My first-year chemistry teacher depicted $ \mathrm{PO_{4}^{3-}}$ as exhibiting the resonance structure inset below this question.

However, I can't figure out what diagramming rules (which is what she was teaching) she used to deduce her answer. I thought that the double bonded oxygen would be single bonded and that it would have six unbonded electrons like the other oxygens.

  1. Why should one oxygen be different than any of the others?
  2. What diagramming rules imply that $ \mathrm{PO_{4}^{3-}}$ should be depicted as my teacher did?


  • $\begingroup$ Actually your teacher omitted two paths. There should be connections between the diagonals of the square shown. This forms a "complete graph" in mathematics. $\endgroup$
    – MaxW
    Nov 19, 2015 at 2:32
  • 2
    $\begingroup$ There is also at least one resonance structure missing, where the charges are separated and the molecule formally obeys the octet rule. Resonance structures with an extended octet are also not really accurate. $\endgroup$ Nov 19, 2015 at 9:14
  • $\begingroup$ @MaxW Are most resonance structures complete graphs? $\endgroup$
    – Hal
    Nov 19, 2015 at 17:58

2 Answers 2


A much better representation of phosphate is the one given below (You need to add three lone pairs to each oxygen atom which I was too lazy to do):

  • It does not expand phosphorous’ octet, which calculations have shown is unlikely.
  • It does not require phosphorous’ 3d-orbitals which are energetically too far removed to participate in hybridisation in a noteworthy manner.
  • It gives you four equivalent $\ce{P-O}$ bonds a priori without needing to resort to mesomery.
  • It gives all oxygen atoms the same charge a priori and also gives phosphorous a charge in better agreement with experiment/calculation.
  • The experimental bond lengths are better explained as single bonds rather than having a bond order of $\frac{5}{4}$.

This can all be summed up as ‘phosphorous obeys the octet rule much like every other main group atom (doublet rule for hydrogen/helium).’

But you probably want answers to your questions on the decidedly minor mesomeric representations your teacher drew. Well:

  • Why should one oxygen be different than any of the others?

    Those four mesomeric structures are entirely equal, just like in benzene. To get a ‘picture closer to the truth’ you would need to put them all on top of each other and look through them. You would see that every $\ce{P-O}$ bond has the same (minute) chance of being a double bond and thus no oxygen is different from the others.

  • What diagramming rules imply that $\ce{PO4^3-}$ should be depicted as my teacher did?

    Well: none, really. If you are going to stick with the double bond representation, the order of the four structures does not matter as long as they are all there. Also, you can go from every single one to every single other one in a single step by appropriate electron shifts. So, as MaxW commented, you really should add the two missing diagonal arrows to be correct.

  • $\begingroup$ What are the calculations which you talk about? I haven't seen many calculations showing octet rule 'more important' than hypervalency-formal charge...; In any case, +1... $\endgroup$
    – user43021
    Sep 24, 2018 at 1:33

Why should one oxygen be different than any of the others?

A very important thing to understand when representing a molecule with its resonance structures are that they do not exist, at least not individually. What the Lewis depiction of a molecule does a semi-poor job of showing is electron delocalization. This is why we have to represent a molecule that shows delocalization with its resonance structures. It is not the case that one oxygen is double bonded while the rest of them are single bonded, in fact all the $\ce{P-O}$ bonds are identical in length, and each have the strength of $\frac{5}{4}$ of a $\ce{P-O}$ single bond.

What diagramming rules imply that $\ce{PO_4^{-3}}$ should be depicted as my teacher did?

Again, this is a Lewis structure. The general rules for this should be in the link I provided. As for what you need to know about phosphorus, or any element below the second period really, is that it has the ability to form more than the usual 8 bonds through hybridization of their $\mathrm{p}$ and $\mathrm{d}$-orbitals.

The most basic thing you need to know for this problem is this: a neutral phosphorus atom has 5 valence electrons. If it forms a single bond with each of the four oxygens, it will be left with one unbonded electron. To explain this in the Lewis structure, it is drawn as having a double bond with one of the oxygens.


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