Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\ce{ClO2}$.

$\ce{ClO2}$ has 2 $\sigma$ bonds, 1 lone pair, 2π bonds and 1 odd electron.

Hybridisation is equal to number of $\sigma$ bonds + lone pairs. Since we consider odd electron a lone pair like in $\ce{NO2}$ therefore hybridisation is coming to be $\ce{sp^3}$. There will be no pπ-pπ bonding as all p orbitals are hybridised and there will be 3pπ-dπ bonds

But this is given wrong according to my textbook. What is my mistake?


2 Answers 2


I actually take issue with the question (which was asked of you not by you), as I think this oversimplifies a molecule, that still isn't well understood in the first place.

That is why I find this question incredibly difficult to answer and why I find that Nuclear Chemist's answer dangerous, because it propagates myths about bonding that have been disproved already.

But enough of the commentary, let's clear up some issues.

Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\ce{ClO2}$.

$\ce{ClO2}$ has two $\sigma$ bonds, one lone pair, two π bonds and one odd electron.

There comes a time, where you should realise that hybridisation is not always a useful concept to understand bonding. This time is now.

Your analysis of the bonding is not wrong, but maybe a bit too simple.

The reason for this is that the molecule has a very high symmetry, but it also has an unpaired electron. The problem here is that orbital symmetry cannot match with spacial symmetry, which means that hybridisation doesn't actually work here.
A bit more precise: The electron density of α-spin electrons is different from that of β-spin electrons. The hybridisation of the orbitals to produce either electron density need to also be different. In the very least you would have to consider two different hybridisation schemes at once.
All possible states (that is independent of any hybridisation model) then will be in superposition to form the actual ground state. This ground state will have a complicated bonding situation.

Of course you can use VSEPR theory to predict the shape of the molecule, that is a very good starting point. You will find out that the molecule is bent, and that matches the crystal structure. A bent molecule of the form $\ce{B-A-B}$ always has $C_\mathrm{2v}$ symmetry, it therefore always has π-orbitals. Whether those orbitals are essentially bonding or not, is an entirely different question.
It also does not tell you which atomic orbitals will be involved.

One obvious fault in the question is that there is no involvement of d-orbitals whatsoever in the molecules. The d-orbitals of chlorine are simply not accessible energetically; at least not in the way that it would form stable bonds. Anyone who is jibber-jabbering about 'octet-expansion' has stopped learning about chemistry about 20 years ago. (There are plentiful discussions about this on chemistry.se alone.)
As long as that last part of the question is not meant as a trap (which would be even more awful), it alone shows that the exercise is not well-thought about.

The π-orbitals will therefore be entirely composed out of p-orbitals of the involved elements.

If you look close enough you will realise, that writing a valid Lewis structure of the molecule is not that hard, and you don't even need π-bonds in the first place.
These are solely dictated by symmetry.

In principle that molecule is of a similar system like allyl-systems (with two extra electrons), or ozone (with one more electron). You will need a better than rudimentary grasp at molecular orbital theory to understand it.
Let's assume that things would be easier, and try tor roll with that for as long as possible. I have written a couple of times that terminal atoms can be (as a rule of thumb) regarded as having sp-orbitals. Therefore the two oxygen will use one sp-orbital to form a σ-bond towards chlorine, the other one is an in-plane-σ-lone-pair. That leaves for every oxygen one in-plane-p-lone-pair, which we will (in the spirit of the approximation) disregard as having any effect on bonding. There is one remaining out-of-plane-p-orbital per oxygen. These will, due to symmetry, participate in π-bonding.
From VSEPR-theory we know the molecule is bent. The approximate hybridisation of the orbitals of chlorine that match that would be sp². Let's just assume that this is the case. Two of these will be the counterpart for the σ-bonds. The remaining will be either a lone-pair, or singly occupied. Already in this very simplified view, the problems start here. Let's further assume, that because it has higher s-character, the orbital will be lower in energy and therefore be doubly occupied. That leaves the odd electron in an out-of-plane-p-orbital, which will participate in π-bonding.
In this simplified and assumption riddled thought-experiment, the π-bonding will form three molecular orbitals pictured below.

pi mo of ClO2

Essentially the stabilisation from the π-bonds itself should be small, and could be described as π-backbonding, or is similar to negative hyperconjugation; maybe negative π-backbonding (!?).

This all was deduced from a very reduced and simplified set of theory; the reality is probably by far more complicated. Some of the greatest minds of (theoretical) chemistry have been puzzled about $\bf\ce{ClO2}$, and we have not even started to discuss the three-electron bond they assumed in between. Let's just say that this is not an undergrad study problem.

I know this doesn't really answer your question, and this is not very satisfactory either, but this is science: It often gives you more questions than answers.


What I would suggest to you is to start with working out how many electrons (valence electrons) are present in the molecule.

Each oxygen atom ($\ce{O}$) brings six electrons and the chlorine ($\ce{Cl}$) has seven electrons. This gives us a total of 19 electrons.

We should aim for having 8 electrons around each atom, keep in mind that the more heavy p block elements can expand above an octet.

If we look at the crystallography of chlorine dioxide ($\ce{ClO2}$) (A.Rehr, M.Jansen, Investigations on Solid Chlorine Dioxide ($\ce{ClO2}$)): Temperature-Dependent Crystal Structure, IR Spectrum and Magnetic Susceptibility, Inorganic Chemistry (1992) 31, (*) p4740-p4742) we find that the chlorine dioxide ($\ce{ClO2}$) molecule is clearly bent. The O-Cl-O angle is 116.2 degrees. Now this is an interesting finding.

Normally if we have a perfect $sp^3$ molecule such as methane ($\ce{CH4}$) we get an angle of 109.5 degrees between the bonds. But in the case of ammonia ($\ce{NH3}$) or water ($\ce{H2O}$) the increased ability of the lone pairs to repel electron pairs decreases the angle.

Thus it looks like a distorted $sp^2$ to me, I think that the bonding is likely to be two lone pairs on each $\ce{O}$, one lone pair on the $\ce{Cl}$.

Two $\sigma$ bonds each holding two electrons connecting oxygen ($\ce{O}$) atoms to the chlorine ($\ce{Cl}$).

Two $\pi$ bonds each holding two electrons connecting oxygen ($\ce{O}$) atoms to the chlorine ($\ce{Cl}$).

A $\pi^{*}$ (antibonding) orbital will have one electron in it, think it would be interesting to do ESR on $\ce{^{17}O}$ enriched chlorine dioxide ($\ce{ClO2}$) and compare the results with ESR on normal chlorine dioxide ($\ce{ClO2}$). This could allow us to get a better idea of where the unpaired electron is.

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    $\begingroup$ -1 for expanding octet, -1 for increased ability of the lone pairs to repel. This answer is propagating those myths in chemistry which gives thousands of students the wrong idea of chemical bonding. The most important part is that hybridisation (if applicable at all) follows the molecular structure. From Coulson's theorem we can actually calculate what hybrid orbitals are necessary. $\endgroup$ Commented Jul 6, 2018 at 11:13
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    $\begingroup$ Martin, it appears that you do not like VSEPR. The problem is that VSEPR does work for a lot of compounds. In this case the use of MO theroy to deal with the pi bonds and VSEPR for the sigma bonds does provide an answer which can be reconciled with the crystallographic evidence. $\endgroup$ Commented Jul 6, 2018 at 15:55
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    $\begingroup$ No you are wrong, I do like the VSEPR model very much. I also know that it works for many compounds, and I know that it offers valuable insights into geometries. However, the claim of expanded octets and electron clouds being more repelling is fabricated garbage to make weak theory fit beyond its means. In this case, if you are already using MO theory, there is no reason to use VSEPR in the first place; it also, while at first glance looks correct, produces the 'right' answer/ prediction, for the wrong reasons. $\endgroup$ Commented Jul 9, 2018 at 8:39
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    $\begingroup$ @Martin-マーチン I agree. I would go so far as to shame whomever made the task. ClO2 needs MO, and 3-electron-2-center bonds to make even close to sense, and the woefully inadequate vsepr theory or hybridisation will not be enough. And don't get me wrong, VSEPR is a great model, but as with Newtons Law - there comes a point where it just doesn't explain what is going on. For some reason, many chemists in teaching positions keep VSEPR as a holy cow for far too long up the grades. $\endgroup$
    – Stian
    Commented Jul 9, 2018 at 12:38
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    $\begingroup$ Well I still love VSEPR, but it has its limits. If we take the crystallogrpahic data then it is clear that the chlorine has three sigma bonds / lone pairs (or single version of lone pairs). I reason that the angle of 116 degrees makes it look like we have a distorted trigonal arrangement of sigma orbitials around the chlorine. I am sure that for the pi system we need to use MO theroy and the the best I have come up with is that a pi star orbitial has a single electron in it. I was always taught to use VSEPR for the sigma bonds and then MO for the pi system. $\endgroup$ Commented Jul 9, 2018 at 13:06

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