Bent's Rule states that $s$ orbitals on a central atom participate contribute more to molecular orbitals directed towards more electropositive ligands. (For a longer explanation, see this question and answers: What is Bent's rule?). Lone pairs are considered to be the extreme limit of an electropositive "ligand".

Many sources claim that this Rule is consistent with LCAO MO theory representations of electron density. It seems, however, that this claim is easily disproved by considering the example of simple molecules of the type $\ce{AH2}$.

We can start with $\ce{H2O}$, whose LCAO MO diagram is well known:

MO Scheme of Water
Source: H2O-MO-Diagram.svg by Officer781 on Wikimedia Commons

The orbitals contributing to the O-H bonds are the lower two in energy. Ignoring orbital mixing for now, these are formed from linear combinations of the hydrogen $s$ orbitals with the oxygen $2s$ (giving $\mathrm{2a_1}$) and $2p_y$ orbitals (giving $\mathrm{1b_2}$). The lone pairs are a pure $p_x$ orbital ($\mathrm{1b_1}$) and a combination of $p_z$ with a small contribution from the hydrogen $s$ orbitals ($\mathrm{3a_1}$). Based on the contributing oxygen atomic orbitals, that gives 100% p lone pairs and $50\%$ $p$/$50\%$ $s$ for the bonds, the opposite of the result predicted by Bent's rule, which is that the lone pairs should have more $s$ character.

If we properly account for orbital mixing, there is no effect on the $\mathrm{1b_1}$ lone pair orbital, while the $\mathrm{3a_1}$ lone pair does in fact get more $s$ character, but this increased $s$ character is not expected to exceed $50\%$, since that would just mean a switch of label between the $\mathrm{2a_1}$ and $\mathrm{3a_1}$ orbitals. $50\%$ $s$ in the $\mathrm{3a_1}$ represents the maximum mixed case. That gives a final result of a maximum of $25\%$ $s$ in the lone pairs ($50\%$ in the $\mathrm{3a_1}$ and $0\%$ in $\mathrm{1b_2}$) and $25\%$ in the $\ce{O-H}$ bonds ($\mathrm{2a_1}$ and $\mathrm{1b_2}$), still contradicting Bent's Rule, which predicts that the lone pairs should have more $s$ than the bonds, not the same amount. [One a side note, we know that $\ce{H2O}$ does not represent the maximum mixing case, as $\ce{SH2}$ has a smaller bond angle consistent with even greater orbital mixing, but that's beside the point here.]

The paper by Clauss et. al.[1] reconciles this result by claiming that the 1b1 orbital is "always excluded from the Bent's rule competition for in-plane p-character", but I've never seen any formulation of Bent's Rule that indicates such an exception. If there is such a qualification to Bent's Rule, how do we know what lone pairs to include or exclude?

I conclude from this analysis that Bent's Rule is much like VSEPR - useful in some cases as a pedagogical tool, but ultimately inconsistent with LCAO MO theory. Or perhaps it's only consistent when applied to actual bonds rather than to bonds and lone pairs? I'm not concerned with why people still use it, but I am confused as to why so many people argue that it is consistent with LCAO MO theory. Am I missing something here?


  1. Clauss, A. D.; Nelsen, S. F.; Ayoub, M.; Moore, J. W.; Landis, C. R.; Weinhold, F. Rabbit-ears hybrids, VSEPR sterics, and other orbital anachronisms. Chem. Educ. Res. Pract. 2014, 15 (4), 417–434. DOI: 10.1039/C4RP00057A. Or as a free preprint

2 Answers 2


Beyond the educational value, Bent's rule has had significant impact on how we understand wave functions of molecules. It's simplicity is almost unique in the world of quantum chemistry; almost everybody can immediately understand it and apply it. Together with the VSEPR model it can lead to quite accurate predictions on the back of an envelope.

Molecular Orbital Theory vs. Valence Bond Theory (MO vs. VB)

In their respective infinite treatment these theories are congruent. For more on this topic please read What is actually the difference between valence bond theory and molecular orbital theory?

Important for this post is that both theories are approximations for the wave function of a molecule, hence they do not describe bonding beyond the stationary point they are applied to, and they operate within the clamped nuclei (Born-Oppenheimer) approximation.

In MO theory it is common to express the molecular orbitals as linear combinations of atomic orbitals. Those MO are usually chosen to be orthonormal. The MO themselves are delocalised, but there are algorithms to transform them into more intuitive VB type orbitals. This will provide us with a Lewis-like bonding picture. Famously used in this regard is Natural Bond Orbital (NBO) theory.

Bent's rule

Unfortunately I cannot find my copy of Bent's paper, so this answer will be based purely on the definition of Bent's rule.

The IUPAC gold book defines Bent's rule (DOI: 10.1351/goldbook.BT07000):

In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

Water is a good example to demonstrate Bent's rule and its interplay with the VSEPR model. In $\ce{H2O}$ you expect oxygen to follow the octet rule, hence there will be four electron pairs around it, and it predicts a generally tetrahedral structure of the molecule. The angle for these shapes is about $\newcommand{\degree}{^\circ}109\degree$. Since hydrogen is more electronegative than no ligand, Bent's rule predicts a smaller angle than that.

Probably due to its simplicity, Bent's rule can be applied within the MO and VB bonding pictures.

Coulson's Theorem

Coulson's theorem is the formal theory which applies to Bent's rule (see Wikipedia).

We choose a set of orthonormal AO, as would be consistent with the LCAO approximation. $$ \langle\chi_i|\chi_j\rangle = S_{ij} = \delta_{ij} \begin{cases} 0, & i \neq j\\ 1, & i = j \end{cases}\tag{1}\label{orthonormalAO} $$

We can transform these orbitals into sets of orthonormal hybrid orbitals. \begin{align} \varphi_a &= \sum_i \lambda_i \chi_i & \langle\varphi_a|\varphi_b\rangle = S'_{ab} &= \delta_{ab} \begin{cases} 0, & a \neq b\\ 1, & b = b \end{cases}\tag{2}\label{orthonormalHybridAO} \end{align}

Let's look at a special case, where we choose atomic orbitals to be $\chi_i$ with $i \in \mathrm{s}, \mathrm{p}_x, \mathrm{p}_y, \mathrm{p}_z$. We can construct following hybrid orbitals (also compare Mathematical form of four hybrid orbitals): \begin{align}\tag{3}\label{orthonormalSP3} \varphi_{xyz} &= \lambda_\mathrm{s}\chi_\mathrm{s} + \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x} + \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y} + \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\ \varphi_x &= \lambda_\mathrm{s}\chi_\mathrm{s} + \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x} - \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y} - \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\ \varphi_y &= \lambda_\mathrm{s}\chi_\mathrm{s} - \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x} + \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y} - \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z}\\ \varphi_z &= \lambda_\mathrm{s}\chi_\mathrm{s} - \lambda_{\mathrm{p}_x} \chi_{\mathrm{p}_x} - \lambda_{\mathrm{p}_y} \chi_{\mathrm{p}_y} + \lambda_{\mathrm{p}_z} \chi_{\mathrm{p}_z} \end{align}

If all $\lambda_i$ are chosen to be $\frac{1}{2}$, then we would come to the famous $\mathrm{sp}^3$ hybrid orbitals.

Let's generalise this a bit and drop the normalisation: \begin{align}\tag{4}\label{hybrid} \varphi_a &= N\cdot(\chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi_{\mathrm{p}_k}) & \leadsto \varphi_a &= \chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi'_{\mathrm{p}_k} \end{align}

From \eqref{orthonormalHybridAO} and \eqref{hybrid}, and ignoring the trivial case: \begin{align}\tag5 \delta_{ab} &= \langle\varphi_a|\varphi_b\rangle\\ 0 &= \langle \chi_\mathrm{s} + \lambda_{\mathrm{p}_k} \chi'_{\mathrm{p}_k} | \chi_\mathrm{s} + \lambda_{\mathrm{p}_l} \chi'_{\mathrm{p}_l} \rangle\\ &= \langle\chi_\mathrm{s}|\chi_\mathrm{s}\rangle + \lambda_{\mathrm{p}_k} \langle\chi_\mathrm{s}|\chi'_{\mathrm{p}_k}\rangle + \lambda_{\mathrm{p}_l} \langle\chi_\mathrm{s}|\chi'_{\mathrm{p}_l}\rangle + \lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l} \langle\chi'_{\mathrm{p}_k}|\chi'_{\mathrm{p}_l}\rangle \end{align}

Given \eqref{orthonormalAO} we can simplify this. Note that $\chi'$ is actually a linear combination of $\mathrm{p}$ orbitals, and therefore they don't necessarily have to be orthogonal. The angle between two such orbitals is given by the inner product, i.e. $\langle\chi'_{\mathrm{p}_k}|\chi'_{\mathrm{p}_l}\rangle = \cos \theta_{kl}$. \begin{align}\tag6 0 &= 1 + 0 + 0 + \lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l} \cos \theta_{kl} \end{align} Therefore Coulson's Theorem is \begin{align}\tag{7}\label{Coulson} \cos \theta_{kl} &= \frac{-1}{\lambda_{\mathrm{p}_k}\lambda_{\mathrm{p}_l}} \end{align}

If you call $\lambda$ the hybridisation index, then you see that the angle between two hybrid orbitals is dependent on the this index. In other words, the angle between two ligands around a central atom determines the $\mathrm{p}$ character of the hybrid orbitals that ideally describe this bond.

What you can also derive from this formula is, given a constant angle $\theta_{kl}$, if you use a hybrid-orbital with increased $\mathrm{p}$ character, the other hybrid-orbital must have less $\mathrm{p}$ character, i.e. $\lambda_{\mathrm{p}_k} < \lambda_{\mathrm{p}_l}$.

This is in principle all that Bent's rule requires in the framework of LCAO-MO theory to be consistent; the argument about the electronegativity of a ligand is based on observation of molecular structure.

I hope the above answers the titular question sufficiently. In the second part I'll try to show where the original arguments fail.

First of all, you are basing your argumentation on even more handwavy mathematics then I did above. You are also not considering the actual molecular structure and the implications of that in terms of the LCAO-MO approximation of the wave function.

The orbitals contributing to the O-H bonds are the lower two in energy.

This is not quite correct. In LCAO-MO (or MO in general) theory all orbitals contribute to bonding.

Ignoring orbital mixing for now, these are formed from linear combinations of the hydrogen s orbitals with the oxygen 2s (giving 2a1) and 2py orbitals (giving 1b2). The lone pairs are a pure px orbital (1b1) and a combination of pz with a small contribution from the hydrogen s orbitals (3a1).

The shown MO scheme does not ignore orbital mixing, and neither do you in your argumentation. If the oxygen's $\mathrm{s}$ and $\mathrm{p}$ orbitals were far enough apart to not mix, then only one sort of these orbitals could interact with the hydrogen orbitals to lead to bonding. This would either lead to a situation where there is no oxygen $\mathrm{s}$-character in the bonds, or all of it.
This basically follows the argument that for the heavier homologues of water the oxygen valence $\mathrm{s}$-orbital becomes the lone pair.

Based on the contributing oxygen atomic orbitals, that gives 100% p lone pairs and 50% p/50% s for the bonds, the opposite of the result predicted by Bent's rule, which is that the lone pairs should have more s character.

I don't have an optimal LCAO-MO solution at hand, but I have the next best thing: a HF/STO-3G wave function with $C_\mathrm{2v}$ symmetry, $\angle(\ce{HOH}) = 100.0\degree$, and $d(\ce{OH}) = \pu{98.9 pm}$. That is off by plenty, but it's Hartree-Fock, so that was expected. The general tenor wouldn't change, if you were to run it on the experimental structure. This is the output of the wave function:

 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-20.252 is O1-s=1.0006
 Alpha occ 2 OE=-1.258 is O1-s=0.7899 H2-s=0.0928 H3-s=0.0928 O1-p=0.0245
 Alpha occ 3 OE=-0.594 is O1-p=0.5365 H3-s=0.2317 H2-s=0.2317
 Alpha occ 4 OE=-0.460 is O1-p=0.6808 O1-s=0.1328 H2-s=0.0932 H3-s=0.0932
 Alpha occ 5 OE=-0.393 is O1-p=1.0000
 Alpha vir 6 OE=0.582 is H3-s=0.3144 H2-s=0.3144 O1-p=0.2947 O1-s=0.0766
 Alpha vir 7 OE=0.693 is O1-p=0.4635 H2-s=0.2683 H3-s=0.2683

Unfortunately this doesn't tell us much, apart from that there is obviously one $\mathrm{p}$ lone pair, which was expected due to symmetry. (In MO theory though, this would still count as a bonding π orbital).
From the above table, one would derive that the majority of bonding comes from MO 3, which is a nearly pure oxygen $\mathrm{p}$ orbital.

I've run an NBO6 analysis on this calculation, which is basically a unitary transformation of the MO, and arrived at the following description:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   1. (2.00000) CR ( 1) O  1            s(100.00%)
                                         1.0000  0.0000  0.0000  0.0000  0.0000
   2. (2.00000) LP ( 1) O  1            s( 71.29%)p 0.40( 28.71%)
                                         0.0000  0.8443  0.0000  0.0000  0.5358
   3. (2.00000) LP ( 2) O  1            s(  0.00%)p 1.00(100.00%)
                                         0.0000  0.0000  1.0000  0.0000  0.0000
   4. (1.99926) BD ( 1) O  1- H  2
               ( 59.24%)   0.7697* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000  0.3789  0.0000 -0.7071 -0.5970
               ( 40.76%)   0.6384* H  2 s(100.00%)
   5. (1.99926) BD ( 1) O  1- H  3
               ( 59.24%)   0.7697* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000  0.3789  0.0000  0.7071 -0.5970
               ( 40.76%)   0.6384* H  3 s(100.00%)
 ---------------- non-Lewis ----------------------------------------------------
   6. (0.00074) BD*( 1) O  1- H  2
               ( 40.76%)   0.6384* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000 -0.3789  0.0000  0.7071  0.5970
               ( 59.24%)  -0.7697* H  2 s(100.00%)
   7. (0.00074) BD*( 1) O  1- H  3
               ( 40.76%)   0.6384* O  1 s( 14.36%)p 5.97( 85.64%)
                                         0.0000 -0.3789  0.0000 -0.7071  0.5970
               ( 59.24%)  -0.7697* H  3 s(100.00%)

This results in two $\mathrm{sp}^6$ bonding orbitals, one pure $\mathrm{p}$ lone pair and a $\mathrm{sp}^{0.4}$ lone pair. From this we would conclude that more $\mathrm{p}$ character is directed towards the hydrogen ligands.

If we properly account for orbital mixing, there is no effect on the 1b1 lone pair orbital, [...]

That is correct, this is due to symmetry.

[...] while the 3a1 lone pair does in fact get more s character, [...]

In your argumentation 3a1 would have been a pure $\mathrm{p}$ lone pair, with mixing it does gain $\mathrm{s}$ character, which is obviously more than none.

[...] but this increased s character is not expected to exceed 50%, since that would just mean a switch of label between the 2a1 and 3a1 orbitals. 50% s in the 3a1 represents the maximum mixed case.

The part with the switch is only true if the energy of these orbitals also switches and the contribution of the hydrogen were equal in both orbitals.
Does this argument also imply that an $\mathrm{sp}$ orbital was the maximum mixed orbital?
I'm very sorry, but I cannot follow the logic as I do not understand what is meant with maximum mixed case. In LCAO you are able to continuously mix $\mathrm{s}$ and $\mathrm{p}$ orbitals, if that actually produces a sensible approximation to the wave function is a completely different topic.

That gives a final result of a maximum of 25% s in the lone pairs (50% in the 3a1 and 0% in 2b1) and 25% in the O-H bonds (2a1 and 1b2), still contradicting Bent's Rule, which predicts that the lone pairs should have more s than the bonds, not the same amount.

This would be exactly the result which you were to expect from tetrahedral coordination: two $\mathrm{sp}^3$ bonds, one $\mathrm{sp}$ and one $\mathrm{p}$ lone pair.

[One a side note, we know that H2O does not represent the maximum mixing case, as SH2 has a smaller bond angle consistent with even greater orbital mixing, but that's beside the point here.]

I am really interested in that side note. I still want to understand what is meant by the maximum mixing case, and especially how a smaller bond angle is consistent with more mixing.

  • $\begingroup$ Thank you. This is what I was looking for. I think one issue is that I am using short-hand "s-p" mixing when it is of course MO's mixing, not AO's, so I need to think through the effect of the relative contribution of the H AO's. But if you are correct, would you say then that Albright et al. are wrong when they say that describing the bonding orbitals in SH2 as nearly pure p is nonsensical? (Their exact words are "It is clear from the shape of 1a1 and 2a1 that this is far removed from reality.") $\endgroup$
    – Andrew
    Aug 28, 2020 at 20:16
  • $\begingroup$ Thinking more, I don't see any reason why I can't pre-mix the s and p orbitals and combine those pre-mixed orbitals with the H orbitals. In that case, the contribution of the H orbitals to the MO should be a function of the energy of the s-p mixed orbital, so it should make no difference whether I start with an unmixed 2a1 orbital and increase p from 0 to 50 or if I start with an unmixed 3a1 and decrease p from 100% to 50%. The contribution of the H orbitals will adjust accordingly, and I will end up in the same place. Mixing any further than 50% sends me back along the path towards the start $\endgroup$
    – Andrew
    Aug 28, 2020 at 21:09
  • $\begingroup$ Orbitals aren't real, but besides that, in $\ce{SH2}$ 1a1 and 2a1 should be core orbitals of sulfur, so I'm not sure where the argument was heading. Apart from that, in MO theory you can't really differentiate between bond orbitals and lone pairs. In that line of reasoning, yes, the bonding does need all valence orbitals from sulfur. If you follow the localised bond picture, i.e. with NBO, then the lone pairs will be (1) pure p and (2) 80% s character, leading to the conclusion that the majority of bonding is from the p orbitals of sulfur (about sp⁸ iirc). $\endgroup$ Aug 29, 2020 at 12:42
  • $\begingroup$ It doesn't matter how you set up the basis functions you put in the MO formalism, if they can be transformed into each other, the solution will be the same. However, AO typically will be eigenfunctions of the hamiltonian of an atom, while linear combinations thereof won't. Only eigenfunctions will have eigenvectors, or energies. In MO theory, the MO we are looking at are eigenfunctions, they have eigenvalues; any linear combination of those loses this quality. The overall expectation value for the energy will stay the same. The rest is interpretation, which is challenging. $\endgroup$ Aug 29, 2020 at 12:53
  • $\begingroup$ I don't if you saw that I updated my answer with explanation of reconciliation and where I believe I was going wrong. The key for me was finding in Levine a representation of the localized bonding orbitals and lone pairs as linear combinations of the delocalized orbitals, which helped make the connection between the two clear. The second case of H2S would be an interesting further discussion, but not necessary for the explanation here.And last, it does seem that Albright et al have made the same error I did in extending the bond and nonbond nature of the delocalized MOs to localized MOs $\endgroup$
    – Andrew
    Aug 29, 2020 at 14:38

tl;dr Bent's Rule should be relegated to the dustbin alongside Pauling's claim that $\ce{H2S}$ and $\ce{PH3}$ bond with nearly pure p orbitals.

Corrected My analysis erred in assuming that the nonbonding 3a1 orbital contributes primarily to the lone pairs in a bonding orbital analysis and that the bonding 2a1 orbital contributes primarily to O-H bond orbitals.

The above statement is deliberately provocative in the hope that someone with a different perspective will chime in. In the meantime, here's my understanding.

see update at end

My first thought was that Bent's Rule was never intended to apply to lone pairs, but in Bent's original paper (Bent, H.A. (1961) Chem. Rev. 61:3, 275-311), the very first sentence after he states his rule is "Lone pair electrons are regarded as electrons in bonds to very electropositive atoms," so there is no doubt that he felt the Rule applied equally well to lone pairs.

A thorough reading of the paper revealed that his reasoning was flawed in essentially the same way as Pauling's reasoning regarding $\ce{H2S}$, which is not surprising given that Bent drew heavily on Pauling's ideas regarding hybrid orbitals. The fallacy in both their reasoning can be summed up in two related errors:

  1. They failed to account for the energy of stabilization when a bond is formed and
  2. They did not factor in the electrons contributed by the hydrogen atoms.

Using $\ce{H2O}$ as an example since that is discussed in the original question, Bent and Pauling would start with a neutral oxygen atom with valence shell electron configuration $(2s)^2(2p_1)^2(2p_2)^1(2p_3)^1$. [I am using numbers to index the p orbitals so as not to suggest any bias towards a specific Cartesian axis.]

In order to form $\ce{H2O}$, they would describe the oxygen atom as rearranging to four hybrid $sp^3$ orbitals, two of which are completely filled:

$(2s)^2(2p_1)^2(2p_2)^1(2p_3)^1\rightarrow (sp^3_1)^2(sp^3_2)^2(sp^3_3)^1(sp^3_4)^1$

Looking at these two electron configurations, they make the following logical argument.

Premise: The energy levels of the $sp^3$ orbitals are such that the total energy of four filled $sp^3$ orbitals is the same as a filled s orbital and three filled p orbitals.

$\implies$ Because two electrons have increased in energy from $s$ to $sp^3$ but only four have come down from $p$ to $sp^3$, the state $(sp^3_1)^2(sp^3_2)^2(sp^3_3)^1(sp^3_4)^1$ is higher in energy than the state $(2s)^2(2p_1)^2(2p_2)^1(2p_3)^1$.

$\implies$ The lowest energy state for the $\ce{H2O}$ molecule is when the oxygen adopts a configuration as close as possible to $(2s)^2(2p_1)^2(2p_2)^1(2p_3)^1$, i.e. with lone pairs in a pure $s$ and a pure $p$ orbital and with bonding to hydrogen involving only $p$ orbitals (creating an H-O-H bond angle of 90 degrees).

They argue that the only factor preventing $\ce{H2O}$ from achieving this configuration and having a 90 degree bond angle is repulsion between the partially positively charged H atoms. As a result, it adopts a configuration in which the lone pairs only have an increased s character rather than completely s:


On the larger $\ce{H2S}$ molecule, there is more space between the H atoms so they can get much closer to the ideal 90 degree angle which (to Pauling and Bent) represented bonding with pure p orbitals.

You will hopefully notice right away that the above analysis does not allow for any effect of bonding on the electron energy levels, nor does it consider the electrons provided by the H atoms at all. Therein lies the problem.

Let us first consider the effect of bond formation. As every introductory presentation of qualitative MO diagrams informs us, a bonding orbital is lower in energy than either of the participating atomic orbitals, even in a valence bond treatment.

Looking again at the hybridized O atom, we need to form $\sigma$ bonding orbitals with the H atom 1s orbitals, leaving the lone pairs as non-bonding orbitals (nb):

$(sp^3_1)^2(sp^3_2)^2(sp^3_3)^1(sp^3_4)^1 + (1s)^1 + (1s)^1 \rightarrow (\sigma_{OH})^2(\sigma_{OH})^2(sp^3_{nb})^2(sp^3_{nb})^2$

Furthermore, because of the stabilization energy of bonds, we can conclude that the $\sigma_{OH}$ orbitals are lower in energy than the $sp^3_{nb}$ orbitals. Already, we have a contradiction with the Bent/Pauling model, since our lone pairs are now in the higher energy orbitals rather than the lower energy. Additionally, we have four electrons that have been stabilized in lower energy orbitals, because we are including the electrons from the H atoms.

Thus, there is no longer any basis for arguing that the s electrons are more likely to be found in lone pair orbitals or, by extension, in bonding orbitals to more electropositive elements.

Extending from this valence bond analysis to a delocalized LCAO treatment, we see again that the lone pairs are found in preferentially p orbitals, rather than s, and that the s contribution to the lone pair orbital only comes about because of the orbital mixing, which cannot occur to such an extent that the bonding orbitals have more p character than the lone pairs.


Thanks to some prompting from Martin, I dug a little deeper, and I believe I have found my error, which is possibly a source of confusion to others. All of the quantitative results are taken from Chapter 15 of Levine's Quantum Chemistry text and are typically from computational work in the 1970's. As such, the numbers are not as accurate as what one would get from a modern computation, but seem to be close enough to be conceptually useful.

First, to address the questions about about degree of mixing. Levine provides coefficients for the contributions of each atomic orbital to the occupied delocalized MO's of $\ce{H2O}$. If we look only at the contribution from oxygen and convert the coefficients into percentage, we find that the 2a1 orbital is 97% s and 3% p, while the 3a1 is 30% s and 70 % p, confirming that there is not substantial loss of s from the bonding 2a1 orbital nor a more than 50% gain in s in the nonbonding 3a1 orbital. That the values do not add to 100% s and 100% p between them is due to differing contributions of the H orbitals and to the fact that the s and p orbitals are also mixed in the unoccupied 4a1 orbital.

The problem, as Martin suggested, is in the description of 3a1 as "non-bonding". This statement is based on the fact that there is very little orbital overlap between the H-centered orbitals and the O-centered portions of the orbital, which is supported by approaches such as Mulliken population analysis, which gives overlap populations of 0.53 for 2a1, 0.50 for 1b2 and -0.03 for 3a1.

Because of the described nonbonding nature of 3a1, I mistakenly assumed that it would contributed almost exclusively to a lone pair orbital in a bond orbital analysis, and likewise that 1a1 would contribute primarily to a bond orbital. Therein lies the error.

Although 3a1 is a non-bonding orbital, it achieves that by mixing a somewhat strongly bonding interaction (Opz with H1s) with a fairly strongly anti-bonding interaction (O2s with oppositely phased H1s).

In the bond orbital analysis, these two competing contributions are separated, and the pz bonding interaction is allowed to contribute significantly to the O-H bond orbitals. The s antibonding interaction acts to negate the bonding contribution of the s bonding interaction that comprised 2a1, such that the net result (as reported by Levine) is that the O-H bonding orbital is 85% p/15% s and the lone pairs are 62%p/38%s, supporting the Bent's rule statement that the lone pairs should have more s character.

A helpful analysis is provided in the construction of the bond orbitals by linear combination of the delocalized MOs, which shows that 3a1 contributes 18% of each O-H bonding orbital and 32% to each lone pair orbital, while 2a1 contributes 32% of each bonding orbital and 17% of each lone pair. Note that the distribution in both cases matches closely to the s/p distribution of 3a1, as we expect if the p interaction of 3a1 is bonding and the s interaction is anti-bonding (ie contributing to lone pairs).

  • $\begingroup$ I'm not sure I can follow this, and I cannot find my copy of Bent's review to read it again. In any case, here is what I remember: Bent's observation (later: rule) was rooted in or derived from (ab-initio) Valence Bond Theory. This is one possibility of constructing the wave function. Like in Molecular Orbital Theory, energy gain due to bonding cannot be included in this description (it can be approximated from considering different states though, which you can also do for VB). Since MOT is congruent to VBT at their respective full treatment, Bent's rule must be consistent with LCAO-MO. $\endgroup$ Aug 17, 2020 at 14:16
  • $\begingroup$ @Martin-マーチン- I'm afraid your memory might be letting you down here. First, Bent himself used the word "rule" and formulated the exact one-sentence wording still used today. Second, he did not use ab initio VBT at all, nor did he derive his rule from theory. In his words, "Mathematical models of molecular structure. . .are not considered in this review." He used VBT only at the descriptive level, eg single bonds means sp3, and a double bond means sp2. Otherwise, he simply conjectured based on experimental data on bond angles, lengths, dipole moments and electronegativities. $\endgroup$
    – Andrew
    Aug 17, 2020 at 17:36

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