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Content: Kimball’s model with 2-electron cloud: , He, H2, and others. Correlation problem.

ES 23 August 2017//22 Feb 2015//2002//1982

Next step: How to put 2 electrons into the Kimball cloud

Again:

Kimball’s equation for the H atom in its stationary groundstate:

R is the expectation value of the electron-proton distance of the H1s state

Kinetic energy Eh:

Potential energy Eh:

The electron density ρ in a Kimball 1-electron sphere with radius R is

Now we add a second electron with the same density into an infinitesimal shell at r and let it interact with the clouds charge within r=0 to r to obtain the contribution to the repulsion energy:

and do this from r=0 to r=R to sum it up to 1 electron:

This is the electron repulsion energy of the superposition of two identical Kimball spheres, in atomic units:

Now we can combine the terms to compute the energy of , the hydride ion in the gas state at 0 Kelvin:

is -0.36 Eh. That is less negative than the energy of the H atom, hence the ion is unstable towards ionization to H + !

As chemists we know the hydride ion in a condensed state, e.g. as {LiH}, {Li[AlH4]} or in solution in aprotic solvents like diethylether. In the high vacuum of a mass spectrometer, it is easily detectable as gaseous ion. Its electron affinity is positive and amounts to 0.0275 Eh (= 0.748 eV), hence it is a stable system, = -0.5275 Eh! Another failure of the Kimball model ?

Since is spherically symmetric it should behave like the H atom in regard to the nuclear density cusp. So, perhaps the electron repulsion energy has not been computed correctly? The way we have treated this above, both electrons have the same probability to be at a certain spot, alone or together. That is certainly not correct! Electrons repel each other. In they try to avoid each other as much as they are attracted to the proton in the center. This leads to a certain amount of correlated motion. We try to calculate the repulsion under the assumption that the two electrons maintain an average distance from each other of the order of R. This would lead to a repulsion energy Ve = 1.0/R instead of 1.2/R. The following computation of He and homologues as well as a long experience with Slater’s screening constants recommends even 0.9/R.

He atom and homologues

He atom, the next 2-electron problem, can easily be formulated. We do it with the whole group in mind:

We take Ve=0.9/R = 3 σ/R and interprete σ below

This is the energy of He like ions , He, , , ..., in Eh, compared to the experiment:

σ=0.3 is remarkably better than σ=0.4. If you look at Ehe = you see that it differs from the energy of H like ions /2 by a factor of 2. That comes from the second electron. In addition, Z is diminished by σ, the number proportional to the electron repulsion energy. Therefore, σ is called screening constant. The second electron screens the nuclear charge from being “seen” entirely by the first and vice-versa, we don’t know how to number electrons ...

Look up slater.htm do learn more about screening constants and their historical development.

Accepting Slater’s σ=0.3 as an improvement over Kimball’s correlation-free σ=0.4, we want to check on again. There is a large improvement from -0.36 to -0.49, but we haven’t made it yet. is still slightly unstable towards spontaneous loss of the second electron. But now we are in good company: Even the Hartree-Fock limit of a quantum chemical computation gives the same result! see e.g. C.J.Roothaan et al.Rev.Mod.Phys. 32(1960)186.

Hydrogen molecule

We have Z=1, 2 protons like , 2 electrons like He. Just copy appropriate terms from those systems, above, and get:

H + H → H2 ΔE = -1.21 + 2*0.5 = -0.21 Eh = -131.8 kcal/mol; Bond distance = 2*r = R = 15/11 a0 = 0.72 Å

That’s not bad at all, considering that the first computation of H2 by Heitler+London 1926 (including the correction of Y. Sugiura) gave

ΔE = -0.115, dH2 = 0.869 Å, while the experimental values are ΔE = - 0.17447, dH2 = 0.7414 Å.

The bond distance is just slightly too small, while the bond energy is 20% too large.

This rings a bell: ? Interestingly, we have not applied any correction to the gospel of St.Kimball. But we like to improve the result, which can always be done. We learn by doing so, how Kimball’s model can be tweaked, and perhaps even understand why. In protag.htm some of the deliberations are presented. The result is, that neither correcting the kinetic energy as for , nor decreasing σ as with He, helps to improve the result. Strangely, a slight increase of σ from the (non correlated) Kimball value of 0.4 to 0.4165 brings H2 into a spendid approximation of both experimental values:

H + H → H2 ΔE = -1.17397 + 2*0.5 = -0.1739 Eh = -109.2 kcal/mol; Bond distance = 2*r = 1.3844 a0 = 0.7326 Å

Now, that's excellent! Explanation? A higher σ creates more screening for the off-center cusps around the protons and thus helps similarly to the recipe for to reduce the attraction by the charge between the nuclei. This trick cannot be applied in , because with one electron there is no screening!

Many more computations of molecules with bonded H atoms have been done to support this ad hoc improvement as a valid hypothesis.

For an analysis of the total energy of H2 and comparison with “first principles” QC computations it is interesting to list the components: Vnn = nuclear repulsion energy, Vee = electron repulsion energy, Vne= nuclear-electronic attraction energy, T kinetic energy, and the Virialratio Vir

Generalized H2 molecule

Let’s imitate J.D Dunitz and T.K. Ha, J.C.S. Chem.Comm.,1972,p.568: JDD gave a talk at ETH, Zürich, 1965, about these results, which I heard. I enjoyed them so much, that on the train home, I did the following calculation with Kimball’s model on a postcard and sent it to him. Below, I show the comparison of the results:

Assume that nuclei of arbitrary fractional charge exist and “construct” a Z2 hydrogen like molecule. Where on the curve EZ2(Z) is the most stable diatom? What does this teach chemistry?

The results of Dunitz & Ha for the Z2 molecules are in the next list. D & H used a basis set which in modern QC packages would be characterized as elementary. They only found -0.1329 Eh for the H2 binding energy, instead of -0.1744. However, their paper does not suffer from this inaccuracy. It has led to very interesting discussions on the influence of polarisation for the stability of the covalent bond, and therefore to Pauling’s electronegativity. We plot it below, too.

The highest bonding energy is with Z ~ 1.2, similar to an early treatment of the H2 molecule by S. Wang (1928), using effective nuclear charges in a variational wavefunction.

The point at Z=2.0, the He2(++) molecule, is strongly positive. Hence, this is not a stable molecule, as experiment proves. H2, Z=1.0: D&H have only 76% of the Kimball bond energy. But, both attempts get the story correctly. For the interesting chemical conclusions I refer you to the original paper, which you can download with the link given above.

D & H also treat the asymmetrical diatomic Z1Z2 with nuclear charges [(1+δ)(1-δ)], 0<δ≤1. I leave this as an exercise for the reader, who can easily make the necessary changes to the formula for the Z2 molecules, above (just a hint: the nuclei with Z1=1+δ, and Z2=1-δ, now have different distances from the center of the electron cloud, r1, r2, and their distance is r1+r2; so, there is an additional variable for the FindMinimum function).