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The quantitative
G.N.Lewis Model

  The protagonists H, H2+, H2, HeH+, He, H-, H3+, H3

Details for the calculations are given in the Tutorial; original Kimball version, experimental numbers in parenthesis.

H-Atom: Variable R, Parameter Z

Ekin = 9/8R2;  Vne = -3Z/2R;   Vee = 0;  Vnn = 0
Etot = Ekin + Vne;  Min[Etot(R)] = -Z2/2 Eh;  R = 3/2Z ao (exact match of experiment)
Extension with principle quantum number n produces the H-atom spectrum.

H2+ molecular ion: Variables R, r

Ekin = 9/8R2;  Vne = -3/2R +1/2*(r/R)2/R;  Vnn = 1/2r;
Etot = Ekin + Vne + Vnn;  Min[Etot(R,r)] = -0.7276 (-0.6026) Eh;  R = 1.2435 ao;  2r = 0.829 (1.058)

H2 molecule: Variables R, r

Ekin = 2*9/8R2;  Vne = 2*2*(-3/2R + 1/2*(r/R)2/R);  Vee = 6/5R;  Vnn = 1/2r;
Etot = Ekin + Vne + Vee + Vnn;  Min[Etot(R,r)] = -1.12 = -1.21 (-1.1745) Eh;  R = 1.5/1.1 ao;  2r = R = 0.7216 (0.7414)

Of course, the two H-atoms need to have opposite electron spin for the penetration of their clouds. It is interesting, that Kimballs model gives one stable H2 molecule and an infinite series of metastable "precursors" until both protons are on the fringe of the other H-atom: At right the last metastable form is shown with about half the binding energy of the stable H2, and a H-H distance 3% larger (optical illusion makes it smaller!). This is an artifact of the homogeneous charge distribution of the model. Check the forces! The protons have to actually penetrate the other H-atom for the stable H2 on the left to be produced. This has equal H-H distance and cloud radius with H-clouds fused completely.

Here is an animation of the formation of H2 from two H atoms starting to penetrate (of course with opposite electron spins!) in 41 frames of nonempirical, converged Kimball computations, see H2stat. The two pink points show the binding energy (the attractive part of the potential curve, shown as black points in the right picture) as function of the distance of the H atoms: from far to touching clouds, the atoms have zero interaction according to Kimballs model. With penetration of a ↑-spin and a ↓-spin cloud the energy slowly falls, then rapidly reaches the stationary minimum (scale can be read in H2stat) when each proton enters the other cloud as well. Watch the small proton disks, which move within the open circles of a cloud center but detach from it when entering the other cloud. Pressing the protons together would produce the repulsive part of the potential curve. Make sure to observe the ~10% shrinking cloud size from the atom to the stationary H2 molecule.


HeH+: Variables R, rHe, rH

similar to H2 molecule, but rHe = rH/2, stable, observed, see Tutorial p. 17/18

He and He like ions (including H-): Variable R, Parameter Z

Ekin = 2*9/8R2;  Vne = Z*(-3/R);  Vee = 6/5R;  
Etot = Ekin + Vne + Vee;  Min[Etot(R)] = -(Z-0.4)2 Eh;  R = 1.5/(Z-0.4) ao
For He we get Etot = -2.56 (-2.91) Eh and for its ionization potential 0.56 (0.91) Eh

H3+: Variables R, r; centered equilateral triangle formed spontaneously from dashed start structure

Ekin = 2*9/8R2;  Vne = -6*(3/2 - 1/2*(r/R)2)/R;  Vee = 6/5R;  Vnn = 3^(1/2)/r
Etot = Ekin + Vne + Vee + Vnn;  Min[Etot(R,r)] = -1.6631 (-1.3289) Eh;  R = 1.1631 ao;  r = 0.7687 ao;  d(HH) = 1.3314 (1.70) ao;  
dashed circle is the H atom cloud for comparison, solid circle the contracted result for the field of three protons. Full computation in H3p.html

H3: Variables R1, R2, r; linear H-H-H formed spontaneously, Pauli exclusion obeyed; not stable against -> H2 + H predicted

Ekin = 9/8R12 + 9/4R22;   other terms see in H3linorig.nb, first computed by G.F.Neumark, thesis loc.cit. p.27(1951)
Etot = Ekin + Vne + Vee + Vnn;  Min[Etot(R1,R2,r)] = -1.7090 Eh;  R1 = 1.7256 ao;  R2 = 1.3001 ao;
r = 0.2353 ao;  d(HH) = 1.5354 ao


1) Electron-electron repulsion of pair in same cloud, spins different:
This produces a screening of the nuclear charge for each electron. Assuming equal density and distribution of both electrons over the cloud gives a screening constant of σ = 0.4 , see derivation. Empirically Slater found σ = 0.3. Applying this to:
  • He gives Etot = -2.89 (-2.91) Eh instead of bad -2.56. The IP rises from 0.56 to 0.89 (0.91) Eh. The He+(1s1) energy is -2.00 = -Z2/2 Eh; σ = 0.3 gives good energies for all Z(1s2) He homologues as shown in table as K' model.
  • H- gives Etot = -0.49 (-0.5275) Eh for σ = 0.3. With σ = 0.4 it is -0.36 Eh. Both cases yield an unstable hydride ion in the gas phase because E(H1s) = -0.5 Eh, i.e. more stable than the hydride. The experiment has an EA = 0.0275 Eh. But even the Hartree-Fock limit produces no better result.

  • Hence, the error originates from neglecting correlation energy: The electrons of the pair do not independently move through the whole cloud (space). They prefer being as close to the nuclei as possible but simultaneously avoid each other as much as possible. There is no simple scheme to correct this gross error in Kimball's model nor in Hartree-Fock theory (see any text of QC)! In order to rationalize Slater's value of σ = 0.3 we try to model it:
    a) e1 is mainly in an inner sphere with r1 while e2 is in an equal volume outer spherical shell R-r1 -> σ = 0.37
    b) the average distance of e1 and e2 is <r> = R -> σ = 0.3333 (best try and better than 0.3 for higher shells)
    c) each electron is predominantly near the center of mass of the other halfsphere -> σ = 0.4444; even worse!
    d) apply E. Wigner's correlation hole function: would destroy the simplicity of Kimball's model
    Conclusion: we accept Slater's screening constant for 1s2 and determine others by parametrization with a good (correlation corrected) QC result. This applies to all electron pairs in Kimball calculations and is shown in the examples of the main page.

    2) High bond energy, low distance in H2+, H2, H3+:

    This is an effect of the inability of Kimballs ansatz to model the density cusps at the nuclei, partially relieved by FSGO, floating spherical gaussian orbitals, which destroy the simplicity of the K-model.
  • H2+: Etot = -0.7276 (-0.6026) Eh;  Bond energy 0.2276 (0.1026) Eh;  Bond length 0.829 (1.058) . Energy minimum reached, Virial theorem correct, Hellman-Feynman force equlibrium true! (the Variation Principle does not apply!). These three pillars of the model cannot guarantee quality of the result. In fact H2+ is the worst case of anything computed hitherto with the K-model, a failure.- What can be done: Two variables: R,r, one parameter Z(=1). Easy: Change the "effective" nuclear charge - often used in QC functions - not applicable here: There is no screening constant to adjust! We have no criterion for how much Z=1 should be lowered and thus destabilize the molecular ion. Next: Tweak the kinetic energy, i.e. R, of the bonding electron: With a kinetic energy parameter of k=1.21 we get Etot = -0.6013 (-0.6026) Eh;  Bond energy 0.1013 (0.1026) Eh;  Bond length 1.0032 (1.058) . Not bad! Ingenious: compose two half clouds: Gives what we hope for, and is even pleasing as a more elliptical molecule, but the sins immediate penalty manifests itself by a bad Virial ratio and non vanishing Hellmann-Feynman forces!

    Conclusion: We can save the best case of a didactical introduction of chemical bonding with an adjustment of the kinetic energy of the K-model from 9/8R2 to 1.21*9/8R2. Remember, that Heisenberg's uncertainty relation requires 9/8R2 as a minimum.
  • H2: Etot = -1.21 (-1.1745) Eh;  Bond energy 0.21 (0.1745) Eh;  Bond length 0.7216 (0.7414) . Bond energy +20%, length -2.7% error. Same trend as in H2+, although much smaller. Remedy: 2-electron problem, hence apply He-adjusted screening constant σ = 0.3 ? Wrong! Both errors larger, because missing cusp modelling increased by higher effective charge. Tweak σ to the other side? Yes: σ = 0.4 -> σ = 0.4165 brings both errors down to 0.3%. Rationale: An increase in the screening constant goes in the direction of modelling a cusp near each proton, and that is, how Kimball should be corrected.