My molecule of interest is a single negatively charged glucose-based molecule that consists of 30 atoms (9 carbon atoms, 8 oxygen atoms and 13 hydrogen atoms). The total valence electrons are 131 with -1 charge. Therefore, I have set -1 charge and a multiplicity of 2 (since it has one unpaired electron) in the input file. However, I faced an error termination that stated this "The combination of multiplicity 2 and 132 electrons is impossible". I have no idea of how the Gaussian09 program calculated a total of 132 electrons. When I changed the setting to 0 charge and a multiplicity of 2, the calculation works fine (Gaussian calculated 131 electrons) with a normal termination. Is there anyone can give some inputs on this? Thank you.
Mr Ong, if there is carboxylate group present, with charge -1, that means that only proton (H+) is missing, so the number of electrons is the same as it was before the deprotonation (in COOH). In other words, the total number of electrons in your anion is 132 (9 carbons give 54 electrons, 8 oxygens give 64, 13 hydrogens contribute with 13 electrons PLUS one that is left after H+ left the COOH group). So, 54+64+13+1=132. This is closed shell system, the multiplicity is 1 (singlet state). I hope this is clear to you now.
I think Gaussian may be right. C9H13O8- has an even number of electrons and so should be a singlet (it's the conjugate base of C9H14O8, if that helps). Your second calculation is C9H13O8* which is the corresponding radical (made by removing one H atom from C9H14O8).
I have a carboxylate group COO- and it has an unpaired electron. Therefore, I have to put -1 charge and at the same time, a multiplicity of 2 (that is 2(1/2)+1 = 2). Is it possible that it could be a multiplicity of 2(-1/2)+1=0?
You're not going to find a "0-plet" anywhere. The multiplicity has to be a positive integer.
Are you sure you have the right number of hydrogen atoms? Usually in Gaussview, one makes a radical center by removing a H atom and then removing the resulting "bond to nowhere". Not removing the "bond to nowhere" will, I've noticed, confuse Gaussian as to the number of electrons in the system, so it's something to check.
It seems you may already have your answer but I wanted to add a tad. Note this is mainly for other users that may see this post in the future.
Gaussian09 determines if a requested charge/multiplicity is correct via electron count (atoms + specified charge) and the standard equation for multiplicity (2S+1, where S = spin angular momentum). Yes it is important to determine the possible states your molecule could exist but it is far more important to find the CORRECT state you are interested in.
For example, consider the neutral state of the monohydrated dioxygen (O2/H2O) molecule specifically, the O2 fragment. Here, Gaussian will complain with the error you are seeing if any even multiplicity (doublet, quartet, etc.) is specified, for obvious reasons. Conversely, entering a odd multiplicity passes Gaussian's test and will run just fine. One may start by specifying a singlet, which Gaussian will allow to run without error. However, in this particular case, singlet oxygen does not correspond to the ground state, but instead a triplet.
So although Gaussian will execute and terminate properly with singlet oxygen, if you were interested in the ground state, your results would be incorrect. This can be determined via a quick UHF (with minimal basis set) calculation on your systems for each state of interest.
I hope this helps and good luck with your research!
Mr Ong, if there is carboxylate group present, with charge -1, that means that only proton (H+) is missing, so the number of electrons is the same as it was before the deprotonation (in COOH). In other words, the total number of electrons in your anion is 132 (9 carbons give 54 electrons, 8 oxygens give 64, 13 hydrogens contribute with 13 electrons PLUS one that is left after H+ left the COOH group). So, 54+64+13+1=132. This is closed shell system, the multiplicity is 1 (singlet state). I hope this is clear to you now.
dear Ong,
Maybe the explanation below will help you: 1- if the charge is due to the carboxylate group (COO-) the multiplicity is 1, because there are no unpaired electrons; 2- If the charge is from a neutral group that received an electron, then it will be triplet ((1/2 * S + 1) - Hund Rule).
for the case 2, you need to inform in Gaussian that the system has open shell or unrestricted.
Your charge is based on the atoms in the QM region of your simulation. The charge and multiplicity need to be specified correctly at the same time. The multiplicity takes into account the spin states of your system, e.g. singlet vs triplet oxygen. Gaussian will not converge unless you have a meaningful setup of both.
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