dihedral angles 0 (minimum) and 90 degrees (maximum). I used the input file as below:

%nproc=1

%mem=1500MB

#b3lyp/6-31g(d) Opt=ModRedundant

test job

0 X

C1 5.0000000 4.9993134 4.3338099

C2 5.0000000 4.9993134 5.6661901

H3 5.0000000 5.9287934 3.7598300

H4 5.0000000 4.0718942 3.7611027

H5 4.0705204 4.9993138 6.2401705

H6 5.9274197 4.9993138 6.2388978

3 1 2 5 F

where X = 1 or 3 as was suggested above by Andras and David.

I got the energies:

                    singlet state                        triplet state

  0 deg         -78.5874577903               -78.4557121714

90 deg        -78.5250265687               -78.4856245833

So, if I take the lower energies in this both cases (both for singlet state) I got the energy barrier of ~39.2 kcal/mol, so much lower than reported in literature.

However, during rotation of ethene molecule at 90 deg the p-bond is broken so there are 2 unpaired electrons (triplet state). So if I take energy of triplet state for 90 deg I got the energy barrier of 0.1 Hartrees = 62.7 kcal/mol so the value that is (moreless) known from literature.

So, ok - this was easy as I know that 90 degrees should be triplet. But what to do in the case of angles that are between singlet and triplet state (e.g. 60-80 degrees as I mainly want to obtain the whole shape of rotational energy not only the rotational barrier) for which I cannot be sure which state I should assume? Taking lower energy might not work as I showed for the case of 90 degrees.

Ok, I found the misteake I made.

I set only one dihedral angle, so the other can change and the geometry of ethene 90 deg singlet was far away from what should be, even if this one dihedral angle was set correctly.

When I set both of them fixed:

3 1 2 5 F

4 1 2 6 F

I obtained:

                              singlet state                                  triplet state

0 deg                 -78.5874577903                      -78.4557121714

90 deg               -78.4356687025                      -78.4856245833

so 90 deg singlet has higher energy as triplet as it should be. And then rotational energy barrier is also as it should be (~63.3 kcal/mol).

Then I wanted to check how the Stable instructions mentioned by Xingyong changes the results and how it will work in practice showing (or not) correct spin multiplicity. I follow the page http://www.gaussian.com/g_tech/afc.htm and did Job1 -> Job4 with the input files for the first two jobs:

Job1

%nproc=1

%mem=1500MB

%chk=s1.chk

#b3lyp/6-31g(d) Opt=ModRedundant

test job

0 1

C1 5.0000000 4.9993134 4.3338099

C2 5.0000000 4.9993134 5.6661901

H3 5.0000000 5.9287934 3.7598300

H4 5.0000000 4.0718942 3.7611027

H5 4.0705204 4.9993138 6.2401705

H6 5.9274197 4.9993138 6.2388978

3 1 2 5 F

4 1 2 6 F

and Job2

%nproc=1

%mem=1500MB

%chk=s1.chk

#b3lyp/6-31g(d) guess=read geom=check stable=opt

test job

0 1

and the bolded energy was improved to -78.487 ... so close to original triplet.

But what is also quite interesting even after doing Job1 in the output file I got the information about the instability (line 268), suggestion that the state is triplet (line 265 and 268) as well as the energy (after assuming correct spin multiplicity?) of -78.4865035399 (line 294). The same energy I got when applied 2 dihedral angles fixed for the case of 90 deg triplet state.

I attached the file after Job1 (edit: this is file after Job2!)

So the question is, if I really need to do stability test (Jobs2-4) as even after assuming wrongly the spin multiplicity Gaussian will suggests me the correct multiplicity and calculates the energy close to this state?

I wrote down the way I did the whole procedure as I am still unsure (as working 3rd day with Gaussian) if I did not make somewhere a stupid misteake.

The underlined text is nor correct as suggestion of the correct spin multiplicity came after Job2, so as was described in link cited above.

Multiplicity M = ( 2s +1) , radical neutre M= 2*1/2 + 1 = 2, charge = 0

radical cation M =2. charge = + 2

radical anion M = 2 , charge = -2.

s=n/2 where n is the number of the unpaired electrons

Meriem Zerrouki

s=n/2 where n is the number of the unpaired electrons and M = ( 2s +1) .

Is it correct then to write M = n + 1

M: multiplicity

n: number of unpaired electrons.

In order to identify the unpaired electrons, you have to refer to the orbital theory and based on the structure of your compound you might be able to determine the number of unpaired electrons.

HF Hartree-Fock can be used. It uses the approximation that Coulombic electron-electron repulsion can be averaged, instead of considering explicitly repulsion interactions (central field approximation). There are two ways to compute molecular orbitals using HF: UHF (unrestricted) or RHF (restricted). UHF uses a separate orbital for each electron, even if they are paired (used for ions, excited states, radicals, etc.). RHF uses the same orbital spatial function for electrons in the same pair (good for species with paired electrons, no spin contamination).