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General Chemistry perspective: Looking at the molecular orbitals of water, we can see that the oxygen is $sp^3$ hybridized. Oxygen forms two sigma bonds with hydrogens, and there are two lone pair orbitals in the molecule. There are five occupied molecular orbitals: The core $1s$ of oxygen, and the four $sp^3$ orbitals, out of which two are lone pair orbitals.

Group Theory and Quantum Chemistry perspective: Water belongs to $C_{2v}$ point group. There are four irreducible representations, and the atomic orbitals belong to these irreducible representations (or linear combinations of these irreducible representations). Hence, we get five occupied molecular orbitals of symmetries $a_1, b_1,$ and $b_2$.

MO picture of water

MO structures

There is only one non-bonding orbital ($1b_1$) in which the AOs of hydrogens do not participate. Where is the second lone pair orbital of water?

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General Chemistry perspective: Looking at the molecular orbitals of water, we can see that the oxygen is $\mathrm{sp}^3$ hybridized.

No. One could say that, but $\mathrm{sp}^2$ is equally possible. In fact, it may seem more likely, we’ll get to that in a second.

Group Theory and Quantum Chemistry perspective: Water belongs to $C_{2\mathrm{v}}$ point group. There are four irreducible representations, and the atomic orbitals belong to these irreducible representations (or linear combinations of these irreducible representations). Hence, we get five occupied molecular orbitals of symmetries $\mathrm{a}_1, \mathrm{b}_1$, and $\mathrm{b}_2$.

Yes, and you did a nice job of drawing them.

There is only one non-bonding orbital ($1\mathrm{b}_1$) in which the AOs of hydrogens do not participate. Where is the second lone pair orbital of water?

And this is where it gets interesting.

I’ll once again point to Professor Klüfers’ web scriptum for the basic and inorganic chemistry course in Munich, section about localising molecular orbitals. If you don’t understand German, all you need to do is look at the pictures and understand that wenig means a little in this context.

You see, to do a discussion such as ‘this is the bonding electron pair’ we need to localise molecular orbitals. Specifically by linear combining them. It’s almost like in the linked pictures: $\Psi_2 (1\mathrm{b}_2) + \Psi_3 (3\mathrm{a}_1)- \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_1$. Invert the signs: $\Psi_2 (1\mathrm{b}_2) - \Psi_3 (3\mathrm{a}_1) + \mathrm{some~} \Psi_1 (2\mathrm{a}_1) = \unicode[Times]{x3C3}_2$. These are your two bonding orbitals to hydrogen. We have some $\Psi_1$ remaining (and, theoretically, also some $\Psi_2$ and $\Psi_3$, since we didn’t use all of them). That (remaining linear combination) is our second non-bonding orbital.


Now why is $\mathrm{sp}^3$ a bad description? You see, we arrived at one lone pair that is definitely $\unicode[Times]{x3C0}$-shaped or p-shaped. The second is somewhat p-ish but also somewhat s-ish, so maybe some sp-hybrid. But there is no mistaking that $1\mathrm{b}_1$ is in no way influenced by all the others. It is antisymmetric to the $\ce{H-O-H}$-plain of symmetry while all the others are symmetric. $\mathrm{sp}^2$ describes this a lot better.

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  • $\begingroup$ How does one rectify $sp^2$ hybridization of the oxygen center with the H-O-H bond angle being much closer to tetrahedral? $\endgroup$
    – Blaise
    May 15 '19 at 7:58
  • $\begingroup$ @Blaise From an angle point of view, both are wrong and the actual situation is more of a ‘p with a little s mixed into the in-plane orbitals’. (Bond angle 104.5° < 109.5°, not larger as tetrahedral or $\mathrm{sp^2}$ would imply.) But that is a higher level of analysis. What I want to make clear is that there is an almost pure p type nonbonding orbital and only two p and one s orbital of oxygen take part in hybridisation at all. Hence ‘$\mathrm{sp^2}$’. $\endgroup$
    – Jan
    May 17 '19 at 2:51
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    $\begingroup$ I understand where you're getting the sp2 from, but that's essentially assuming that the second lone pair and the O-H bonds are equivalent (so each is sp2). But they aren't. Based on calculated s and p content of localized bond orbitals, the O-H bonds are approximately sp4. The combined lone pair electron density can be described as one p and the other something like sp(2/3) or as two equivalent ~sp(8/3) hybrids. So with respect just to the O-H bonds, the general chem "sp3" is closer to reality than your description of sp2. $\endgroup$
    – Andrew
    Jan 19 at 13:42
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    $\begingroup$ Hmm, this is a bit of a contentious issue: pubs.rsc.org/en/content/articlelanding/2015/rp/c5rp00061k As long as we limit ourselves to the gen chem sp^n models, then the question IMO isn't really which is better, it's more like which one out of a set of flawed models is least flawed. $\endgroup$
    – orthocresol
    Jan 19 at 18:39
  • $\begingroup$ @Andrew In the end, the closer we look, the less useful the simple $\mathrm{sp}^n$ (where $n\le 3$ and a whole number) becomes. The bond angle (104.5°) suggests a higher p content than both $\mathrm{sp^2}$ (120°) and $\mathrm{sp^3}$ (109.5°) – an estimation of $\mathrm{sp^4}$ might be good but I’m too lazy to look up and do the calculations. But the p-shape of one lone pair is obvious. General, especially introductory chem needs to strike this balance between an accurate and an easy to understand explanation. I feel that the white lie of $\mathrm{sp^2}$ is not too wrong for this area. $\endgroup$
    – Jan
    Jan 21 at 7:10
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Although this is an old question with an accepted answer, I think it's worth providing a second perspective rather than continuing extended discussion in the comments.

First off, let's consider the purpose of even considering hybridization. The purpose of constructing hydrid orbitals is to create localized bonding orbitals and to see how they can be formed as linear combinations of atomic orbitals. By determining the contribution of different s and p orbitals to the hybrids, we can draw some conclusions about the bonding in a molecule.

Since delocalized orbitals such as those shown in the question are also constructed from linear combinations of atomic orbitals, one approach (as described in the other answer) is to construct our localized bonding orbitals by linear combinations of the delocalized MOs. As shown in the other answer, we can create two equivalent localized O–H bonding orbitals by the combinations

$$\Psi_2(\mathrm{1b_2}) + \Psi_3(\mathrm{3a_1}) − \text{some } \Psi_1(\mathrm{2a_1}) = \sigma_1$$

and

$$\Psi_2(\mathrm{1b_2}) - \Psi_3(\mathrm{3a_1}) + \text{some } \Psi_1(\mathrm{2a_1}) = \sigma_2$$

Since the question is about the nonbonded orbitals, however, we need to complete the process and create two equivalent localized non-bonding orbitals. These necessarily are comprised of the MOs we didn't use to make the bonding orbitals.

That means combining $\Psi_4(\mathrm{1b_1})$ [pure p] with the remaining $\Psi_1$ [mostly s] and a little bit of $\Psi_2$ [mostly p] and $\Psi_3$ [mixed s and p], since we didn’t use all of them. The result is two orbitals that are both a mix of s and p.

The problem with this approach is that it's hard to deduce the quantitative contribution of the atomic orbitals to either the localized bonding or the localized non-bonding orbitals without extensive calculations, so we don't know how well this picture matches to the general chem "water has $sp^3$ hybrid orbitals" description. I've been misled myself trying to draw conclusions from this approach (as evidence by some wrong answers I've written on this site).

Instead, it's much simpler to build the localize hybrid orbitals from atomic orbitals independently of the delocalized orbitals.

When doing so, Coulson's Theorem is very helpful for translating a bond angle into a degree of hybridization. For two equivalent bonds (as the O-H bonds here are expected to be) with a bond angle $\theta$, the hybridization of the localized bond orbital is given by $sp^\lambda$ where $$\lambda = -\frac{1}{\cos\theta}.$$

Applying this to the known bond angle of water, we find that the O–H bond orbitals are $\approx sp^4$, that is, 80% p and 20% s. So they are slightly more p in character than the simple $sp^3$ model (ie 75% p). This is consistent with our intuitive expectation that a bond angle less than 109.5 indicates more than 75% p character.

Although we don't have any bond angle between the lone pairs to work with, we can conclude by simple subtraction that if the bond orbitals are 80% p/20% s, the lone pairs must be 70% p/30% s, so $sp^{2.3}$.

The reason that these values differ from the Gen Chem simplification is that the Gen Chem approach treats the bonding and non-bonding orbitals as equivalent and provides only a single hybridization value. Is that incorrect? Yes. Does it really matter? Probably not. It's a halfway decent approximation, and most Gen Chem texts follow shortly with a discussion of $\ce{H2S}$, where the smaller bond angle indicates even further deviation from $sp^3$, to the point that the increased p contribution is far too great to be ignored.

One last important point is that the electron density overall on the molecule that is described by these localized orbitals is necessarily identical to that total electron density described by the delocalized orbitals. Much as we can describe the number 12 as 1+2+4+5 or as 3+3+3+3, the two sums are equivalent even though the component parts are completely different from each other.

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  • $\begingroup$ "By determining the contribution of different s and p orbitals to the hybrids, we can draw some conclusions about the bonding in a molecule." - I say that no, we can't, because the resulting hybrid orbitals have neither an energy, nor a fixed shape to them. I've made that point in detail here. $\endgroup$
    – Antimon
    Jan 23 at 0:11
  • $\begingroup$ I completely agree they have no energy associated with them, but disagree that they are not stringently mathematically defined, since Coulson's theorem does that, admittedly in a time-independent way, but pretty much all basic models of chemical structure and orbitals are treated as time independent, so I don't see a problem with that. If we consider time, then the canonical delocalized basis orbitals do not have defined shape either. $\endgroup$
    – Andrew
    Jan 23 at 2:11
  • $\begingroup$ They may very well be treated as time-independent, but that doesn't mean that it would be correct. Also, if we're talking about canonical orbitals as they come out of e.g. a Hartree-Fock calculation, then their associated probability density for sure is time independent. (Granted, I didn't talk about the probability density before, but my point still stands.) $\endgroup$
    – Antimon
    Jan 23 at 3:39

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