$\ce{SO3}$ molecule has three double bonded oxygen to the central sulfur atom.
Sulfur has $\ce{sp^2}$ hybridization and it has 6 outer electrons which make the bonds with the oxygen.
So shouldn't the bond order be 2?
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3$\begingroup$ I just don't have time to write a complete answer, so instead I leave this link to a previous answer of mine that somewhat deals with this question. The doubly bonded picture is a very crude oversimplification, though. $\endgroup$– Martin - マーチン ♦Apr 21, 2015 at 10:06
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$\begingroup$ Aaaah! I'm used to drawing the structures one of these three ways though. $\endgroup$– M.A.R.Apr 21, 2015 at 16:32
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$\begingroup$ @MARamezani This is essentially the concept of resonance, and it's intended purpose. You need all three structures though, not only one. Because one alone does not describe delocalisation. $\endgroup$– Martin - マーチン ♦Apr 23, 2015 at 3:31
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$\begingroup$ Indeed @Martin. The resonance hybrid is the only correct structure for this. $\endgroup$– M.A.R.Apr 23, 2015 at 8:16
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3 Answers
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The bonding situation in $\ce{SO3}$ is a tough nut to understand. In a historical context this molecule belongs to the species of hypervalent molecules, that disobey the octet rule. The concept of hypervalence is still very much debated. Recently there was a question raised by ron, seeking for more guidance: Hypervalency and the octet rule.
I stated in Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?
The bonding picture is usually trivialised as each $\ce{S−O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=−\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three π bonding orbitals, resulting in electron density deficiency.
There are a couple of molecules that belong roughly to the same class. I call them Y aromatic systems: $\ce{CO3^2-}$, $\ce{NO3^-}$, and $\ce{SO3}$. The description is best carried out in the framework of resonance. (Please also see Why is there a need for resonance?)
I have explained the delocalisation in the nitrate ion, but I am happy to repeat it here for $\ce{SO3}$. The molecule has an overall symmetry of $D_\mathrm{3h}$, hence the central sulfur can be regarded $sp^2$ hybridised. Since oxygen has only one bonding partner, it can sufficiently be described as $sp$ hybridised (compare here). The combination of these orbitals form a $\sigma$ bond each. The remaining $p$ orbitals perpendicular to the molecular plane form the delocalised $\pi$ system. Since delocalisation is not part of the original Lewis description it is impossible to address this with such structures, which ultimately lead to the concept of resonance. It is important, that the bonding situation can only be described by all resonance structures, as there is no predominant. The blue structure tries to emulate delocalisation, but here formal charges and charges are quite messed up.
However, we can take home an approximate bonding order for the $\ce{S-O}$ bond. Since all of the resonance structures are equally dominant, we can see that all of these bonds consist of a $\sigma$ bond. We also see, that there is a $\pi$ bond in the first three structures. Each of them contributes equally, therefore it makes a third of a full contribution per bond, hence the bond order is expected to be about $1\frac13$.
Now the molecular orbital scheme is quite similar, but it suffices with one structure. The explanation of the $\sigma$ system remains unchanged. From the four $p$ orbitals we can form four symmetry adapted linear combinations. Two of the are non-bonding degenerate, the completely symmetric one accounts for the huge stabilisation. The antibonding orbital is not occupied. See below for the schematics of $\ce{NO3^-}$, which is analogous to $\ce{SO3}$. You can find the complete MO scheme here.
Since each of the $\ce{S-O}$ bond contributes equally to the lowest lying $\pi$ orbital, the bond order is found to be about $1.33$. The calculated bond order is a little bit higher, most likely due to attractive electrostatic interactions.
answered Apr 24, 2015 at 10:27
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Great question, I had to do a little reading to piece together an answer as I haven't worked with sulfur oxides in quite some time:
Sources:
- Wikipedia: Sulfur Trioxide
- Shriver & Atkins, Inorganic Chemistry, 4th ed., p388
- Some Yahoo Answer
- Greenwood & Earnshaw, Chemistry of the Elements, 2nd ed., p703
As a simplistic explanation, the above sources state that the lewis structure of $\ce{SO_3}$ contains a 2$^+$ charge on the central sulfur and negative charges on two of the three bonded oxygen atoms. In that case, $\ce{SO_3}$ contains one double bond and two single bonds, which is why people tend to list the overall bond-order as 1.33. The actual bonding structure of $\ce{SO_3}$ is a little more complicated than that, as J. LS points out, so you might need to brush up on molecular-orbital theory to get into the nitty-gritty of its bonding structure.
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1$\begingroup$ There is no need to invoke d-orbitals in bonding for early p-block elements; it's been theoretically demonstrated to be unimportant. Really you need to use MO theory to understand how the bonding in compounds like this actually works. $\endgroup$– J. LSApr 21, 2015 at 10:18
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1$\begingroup$ I was giving the basic historical interpretation of why people tended to draw it like that, followed by the straightforward explanation of why it isn't. I highly anticipate your (correct) MO-based explanation, though. $\endgroup$ Apr 21, 2015 at 10:24
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$\begingroup$ The simplest correct explanation is as you say, simply having three sigma bonds, one pi bond and two ionic interactions, giving a bonding order of 1.33. Invoking d-orbitals is unnecessary. $\endgroup$– J. LSApr 21, 2015 at 10:31
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$\begingroup$ Fair enough, I agree, it's a bit of fluff that I carried over. I will edit the answer to reflect that! $\endgroup$ Apr 21, 2015 at 10:34
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3$\begingroup$ The most important feature of this compound is, that it is an Y aromatic system (I have done calculations on it and linked the post to the question). There is one pi orbital delocalised over three bonds, hence one third per bond (the calculated bond order as a little higher though). The description with one SO-pi-bond and two ionic interactions is only half correct in the framework of the symmetry equivalent resonance structures - but they all have to be present. $\endgroup$– Martin - マーチン ♦Apr 21, 2015 at 10:56
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During the reaction of $\ce{SO2}$ and $\ce{O2}$, there's a formation of nascent oxygen which oxidises sulphur. In doing this sulphur co ordinates it's lone pair to the nascent oxygen. Since only 2 electrons take part here in this co-ordinate bond formation it is just a single bond. Now the $\ce{SO3}$ thus formed has 4 bonds shared between 3 places. Hence the bond order.
