I was looking at the chemical structure of $\ce{H2SO4}$.

H2SO4 chemical structure H2SO4 ball-and-stick model

Intuitively, I would have expected this molecule to be square planar in accordance with $p^2d^2$ or $sp^2d$ hybridization, but instead it is shown to be in a tetrahedral geometry consistent with $sp^3$ hybridization. Why is this?

I guess that an alternative way of asking this question is: What is the hybridization of the sulfur atom in $\ce{H2SO4}$?

  • $\begingroup$ From reviewing my p-chem books, I believe that sd3 orbitals are certainly possible. Then again I last looked at molecular orbital theory 6-7 years ago. $\endgroup$
    – bobthejoe
    Commented Aug 27, 2012 at 4:56
  • $\begingroup$ Sorry, I should have been clearer. I don't rule out these hybridizations in an absolute sense, but in the precise case of sulfur. I don't know much transition metal chemistry, but I'm pretty sure sulfur doesn't hybridize sd3. So considering its sp3, can all the electrons fit? $\endgroup$
    – CHM
    Commented Aug 27, 2012 at 5:54
  • 1
    $\begingroup$ You may find the wikipedia article on hypervalency interesting. Suffice to say that orbital hybridisation is not a cogent description of many molecules. $\endgroup$ Commented Aug 27, 2012 at 8:14
  • $\begingroup$ Good read. I'm asking the question primarily because I definitely don't see how the electrons can fit yet, this is how the molecule is displayed. So either my intuition is completely off or the interwebs are wrong. $\endgroup$
    – bobthejoe
    Commented Aug 27, 2012 at 10:21
  • $\begingroup$ Orbital hybridizations were invented to make valence bond theory place nice with known geometries. You should use VSEPR to determine geometry, and then extrapolate to hybridization, if you believe such a thing. $\endgroup$
    – Ben Norris
    Commented Aug 27, 2012 at 10:42

5 Answers 5


It is much easier to explain it on a simpler example – the $\ce{O3}$ molecule. It has the structure of resonance hybrid of $\ce{O=O+-O-}$ and its mirror. And of course, the central atom has hybridization state $sp^2$. One bond here is a normal covalent bond, and another bond is dative: an electron pair is donated onto a vacant orbital of an $\ce{O}$ atom with all electrons paired. In the $\ce{H2SO4}$ molecule, two bonds are simple covalent ($\ce{S-OH}$ ones), and two are dative ($\ce{S-O}$ ones). A common concept of electron unpairing to my knowledge is proved broken by quantum chemistry calculations and spectral experiments for hypervalent compounds of $\ce{P}$ and $\ce{S}$.

A little more interesting as an example is the $\ce{XeF2}$ molecule, where three-atom four-electron bonds $\ce{F-Xe-F}$ are formed; that can be thought of as a reasonance hybrid of structure $\ce{F-Xe+\ F-}$ and its mirror $\ce{F- \ Xe+-F}$

Of course, this scheme is still far from perfect, as reality is much more complicated, but if you do not wish to take a course in quantum chemistry, it should be enough. However, I'll recommend to search for "MO LCAO model": it is quite simple and very useful. It is also often used in advanced chemistry books and articles.


Permeakra's explanation is spot on but I didn't really get a good understanding of the result until I visited the Wikipedia page on Sulfate.

The central hallmark of the question and the confusion was why a ten electron configuration would adopt a sp3 configuration more reflective of an octet. This was a source of confusion that goes back to Gilbert Lewis who originally proposed the below structure fitting of a sp3 hybridization (model 2):

Two different models

Linus Pauling steps in to propose that two of the d orbitals play a role suggesting that there should be a sp3d2 hybridization. The issue then arises with the role of Pi bonds and how they fit in the structure with the proposal that the occupied p orbitals overlap with the empty d orbitals.

Ultimately from computational calculations it turns out that the sulfur is indeed charged and there is very little pi character. As concluded in How relevant are S=O and P=O Double Bonds for the Description of the Acid Molecules H2SO3, H2SO4, and H3PO4, respectively?, model 1 with S=O bonds is incorrect even though this is how we teach highschool and college chemistry.


I guess that an alternative way of asking this question is what is the hybridization of H2SO4?

A better way to ask this might be: "What is the hybridization of the sulfur atom in Hydrogen Sulfate?"

The sulfur atom has a bond with four other atoms in this molecule. Because that the electrons try to stay as far away from each other as possible according to Valence Shell Electron Pair Repulsion Theory (VSEPR), the other atoms will seperate as far as possible, which results in bonds that are 109.5 degrees apart from one another. If this were a square planar arrangement, the bonds would only be 90 degrees apart, which is not as satisfactory as the tetrahedral arrangement would provide.

According to my book, if the number of effective pairs is 4, the arrangement is tetrahedral, and therefore the hybridization required is $sp^3$

Source: Zumdahl, Zumdahl, Chemistry: An Atoms First Approach

ISBN-13: 978-0840065322

This certainly answers the question, but why?

It has to do with molecular orbitals, and how they hold only a certain number of electrons, and they prefer to spread themselves out evenly because in the quantum world, electrons tend to prefer the lowest possible energy state. Once the sigma bond is occupied, which is the first orbital referred to as the s orbital, the electrons end up occupying the p orbitals, and when electrons are shared through these orbitals they are called pi bonds, they actually form a whole new orbital, a molecular orbital, which allows the sharing of electrons between two atoms. Let me refer you to this website to help visually show you what is happening.

  • $\begingroup$ This certainly answers the question but why? $\endgroup$
    – bobthejoe
    Commented Sep 3, 2012 at 0:16
  • $\begingroup$ Going to follow up even further. The SP3 configuration only accounts for 8 electrons. From my count, H2SO4 has 10 electrons. While I agree that there are only 4 effective pairs, where are the extra ones? $\endgroup$
    – bobthejoe
    Commented Sep 6, 2012 at 0:18
  • $\begingroup$ I ended up looking it up. Sulfate adopts a resonance structure. $\endgroup$
    – bobthejoe
    Commented Sep 6, 2012 at 0:18

In the end there is no way to predict the structure of a given molecule. Yes, with experience one can guess correctly in many cases, but not in all of them.

The way all this works is to first find the structure and then second work out the bonding scheme that gives this structure. In principle one can do an ab initio quantum mechanical calculation, but that's not really practical for molecules with a fair number of electrons such as $\ce{(NH3)2SO4}$. So most often an approximation procedure is used.

There are two simple approximations. One is molecular orbitals, the other is the LCAO (linear combination of atomic orbitals) approximation. They describe the same thing but use different language. The choice of which to use is up to the person wanting the answer.

As an example of all this, try predicting the bonding and structure of $\ce{H3PO4}$ without knowing the answer.


Sulfur has four $sp^3$ orbitals (four $\sigma$-bonds) with oxygen and two $\pi$-bonds ($d$ orbitals). In the case of $\ce{SO4^{2-}}$ ion, also sulfur also has four $sp^3$ orbitals and four $\pi$-bonds with $d$ orbitals. The electron from hydrogen is moves to sulfur's d orbitalsenter image description here

  • 2
    $\begingroup$ A structure with four double bonds in the sulphate ion would give the sulphur atom a formal charge of -2. This is not expected to be more stable than a configuration with two double bonds and two single bonds, because in the latter case the same amount of formal charge is more distributed, and lies mostly on a more electronegative element (oxygen). However, due to mesomerism, the bonding is more accurately described as a combination of six similar canonical structures, creating four bonds of order 1.5, with all oxygen atoms having a formal charge of -0.5 and perfect tetrahedral symmetry. $\endgroup$ Commented Nov 10, 2013 at 16:56
  • 1
    $\begingroup$ Your d-orbitals are waa~y to close to the p-orbitals energeticly. Remember that 4s has a lower energy than 3d in the ground state of pretty much every single atom except hydrogen. $\endgroup$
    – Jan
    Commented Nov 19, 2016 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.