Hi, my apologies if this is quite rudimentary, I am having to learn a lot whilst racing against the clock. I am trying to calculate the charge and spin multiplicity for Cu(+2) ion sitting between the COO- of two backbones (the actual side chains are then bound to an intermediate molecule which has a +1 charge (due to mass spec protonation) holding them together). Mass spec data is showing the removal of the 2H from either backbone (hence why I assume that there is now a COO- group). From reading literature I am thinking the interaction between the copper ion and the carboxylate groups are electrostatic in nature. However, a colleague has informed me of cases where it can be covalent. So I am now of two minds how to model the spin multiplicity and charge in Gaussian.
1) Electrostatic assumption:
The total unpaired electrons are 3, one on each COO- and one on the Cu(+2) as [Ar]3d9. Total spin of 4. The total charge is 1 (-2 from the two COO-, and +2 from Cu, and +1 from the mass spec protonation of an NH elsewhere). Gaussian input is 1 4. Is this correct, or should I be treating the spin on COO- differently?
2) Covalent assumption:
I am not convinced with this case, although my experience is limited, and I am not sure what the overall charge or spin is.
I would really appreciate the help of a chemist.
Many thanks
Hi Anthony,
I don't understand what you mean with: "+1 from the mass spec protonation of an NH elsewhere". Are you trying to include the whole chain in the calculation or just a part of it?
In any case, I'd say that the COO- Cu -OOC group is a neutral doublet. The COO groups do not have an unpaired electron. If the remaining part of the chain is +1 then you'll have a cationic doublet. "1 2" in gaussian input line.
Hope it helps...
Hi Anthony,
I don't understand what you mean with: "+1 from the mass spec protonation of an NH elsewhere". Are you trying to include the whole chain in the calculation or just a part of it?
In any case, I'd say that the COO- Cu -OOC group is a neutral doublet. The COO groups do not have an unpaired electron. If the remaining part of the chain is +1 then you'll have a cationic doublet. "1 2" in gaussian input line.
Hope it helps...
Hi Sergi,
Could you explain (seems a little simple I am afraid) why RCOO- does not have an unpaired electron. Is this to do with the delocalisation of the charge between the two O atoms?
I assumed you would have:
R C O[1s2 2s2 2p6] O [1s2 2s2 2p5]
Sure, it is not simple, specially if you're not chemist.
Even then it is difficult to explain. Consider you have RCOOH. It is a closed-shell specie. When you remove the H, you're actually removing H+, so the electron remains in the RCOO- unit. Then, since there is the same number of electrons, the molecule is still a closed-shell specie. It has an even number of electrons. You'd have to remove H0 to form an open-shell specie (or radical).
I do not like, although I understand, to descibe the electronic structure using "R C O[1s2 2s2 2p6] O [1s2 2s2 2p5]" I would prefer Lewis notation but maybe you're not used to it. You can find an entry on wikipedia about lewis notation, and I attach here the structure of formic acid, the case with the simplest possible R.
Anyway, for any future doubt you may have, keep in mind that a radical is EXTREMELY reactive so you'd hardly keep RCOO- as an open-shell (or radical) specie. In the end, RCOO- is not that different from CO2. It is very difficult to preserve a radical especially if there is no metal atom involved...
Don't hesitate to ask me further
Hi Anthony,
Interested to read your question as we have been doing almost these exact calculations recently. I can't add much on what Sergi said except to agree with him.
You might like to take a look at the popular review article "Relationships between the carbon-oxygen stretching frequencies of carboxylato complexes and the type of carboxylate coordination" to get an idea about how carboxylates coordinate with metals. Covalent bonding is indeed common. In your case you likely have both carboxylates bound to the metal atom in a monodentate (unidentate) fashion.
I agree that the RCOO- Cu -OOCR group would be treated as a neutral doublet (0 2 in Gaussian) and the protonated form picked up in the mass spec would be a radical cation (1 2).
Hi Gabriel,
Thank you very much for that information, especially the pointing out the review article. This is a new area of research for me so every little helps.
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