I am uncertain about the structure of $\ce {H_3PO_4}$ because different sources show different ones.

The structure shown here is:

Phosphoric acid structure as shown on purdue.edu

while Wikipedia gives this one:

Structure of phosphoric acid on Wikipedia

Arguments in favour of the 2nd structure:

  1. Structure having filled octet for second row elements (C, N, O, F) are stable.
  2. Structure having minimum number of formal charges and maximum number of bonds.
  3. Structure in which there is minimal charge separation while keeping the formal charges closer together.

Arguments supporting the first structure:

  1. Structure in which negative charge appears on the most electronegative atom should be chosen.
  2. I feel that this could be more contributing because in the first structure sulphur has an expanded octet while $d$ orbitals are of higher energy so it's more favourable to form bonds without using them.
  3. Structure having filled octet a for second row elements ($\ce {C, N, O, F}$) are stable. (followed by both structures)

$\ce {H_3PO_4}$ is different from phosphate ion.None of the answers in the linked question directly state what would be the more contributing structure. They focus on the hybridisation of P in phosphate.

  • $\begingroup$ Why the downvote? Please let me know. $\endgroup$
    – Archer
    Sep 30, 2017 at 20:53
  • 1
    $\begingroup$ "Expanded octet" exists mainly in outdated textbooks. $\endgroup$
    – Mithoron
    Sep 30, 2017 at 21:08
  • 1
    $\begingroup$ Well, the first structure isn't correct either, formal charges are missing. Please don't use MathJax in titles. $\endgroup$ Sep 30, 2017 at 21:11
  • 1
    $\begingroup$ Moderators cannot tell you why a question is down-voted because thats every users own decision. I agree that your question itself is different, but the answer will still be extremely similar. The core remains the phosphate ion, only that three of the four oxygen bond to protons now. The so called 'octet expansion' utilises d orbitals which has been disproved long time ago. The guideline for fewer formal charges being better is also incorrect as it often does not reflect the actual bonding situation. The first structure would be correct, if formal charges were added. $\endgroup$ Oct 1, 2017 at 12:10
  • 1
    $\begingroup$ You are asking a question about phosphoric acid, there is no mentioning of sulfur whatsoever. I guess that sparked the comment. The whole concept of extended octet is deprecated, obsolete, wrong. Better forget it completely. $\endgroup$ Oct 1, 2017 at 12:57

1 Answer 1


Both structures are not the main contributing structures. In fact, the first structure is outright wrong and should be marked as an error as the electrons of phosphorus do not add up and the non-protonated oxygen is implied to be a radical.

However, the correct most contributing structure can be derived from the first structure simply by adding formal charges: a formal positive charge on phosphorus and a formal negative charge on the non-protonated oxygen. This structure satisfies the stability conditions well:

  1. All main group elements have octets (dublets for hydrogen)

    The second proposal violates this most important principle

  2. The number of formal charges is minimalised

    We cannot go any lower without violating the more important first point

  3. The type of formal charges matches the electronegativity difference

    And also we cannot distribute them differently

  4. The distance between formal charges is minimal

    But again we have no choice.

  • $\begingroup$ (+1) Can you add a diagram of the correct structure too? $\endgroup$
    – Archer
    Oct 1, 2017 at 15:37
  • $\begingroup$ "Most important first point". Well, today itself I read an answer on Chem.SE that octet isn't essentially important for all atoms. Let me see if I can find it. $\endgroup$
    – Archer
    Oct 1, 2017 at 15:39
  • $\begingroup$ @Abcd I wrote more important. And I also added the caveat for hydrogen. Most other candidates for non-octet structures are either not main-group elements or are rarely found in molecules (although lithium, magnesium and aluminium may be counterexamples, I agree) $\endgroup$
    – Jan
    Oct 1, 2017 at 15:44
  • $\begingroup$ Apologies. Yes you said more. By main group elements are you referring to representative elements? How can we ignore the transition elements when they form the bulk of the periodic table? $\endgroup$
    – Archer
    Oct 1, 2017 at 15:49
  • $\begingroup$ @Abcd When a rule is only valid for the elements in groups 1, 2, 13, 14, 15, 16, 17 and 18 (the main group elements) then of course we can ignore the transition metals in this rule. They follow entirely different principles, more subtle, harder to explain. Other rules are made for even less elements, for example the rule that only the second period prefers forming double bonds rather than maximising coordination numbers. $\endgroup$
    – Jan
    Oct 1, 2017 at 16:14

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.