Which d orbitals of sulfur take part in the pi bonds of SO3?

Question

In $\ce{SO3}$, 2 $p\pi-d\pi$ bonds are present. But which $d$ orbitals of sulfur take part in these $\pi$ bonds?

The answer says $d_{xy}$ and $d_{yz}$. Someone also told me that crystal field theory supports it as while splitting $d_{x^2-y^2}$ and $d_{x^2}$ are raised in energy compared to the other 3, but that theory has nothing to do here.

It is a long-standing myth that $d$ orbitals partake in bonding of most hypervalent compounds in any appreciable amount. There is little to no $d$-orbital participation in the bonding in $\ce{SO3}$ (see here). There are almost only $\mathrm{s}$ and $\mathrm{p}$ orbitals of sulphur involved. — Philipp, Aug 30 '14 at 12:03
@permeakra Its specified. I just need one reason why only $d_{xy}$ and $d_{yz}$ participate instead of other 3. — , Aug 30 '14 at 15:56
@Philipp Maybe you are right, but I read that in "J.D. Lee Concise Inorganic Chemistry" which I think is more trustworthy than the link you provided. — , Aug 30 '14 at 15:58
@Hardik I've seen this issue come up several times here on Chemistry.SE. A few other users mentioned that the supposed d-orbital participation in hypervalent molecules is a relic from the past which is contradicted by actual quantum chemical calculations. Maybe people like Martin can chime in an comment on that. — Philipp, Aug 30 '14 at 16:06
If your textbook still needs to use d orbitals to explain the hypercoordinate molecules, then you should get a new one. For the complete molecular orbitals of $\ce{SO3}$ have a look at this question and answer. Apart from this the question is incomplete. In order to determine which orbitals take part in bonding, you have to know the position of the symmetry defining elements. Otherwise the naming of the orbitals is completely interchangeable. (@Philipp thanks for raising awareness.) — Martin - マーチン, Aug 30 '14 at 16:16
@Martin Thanks for giving some support. And once again, nice answer of yours in the link you provided. Really like those orbital pictures of yours. — Philipp, Aug 30 '14 at 16:20
@Philipp Yes, its specified ".....these models still appear in some beginning-level college texts" . And the book I am following is a also for a beginner in college. I am still learning chemistry, just passed high school. Sorry if the question was not upto the level of CHEM-SE. — , Aug 30 '14 at 17:26
@Hardik There is nothing to feel sorry about. It's still a common thing to be told in universities even though it is an outdated view. I've been told it that way too before diving a little deeper into the more theoretical aspects of chemistry. And some of the historical concepts in chemistry still have their uses today, because they give you some easy to grasp guidelines and help you to develop a good intuition, although there might be some limitations and exceptions. But it is always good to be aware of the pitfalls and to know that there are more advanced theories out there. — Philipp, Aug 30 '14 at 17:37
@Philipp I just realised that the first link you gave was a bout sulfite, when I answered, I was like... there is something missing there... where is the pi bond... and then i noticed that there are too many electrons ;) Unfortunately, no one published an article about the electronic structure at the basis set limit. — Martin - マーチン, Apr 24 '15 at 12:55

score 9 · Answer 1 · edited Apr 13 '17 at 12:57

Unfortunately this is an issue that is, even though thoroughly disproved, still taught in many books and schools.

As Philip already pointed out in the comments: There is little to no contribution of the $d$ orbitals to the bonding in $\ce{SO3}$, as modern quantum chemical calculations prove (compare Wikipedia). This bonding scheme is a relic from a time where quantum theory was not yet widespread available. The little contribution of the $d$ orbitals (usually below 1%) we find is due to polarisation functions, that are added for a more accurate description of the molecule.

There are partial answers to the overall bonding in $\ce{SO3}$ in those two question:

But I will go a little further here. In the second post linked, you will find the result of a quantum chemical calculation for $\ce{SO3}$ on the BP86/cc-pVTZ level of theory. Here is the part of the output, containing the major contributions to the molecular orbitals (including the LUMO):

 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-88.394 is S1-s=1.00
 Alpha occ 2 OE=-18.855 is O2-s=0.67 O3-s=0.17 O4-s=0.17
 Alpha occ 3 OE=-18.855 is O3-s=0.50 O4-s=0.50
 Alpha occ 4 OE=-18.855 is O3-s=0.33 O4-s=0.33 O2-s=0.33
 Alpha occ 5 OE=-7.922 is S1-s=1.00
 Alpha occ 6 OE=-5.945 is S1-p=1.00
 Alpha occ 7 OE=-5.941 is S1-p=1.00
 Alpha occ 8 OE=-5.941 is S1-p=1.00
 Alpha occ 9 OE=-1.113 is S1-s=0.39 O2-s=0.17 O3-s=0.17 O4-s=0.17
 Alpha occ 10 OE=-0.993 is O3-s=0.37 O4-s=0.37 S1-p=0.18
 Alpha occ 11 OE=-0.993 is O2-s=0.49 S1-p=0.18 O4-s=0.12 O3-s=0.12
 Alpha occ 12 OE=-0.585 is S1-s=0.24 O3-s=0.14 O4-s=0.14 O2-s=0.14 O2-p=0.11 O4-p=0.11 O3-p=0.11
 Alpha occ 13 OE=-0.502 is S1-p=0.55 O3-p=0.14 O4-p=0.14 O2-p=0.14
 Alpha occ 14 OE=-0.482 is O2-p=0.31 S1-p=0.20 O4-p=0.15 O3-p=0.15 O2-s=0.12
 Alpha occ 15 OE=-0.482 is O4-p=0.26 O3-p=0.26 S1-p=0.20
 Alpha occ 16 OE=-0.374 is O4-p=0.36 O3-p=0.36 O2-p=0.13
 Alpha occ 17 OE=-0.374 is O2-p=0.43 O3-p=0.21 O4-p=0.21
 Alpha occ 18 OE=-0.358 is O2-p=0.60 O4-p=0.15 O3-p=0.15
 Alpha occ 19 OE=-0.358 is O3-p=0.45 O4-p=0.45
 Alpha occ 20 OE=-0.305 is O4-p=0.32 O3-p=0.32 O2-p=0.32
 Alpha vir 21 OE=-0.141 is S1-p=0.45 O2-p=0.18 O4-p=0.18 O3-p=0.18

Now it is important to realise, that these numbers have to add up to 1 (accounting for rounding issues 0.99-1.01). So the reminding contributions are those of polarisation functions. Most notably missing is in orbital 15 the contribution of the O2-p orbital. The missing parts in 16-18 can be attributed to the diffuse and polarisation functions, i.e. higher $p$ and $d$-functions, 0.15; 0.15; 0.10; 0.10. As you can see, the involvement of $d$ orbitals in the total bonding is very little, almost negligible.

If you, however, consider $d$ orbitals to play a role in the bonding, then you have to consider, that all will participate.

Which d orbitals of sulfur take part in the pi bonds of SO3?

1 Answers1