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I am trying to understand how a phosphate ion looks like using the VSEPR theory.

The phosphorus element has an atomic number of 15, and therefore it has a electron configuration as follow by the Aufbau principle.

$$\pu{1s2 2s2 2p6 3s2 3p3}$$

It needs to have one π-bond with one oxygen atoms and 4 σ-bonds with all of the 4 oxygen atoms, so it is hybridizing 4 orbitals to form some hybridized orbital and leaving one p orbitals around for the pi bond? How does that work?

Martin - マーチン
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Andrew Au
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    As a side note: there is no double bond in phosphate, only four $\ce{P-O}$ single bonds. Each oxygen carries a formal negative charge while phosphorus carries a formal positive charge. This leaves us with phosphorus being $\mathrm{sp^3}$ hybridised as (hopefully) expected. – Jan Sep 13 '17 at 07:27
  • @Jan - this is not a duplicate. That post you mentioned is about counting electrons, mine is about molecular shape. – Andrew Au Sep 13 '17 at 13:50
  • I am confused, Wikipedia says phosphate has a P double bond O. https://en.wikipedia.org/wiki/Phosphate Are you saying Wikipedia is wrong? It could be ... – Andrew Au Sep 13 '17 at 13:51
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    Wikipedia is not wrong, it is just... a simplified representation, perhaps? It is tedious to draw the correct thing every time, so sometimes we fall back on simplified representations; as long as we know what it really means, then it is fine. – orthocresol Sep 13 '17 at 16:02
  • I marked as a dupe because once you get the bond counting right (you only need four single bonds) the structure explains itself and there is no need to look for a hypothetical orbital to make a hypothetical fifth bond. – Jan Sep 14 '17 at 04:20
  • It is kind of sad that the solution is removed. It was wrong, but a lot of discussion associated with it was useful.

    I realize now that VSPER is an outdated theory and it cannot describe phosphate, what is a better one?

     – Andrew Au Sep 29 '17 at 14:01

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