Total spin and/or multiplicity for transition metal ions?

Question

I am trying to input various charge states of various transition metal ion species into DFT calculations. The program requires an input of multiplicity = 2*(total spin) + 1.

I have an undergraduate-level understanding of chemistry, up to the point where I understand orbital filling (1s,2s,2p,...) for a given element and its ion of varying charge state.

However, I am having a difficult time understanding how to get the total spin and/or multiplicity for each ion. Does anyone have recommendations where I can find an explanation for how to do this? Would these values be tabulated somewhere?

Answer 1

Even for the simplest transition metal diatomic molecules, the most reliable way to know the ground-state spin configuration is often by doing experiments. I will give an example where it's easy to correctly determine the ground state spin configuration, and then an example where it has remained impossible as of the year 2020.

Cr$_2$: Here we can accurately guess the ground-state spin configuration of the molecule with your undergrad-level knowledge about orbital filling: Each Cr atom has 1 lone 4s electron and 5 lone 3d electrons, and when two Cr atoms come together we will get a bonding of the 2 lone 4s electrons, and the 10 lone 3d electrons, which results in the famous sextuple bond with all 12 orbitals that used to be half-filled, now being doubly-occupied, leaving zero unpaired electrons and a spin-multiplicity of 1. This singlet state turns out to be the correct ground-state configuration as far as we know.

Fe$_2$: In this case the idea of just bonding together every half-filled atomic orbital to create doubly-occupied (spin-0) molecular orbitals, predicts a configuration that disagrees with the present best experiments (we even know this from molecules as simple as O$_2$ whose ground state is a triplet). Even wet-lab experiments have not yet been able to confirm unanimously the ground-state of Fe$_2$: It is either $^7 \Delta_u$ or $^9 \Sigma_g^-$ (good luck even arriving at those 2 final candidates using theory alone!).

While doing the research to write this answer about the ambiguity of the ground spin-state of Fe$_2$, I found that the 2015 paper "Fe$_2$: As simple as a Herculean labour." gives a good history of the pursuit to find the ground spin-state of Fe$_2$, which I have tried to summarize even more compactly here:

\begin{array}{cccc} \text{Year} & \text{First author}& \text{Type} & \text{Ground state}\\ \hline 1975 & \text{Montano} & \text{Experimental} & \text{No conclusion}\\ 1982 & \text{Shim} & \text{Computational} & ^7\Delta_u\\ 1983 & \text{Nagarathna} & \text{Experimental} & ^7\Sigma_g \\ 1984 & \text{Baumann} & \text{Experimental} & ^7\Delta \\ 1988 & \text{Tomonari} & \text{Computational} & ^7\Delta_u\\ 2002 & \text{Huebner} & \text{Computational} & ^9\Sigma_g^- \\ 2003 & \text{Bauschlicher } & \text{Computational} & ^9\Sigma_g^- \\ 2003 & \text{Bauschlicher } & \text{Combined} & ^7\Delta_u \\ 2009 & \text{Casula} & \text{Computational} & ^7\Delta_u \\ 2011 & \text{Angeli} & \text{Computational} & ^9\Sigma_g^- \\ 2014 & \text{Hoyer} & \text{Computational} & ^9\Sigma_g^- \\ 2015 & \text{Kalemos} & \text{Computational} & \text{No conclusion}\\ \hline \end{array}

Conclusion: You cannot always know the ground state spin-multiplicity, even of some simple transition-metal diatomics, without doing careful experiments (either in a wet-lab, or on a computer). If you're dealing with a much larger system (which I assume is the case for you, because you mentioned ligands), perhaps you can try DFT with various functionals and various basis sets and see if there is one spin-symmetry that is always coming up as having the lowest energy.

Answer 2

If you're studying transition metal complexes, in short, there is no simple way to know except for to try the relevant physically plausible spin multiplicities and take the lowest energy solution as the ground state. For instance, an Fe(II) complex can have up to 4 unpaired electrons since it is $3d^6$. This is what we call the high-spin state and would be $2S+1=5$. However, you could also have 2 unpaired electrons for an intermediate spin state of $2S+1=3$ or 0 unpaired electrons for a low-spin state of $2S+1=1$. The only way to know for which is best for your system is to optimize the geometries of all three structures and compare their relative energies. If you refer to any standard inorganic chemistry text, you will likely find various sections devoted to comparing high- vs. low-spin complexes so you can gain an intuition for which is likely to be the ground state for your system, but best-practice is to try all relevant spin states. Note that the density functional you choose will greatly influence the spin splitting energies, so you should be sure to use an appropriate level of theory for your problem.