# Why is it so hard to find the ground state of molecules with d-block or f-block elements?

Based on some of my previous reading and questions that I have asked I am aware that determining the ground state electronic configuration for molecules containing d-block elements theoretically is difficult. The method that is usually suggested is to calculate many energies of the molecule, each time assuming a different spin multiplicity, until all possible candidates have been considered before picking the one that has the lowest energy.

I was just wondering what are some of the theoretical difficulties encountered when trying to deduce the ground state electronic configuration using this method specifically when dealing with d block element containing molecules and what extra difficulties are encountered when trying to deal with the f block elements.

• +1. But I'm curious why you want to know the difficulties? It's a good question but I'm curious about the motivation behind it :) Aug 19 at 18:41
• So I’m trying to calculate scattering ionisation cross sections for molecules containing d and f block elements. For instance WO2. To get accurate results I need to know the multiplicity of ground state and when I look online for papers I see there is a lack of consensus. I was interested to see what difficulties theoretical chemists encounter when trying to deduce the ground state, just so I understand the field a bit more and write a more comprehensive report!
– DJA
Aug 20 at 0:04

Difficulty in capturing enough electron correlation: The difference between the lowest states belonging to different electronic configurations can be small (for example, on the order of 1 kcal/mol). Traditional methods for full configuration interaction (FCI) calculations such as the Davidson method are unable to be used in any remotely accurate basis set for a molecule beyond a handful of electrons, but d-block elements have far more than a handful of electrons.

Take scandium for example, which is the first d-block element. Even if you freeze the neon core, there's 11 electrons. You are asking about molecules, so we must at least have a hydrogen attached to the scandium, so now we have 12 electrons to correlate. Even if we use the cc-pVDZ basis set (which is often not enough to get 1 kcal/mol accuracy in the difference between two electronic ground state energies), we would have 48 spatial orbitals. How many ways can you put 6 spin-up electrons and 6 spin-down electrons into 48 spatial orbitals without imposing any spatial symmetry restrictions? The answer is:

$${48\choose6}^2 \approx 1.5 \times 10^{14}. \tag{1}$$

If each FCI coefficient were to be represented using double-precision arithmetic (8 bytes), you would need more than 1 petabyte to store all coefficients, and even taking advantage of sparsity won't help you get to size where traditional FCI methods will be convenient. More recently developed methods such as FCIQMC, SHCI and DMRG would be able to treat systems with $$10^{14}$$ FCI coefficients fairly well, but these methods are less black-box and have a bit more of a learning curve in order to understand how to determine that the calculations are properly converged. Furthermore, if you switch scandium with an element that has only 3 more electrons (chromium), and you switch hydrogen with an element that has only 2 more electrons (lithium, whose cc-pVDZ basis set is bigger than it is for hydrogen) the calculation in Eq. 1 becomes:

$${63\choose9}{63\choose8} \approx 2 \times 10^{18}. \tag{2}$$

The calculation just became more then 10000 times bigger simply by switching scandium to an element 3 electrons larger and hydrogen to an element 2 electrons larger!

Finally, molecules containing d-block elements tend to have several FCI determinants which are nearly equal in importance, meaning that either FCI or multi-reference methods are required, as opposed to the highly efficient single-reference methods like those based on a coupled cluster model (this is what Roma's answer says).

### Extra difficulties when involving f-block elements

Relativistic effects: The heavier the elements involved, the more that relativistic effects become important, which means that you need to know a bit more in order to know how to do the calculations properly. Furthermore, spin-orbit coupling starts to become more and more important, and this will make calculations more expensive, and will reduce the number of computer programs available for you to do the calculations (since not all electronic structure software packages have all methods implemented for the highly relativistic cases).

Even more electrons: If you thought that the examples in Eqs. 1 and 2 were hard, then just imagine $$\ce{LaH}$$ which involves the first f-block element and the lightest element in the periodic table. The cc-pVDZ-X2C basis set has 84 spatial orbitals and even with the krypton core frozen, you would have 21 electrons from the lanthaninum and 1 from the hydrogen, which means Eq 1 becomes 100000000 (10 million) times larger:

$${89\choose11}{89\choose10} \approx 2 \times 10^{26}. \tag{3}$$

Lack of basis set choices available: Basis set choices are far more limited for the f-block elements compared to the smaller elements in the periodic table. This has improved a bit in recent years, but there is still some more progress to be made in this direction. You may need to optimize your own basis sets, or to resort to using pseudopotentials (which can be considered under the list of extra things you'd have to learn in order to do relativistic calculations).

Lack of programs that can treat basis sets with k-type orbitals: Most electronic structure software does not allow you to do post-HF calculations with k-type orbitals in the basis set, so you wouldn't be able to use a 5Z basis set on uranium, for example, in , , , the public release of , or the publicly purchasable version of .

## d-block

The extra difficulty of studying a d-block element/transition metal complex is that the "breaking of degeneracies of electron orbital states" can result in different stable configurations depending on the geometry and symmetry of the molecule. This is the simple linear combination of atomic orbitals (LCAO) picture, or if you want more details see Crystal Field Theory. Furthermore, because of the Aufbau Principle we fill these orbitals one electron at a time, we can end up with different electronic spin multiplicity.

Therefore, as you said in your question, for d-block complexes you need to check different spins (and geometries).

The question is why? Why do similar complexes in the same column of the periodic table adopt different spins and geometries? For example, the molecule $$\ce{[PdCl4]2−}$$ is diamagnetic, which indicates a square planar geometry as all eight d-electrons are paired in the lower-energy orbitals. However, $$\ce{[NiCl4]2−}$$ is also d8 but has two unpaired electrons, indicating a tetrahedral geometry. The answer to this question is the subject of Crystal Field Theory.

Any theory that does not properly allow for an open-shell solution is no good (e.g. RHF). You must at the minimum allow for unrestricted calculations. Furthermore, you should allow for a multi-reference solution (for more details see Nike's answer in this thread) since these electronic states can be quite close together and can actually interconvert in what is known as a Jahn-Teller distortion.

## f-block

Finally, f-block elements are much different and pose a far greater challenge. Unlike in d-block where the molecular orbitals are split first and then those orbitals can be further split by a magnetic field into different levels, for f-block elements the levels are first split by spin-orbit coupling and then by crystal field splitting. It's very different.

For these you need a theory that correctly describes spin-orbit coupling. The exact solution requires relativistic treatment.

Very briefly, in many cases d- and f- orbitals are (quasi)degenerate. So to accurately calculate energies we need to take into account all permutations of electrons over those orbitals (multireference calculations).

If you want to know more there is a nice tutorial.