Tutorial 5             Kimball Tutorial 6            Tutorial 6a

Content: Chemical structure, 2nd part: What makes a three atomic molecule, like H2O, isosceles and not linear like H-Be-H (g)? Or NH3 pyramidal not like flat BH3 ?
ES 12 Oct 2017//2002//1982

Series CH4 to HF in one cloud

We start a computational experiment, similar to Tutorial5: Given a sphere homogeneously filled with the charge of 10 electrons and their kinetic energy (and accounting for the electron-electron repulsion within the sphere). Now put the nuclei of Ne, HF, H2O, NH3, or CH4, nearby and hopefully build a neutral “molecule”. What is the equilibrium arrangement of the nuclei, i.e. the “shape”, and the size of the sphere?

Further down we do the experiment again, now obeying Pauli’s exclusion principle, i.e. constraining the roaming space of the electrons into 5 spheres occupied by a pair each, and non overlapping:

Input and Definitions

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The cyan initial clouds in these five boxes have all R=0.4 a0 and an assumed central charge of 6.4 au. Note the change of the sizes from start to equilibrium. The highest charged nucleus near the center determines predominantly the final size of the orange mean spheres. Hence, the change is largest for Ne. The red dots “hx” are the initial locations of the protons. The electrostatic forces put them as “Hx” into their final coordinates as small orange spheres. The “heavy” nuclei are started at the origin (black points) but end up eccentrically except Ne, of course.

Do you realize that we do not need any sort of trick, e.g. hybridization, to produce an angular H2O and a pyramidal NH3? This is all accomplished by simple electrostatics as has already been proven ~1949 by theoretical physicist Walter Weizel ("Elektronen, Atome, Moleküle", Berlin, 1949). A linear H2O is in a "metastable state" and less stable than the angular one. A slight deformation, already by zero point vibration, lets it snap into the more stable angular form (check this as an assignment!). Hence, except for the last plot, the structures are superficially looking correctly. But C(+6) plus four H(+) nuclei produce a surprize: The equilibrium structure is a quadratic pyramid not a tetrahedron! Again, that is the correct energy minimum in a homogeneously charged electron sphere. But, perhaps we had better look into the exclusion principle:

Now with Pauli’s exclusion principle

We know from tutorial5 that we have to deal differently with these 5 10 electron problems. Father Pauli commands us to use 5 2-electron clouds to accommodate these molecules with the full nuclear charge and electrical neutrality: the highest charged nucleus goes to the center, while the 4 outer clouds touch the center one and move as far away from each other as the prevailing fields require. They are completing the electron number of Ne.

We show this for Neon and demonstrate three snapshots on the way: 1) start with 5 2-electron clouds in arbitrary positions and size; 2) Bring a Ne(+10) nucleus into the neighborhood and document its penetration to the center of one cloud which shrinks to the Ne(+8) He-like core; 3) show the contraction of the four remaining clouds to their equilibrium size under the influence of the positive core charge of +8, forming a neutral Ne atom with an octet of four doubly occupied clouds.
Here are the computational steps and the snapshots, below:

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Graphics:Equilibrium projected to xy-plane

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Graphics:Equilibrium projected to xz-plane

For CH4 the demonstration is similar: Exchange Ne(+10) for C(+6) and add 4 protons. Run the program with Z=6: The result is qualitatively the same, as shown above with the second triple picture. The smaller nuclear charge just generates a larger core cloud and four larger outer clouds. They form a C(-4) Ne-like anion. Add the protons. They cannot penetrate the core cloud because the C(+6) nucleus pushes them out immediately. There, they are attracted by the outer clouds. Every proton penetrates the cloud nearest to it and lets it shrink to a non-overlapping size while it stabilizes itself off-center, owing to the repulsion by the central C nucleus. Seeking an energy minimum, no proton stays in a cloud already occupied. Hence, each outer cloud gets just one proton. Its final locus is on a radius connecting core-center - outer cloud-center - proton, as you can easily comprehend considering the electrostatic forces. Thus, the protons and the outer, bonding clouds become tetrahedral substituents on the C-core, again, solving the "mystery" of the square pyramid by the commandment of prophet Pauli. Klick methane. Compare F(-), O(-2), N(-3), and C(-4) to Ne, and the following pictorial summary:



This demonstration proves, that the experimentally known structure of methane cannot be built, nor understood, without Pauli’s exclusion principle. It also makes clear, that electrostatics plus Pauli (and electron clouds with zero point kinetic energy) are sufficient to derive the tetrahedral shape. Nothing else is necessary.

For the other 10 electron molecules, NH3, H2O, and HF, the same procedure can be used. It implies that 1,2, and 3  clouds, resp., are empty, i.e. without nucleus and attached to the central core cloud only by the attraction of the heavy nucleus therein. Gilbert Newton Lewis named these empty, single, clouds: Lone Pairs, as you are aware. We'll later deal with empty bonding clouds, "touching" two or more He-like core clouds, e.g. the C-C cloud in ethane.

The existence of two different types of outer clouds, protonated and empty, already breaks the symmetry Td of the tetrahedral shape of the electron clouds of Ne. Assuming that multiple lone pairs are equivalent, Td is reduced to C3v for HF (C∞v for the nuclei alone), C2v for H2O (same as for the nuclei), and C3v for NH3. In Kimball's model different types of electron clouds must have different radii, i.e. size. How do we get information on the size of lone pairs? There are no direct observations available for them, just secondary effects, well known to every chemist e.g. as nucleophilic or "electron-donating" sites (Lewis bases). If they had the same expected radius as the X-H clouds, the angles ∠H-N-H and ∠H-O-H should be the same tetrahedral angle as in CH4, 109°28'16.4'' (excluding the angular deformation, already produced by electrostatics alone!). But ∠H-O-H is 104.52°, ∠H-N-H in ammonia 106.67° (∠e, see Tutorial 6a). This means, that lone pairs indeed have space requirements different from those of the proton occupied clouds. We just look at what a computation of the energy minimum by Kimball's method tells us, when we leave the angles between clouds as free variables to be optimized:

NH3: Here's the step-by-step Kimball computation for NH3.We have 5 variables, three cloud radii, the proton off-center distance in the N-H clouds and one angle, e.g. ∠H-N-LonePair.  

H2O: There's the step-by-step Kimball computation for H2O.We have 6 variables, three cloud radii, the proton off-center distance in the O-H clouds (assuming they are equal), and two angles, e.g. ∠LonePair-O-LonePair, ∠H-O-H.

HF: The HF computation is here. It is similar to NH3, the roles of X-H and lone pairs is just reversed. A straightforward exchange of these roles and of the Z value from 7 to 9 gives you HF with NH3 as template.

In NH3 the ratio of the mean radii of lone pair to N-H cloud is 1.35 to 1.12, in H2O 1.125 to 1.017, and in HF 0.829 to 0.983. Since no tables of lone pair radii exist, we use these radii in other Kimball computations of this site, e.g. in molecules with amino groups like glycine, hydroxy and C-O-C groups in saccharose or -F in organic fluorides like Flunitrazepam.

As a new challenge we ask ourselves: What happens, if we offer an O(+6) core five outer 2-electron clouds instead of the four building the O(-2) anion? Here you see what's up: The fifth electron pair is expelled to infinity - the physical reason of the Octet Principle!


For comparison look at the table of all the gaseous hydrides of the Li...Ne period in simplehydrides.

The general rule, as used for decades in the VSEPR called qualitative discussion of molecular structures, is: Lone pairs (LP) are only in the electric field of one neighbor nucleus X and have, therefore, a larger mean radius than X-H clouds. Hence, they use more “space” on the surface of the small He[1s2] core and force the bonding clouds nearer to each other. For the same reason LP’s repel LP’s stronger than LP-Bonding pair, or BP-BP. This qualitatively explains the decrease of the bonding angles from the tetrahedral value to 105.8° in NH3 (exp 106.67°) and 103.3° in H2O (exp 104.52°). However, in HF the rule breaks down! The LP’s are found smaller than the H-F cloud in the Kimball computation, only we don’t see anything of that, because ∠H-F-LP angles are not observable. Be aware that rules have exceptions! Do not trust rules blindfolded! Note that Kimball’s computations give a bit smaller ∠HNH and ∠HOH than experiment. This is in good company with the QC programs (using small sets of basis functions), however.

Now is the time to look at some critical remarks concerning the concept of the lone pair. Although our arguments and most of our results in the list pertain to the criticized concept of "rabbit ear" lone pairs (Weinhold's paper, but not "rabbit ear Hybrids", o.m.g.!), we just use this representation with Kimball's model. Group-theoretically there is nothing wrong with two equivalent lone pairs in the C2v water molecule (and all doubly substituted O-atoms), otherwise two equivalent O-H bonds would be wrong, too! Frank Jensen, Introduction to Computational Chemistry, 2nd ed., Wiley 2006, e.g., contradicts Weinhold and has shown on p. 304 ff. that the two models, 1) with σ-π lone pairs, and 2) with equivalent rabbit ear lone pairs, are similar to the treatment of the σ-π double bond (Pipek-Mezey procedure, initially Hund and Mulliken) versus localisation into two equivalent banana bonds, using the Edmiston-Ruedenberg or Boys algorithms. We have shown here that the model is easily capable of representing (group-theoretically) unequal lone pairs of the O-atom at the carbonyl group >C=O> and similarly in H2O or in 32-water cluster. However, this complicates Kimball computations with no improvement. We do not want to mimic canonical molecular orbital theory and are of the opinion that todays chemistry texts teach many concepts that need "unteaching" for those who want to understand modern theoretical chemistry. Some of these concepts (not only rabbit ear lone pairs) are hybridization (qualitative use), resonance, ringstrain, and the qualitative use of atomic orbitals to "construct" molecular orbitals. In a later tutorial we shall come back to this question, already begun here.

More to explore: As an adept of chemistry you certainly have noticed that you now have learned how to compute some active species:  BH4(-), CH4, NH4(+) can all be computed with the template for CH4 by just changing the value of the Z. The same is true for CH3(-), NH3, OH3(+); NH2(-), H2O, H2F(+); and OH(-), HF, NeH(+), yes even this very strong Brønsted acid exists, as well as the even stronger HeH(+). Most of the ions mentioned exist well characterized in salts, like {Li[BH4]}, {Mg(CH3)2}, {NaNH2}, {[H3O][ClO4]}, or {[H2F][BF4]}.

I have not gone here into the discussion of the first three gaseous hydrides of the series: LiH, BeH2, BH3. They are all stable molecules but very reactive and not items to be bought in the chemistry stockroom. We have amply discussed LiH, mentioned BeH2 which also condenses spontaneously into solid salts. BH3 dimerizes to B2H6 which you can buy in closed glas containers or as solutions. All this can be explained and computed with the Kimball method and is shown here. A computation similar to that at the beginning of this tutorial6 is shown for LiH, BeH2, BH3 here: Now we have to explain the angular BeH2 and the pyramidal BH3 being the optimal structures by electrostatics alone! But, Pauli's principle puts this in order with a linear BeH2 and a flat BH3 as known from experiment.

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