On the fluorine end the fluoride atoms are simply completing a $3p$ orbital.

On the sulfur end one could posit a hybrid of one $3s$ orbital, two $3p$ orbitals and a three inner $2p$ orbitals. However, I think this would result in a geometry which sulfur hexafluoride does not exhibit.

The explanation I read online is that the orbitals are a hybrid of an $s$ orbital, three $p$ orbitals and two $d$ orbitals.

This confuses me because the $3d$ orbital is quite high energy and I don't think it is possible to have a $2d$ orbital even in a hybrid.

The ground state of a neutral sulfur atom is:

$$ [\ce{Ne}] \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \downarrow \vert \uparrow \vert \uparrow]}$$

With this hybrid the state would be something like the following but with the electrons paired up with the orbitals they are bonded with (also the orbitals are hybridized):

$$ [\ce{Ne}] \underset{3s}{[\uparrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \uparrow]} \underset{3d}{[\uparrow \vert \uparrow \vert \; \; \vert \;\;\vert \;\;]}$$

One approach is that sulfur hexaflouride could be thought of as a less extreme case of $\ce{S^{+6} + 6F^-}$.

Another approach is that sulfur hexaflouride could be a hybrid of a bunch of $\ce{2F2 + SF2}$ arrangements. The $\ce{SF2}$ part has a tetrahedral geometry and so $sp^3$ character. The bonds are then $sp^3-p$ bonds. The total result would be a bunch of partial $sp^3-p$ bonds between the center sulfur and outer flouride atoms and some outer partial $p^3$ bonds between flouride atoms. But I'm not sure what the actual geometry this interpretation would predict or how it would differ in properties.

How are the hybrid orbitals of sulfur hexaflouride shaped?

  • $\begingroup$ It is a $sp^3d^2$ hybrid, the $3d$ orbital is also involved. And the energy if 3d is high but since the flourine atoms are highly electronegative so the energy released from formation of bonds is enough to promote electrons to 3d. $\endgroup$
    – Kartik
    Commented Jul 12, 2016 at 1:53
  • 1
    $\begingroup$ You need to revise your understanding of orbitals. The $2d$ orbital doesn't exist and you can't have a hybrid orbital formed from four $3p$ orbitals because there are only three of them. Also, $\ce{SF6}$ is poorly described by hybridization. $\endgroup$
    – bon
    Commented Jul 12, 2016 at 8:45

2 Answers 2


If you limit consideration to hybridization of atomic orbitals, a good reference to see is On the role of d orbitals in sulfur hexafluoride J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593.

First, it is found that of the 6 valence electrons that atomic sulfur has (two 3s and four 3p), in SF6 a total of 3.1 electrons worth of electron probability density remain in sulfur atomic orbitals.

Of the 3.1 electrons:

0.97 are in the 3s orbital

1.82 are in 3p orbitals

0.24 are in 3d orbitals

and small fractions are in 4s, 4p and 4 d orbitals.

For a more recent study, see Theory of Hypervalency: Recoupled Pair Bonding in SFn (n = 1−6) J. Phys. Chem. A, 2009, 113 (27), pp 7915–7926

  • $\begingroup$ I like your answer, it probably would be more useful if you provide some short and simple hints for conecting these ideas with the mind set or the theoretical background of the one who asked the question. $\endgroup$ Commented Jul 15, 2016 at 13:38
  • $\begingroup$ @user1420303 probably "Uses and Misuses of Hybrid Orbitals" spq.pt/magazines/RPQ/280/article/736/pdf would be the most helpful. Basically the OP question presupposes that there is a unique set of hybrid orbitals, and therefore a shape, but there is no such unique set of hybrid orbitals. $\endgroup$
    – DavePhD
    Commented Jul 15, 2016 at 13:57
  • $\begingroup$ In a technical sense I completely agree with you. Probably the diffusion of articles like the one you linked in the comment would be better. Truly, I normally disagree with teaching this kind of concepts in elementary courses. $\endgroup$ Commented Jul 15, 2016 at 14:46

Hybridisation is determined by geometry!

This is the number one thing that you should learn from this answer. If you know the geometry of a molecule then you can work out the hybridisation of the atoms within it because the hybrid orbitals must have the correct geometry to account for the molecular structure.

The second important thing that you should understand is that hybrid orbitals have no physical existence. They are a mathematical trick that we can use to provide a simpler description of bonding that fits well with the two-centre two-electron bond idea. When atoms come together to form bonds they do not promote electrons into higher energy states and form actual hybrid orbitals.

With this is mind we can look at the case of $\ce{SF6}$. The molecule has octahedral geometry around the central sulfur and so the sulfur hybrid orbitals which we will use to describe the bonding must also have this geometry. The hybridisation scheme which fits this is $sp^3d^2$ - in this case the $3s$, $3p$ and $3d$ orbitals. However, as you have noted this runs into problems when we consider that the $3d$ orbital is really rather high in energy compared to the $3s$ and $3p$ orbitals and so it's contribution to the bonding will be minimal. This is borne out by the figures in @DavePhD's answer which show that the $d$ orbital contribution to bonding is relatively small (although certainly not negligible).

$sp^3d^2$ is not really a good model for $\ce{SF6}$. The $d$ orbital occupancy corresponds to only about 6% of the total sulfur valence electrons, compared to 33% in the $sp^3d^2$ model. The bonding in $\ce{SF6}$ contains a strong ionic component with a calculated charge on sulfur of a little less than +3. Additional bonding contribution comes from donation of electron density from the fluorine $p$ orbitals into the sulfur $d$ orbitals, mainly in the form of $\sigma$ bonding between the fluorine $p_{\sigma}$ orbitals (those pointing towards the central sulfur atom) and the sulfur $d_{\sigma}$ orbitals (the $3d_{x^2-y^2}$ and $3d_{z^2}$), but with some $\pi$ bonding contribution as well from the $p_{\pi}$ and $d_{\pi}$ orbitals.

In general you should be cautious when applying hybridisation to atoms outside the second period of the periodic table. Many compounds such as $\ce{SF6}$ and $\ce{PCl5}$ are poorly described by hybridisation and require theories such as hypervalency to explain adequately.

This topic and related topics have been discussed extensively on this site before. Below are some relevant questions that might help you to understand more about the topic:

Why an asymmetric geometry with sp3d and sp3d3 hybridization?
Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?
Why does F replace the axial bond in PCl5?
Hypervalency and the octet rule

  • $\begingroup$ So, the sulfur atom is more like a +3 ion and has a valence electron configuration like: $[\ce{Ne}] \underset{(3s)^{0.97}(3p)^{1.82}(3d)^{0.24}}{[\uparrow \vert \uparrow \vert \uparrow ]}$? Is this set of hybridized orbitals completely symmetrical or does it have empty spots in it? $\endgroup$ Commented Jul 13, 2016 at 20:52
  • $\begingroup$ @StevenStewart-Gallus More or less. I read the paper provided by DavePhD and added some more detail in my answer. What do you mean by 'does it have empty spots in it'? $\endgroup$
    – bon
    Commented Jul 14, 2016 at 9:11
  • $\begingroup$ @bon But the article points that sulphur d-orbitals are mostly participating in pi-bonding ! $\endgroup$
    – permeakra
    Commented Jul 14, 2016 at 18:24
  • $\begingroup$ @permeakra No it says that they mostly participate in sigma bonding. Read the section on the fourth page about $d_{\sigma}$ and $d_{\pi}$ contributions. pubs.acs.org/doi/pdf/10.1021/ja00273a006 $\endgroup$
    – bon
    Commented Jul 14, 2016 at 19:26
  • $\begingroup$ @bon Apologies. This work dx.doi.org/10.1071/CH04113 suggests that there is some pi-interaction involved and it has to do with d-orbitals. $\endgroup$
    – permeakra
    Commented Jul 14, 2016 at 19:50

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