When discussing the Wittig reaction, Clayden's Organic Chemistry (2nd ed.) cites the strength of the P=O bond formed in triphenylphosphine oxide as a driver of the reaction through enthalpy:

The P=O bond, with its bond energy of 575 kJ mol−1, is one of the strongest double bonds in chemistry, and the Wittig reaction is irreversible and is driven forward by the formation of this P=O bond. No need here for the careful control of an equilibrium necessary when making acetals or imines.

Wittig reaction scheme

According to this table of bond dissociation energies, the value for the P=O bond is similar to the bond energy values for bonds such as the Be–F bond (577 kJ mol−1) and the H–F bond (568.6 kJ mol−1).

However, taking into account the large differences in the p-orbital sizes of phosphorus and oxygen atoms, and the sizes of the phosphorus and oxygen atoms, I fail to understand the similarity in the strength of the P=O and other bonds listed above.

  1. The oxygen atom is larger than the hydrogen, fluorine, while the phosphorus atom is larger than hydrogen, fluorine, and beryllium. Thus, I would imagine that the sigma orbital overlap would be weaker due to a larger distance of interaction in P=O than in the H–F and Be–F bonds.

  2. Due to the differences in the sizes of the p-orbitals of oxygen and phosphorus, I would naturally infer that the p-orbital interactions and, thus, the pi bond in P=O would be weak.

Are there other factors that I am failing to consider that strengthen the phosphorus–oxygen double bond?


Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

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    $\begingroup$ (+1) I'll throw it out there that it's not a true double bond. If you discount d-orbital involvement in bonding, then it's better represented as $\ce{P+-O-}$. Unfortunately, I don't have anything useful to add beyond that. $\endgroup$ Jan 7, 2019 at 20:11
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    $\begingroup$ I also believe that this representation of the bond also agrees with the weak p-orbital interactions in the bond, as stated in the question. $\endgroup$
    – Ethiopius
    Jan 7, 2019 at 20:47
  • $\begingroup$ The d-orbitals of phosphorus help delocalise the O electrons as well, right? $\endgroup$ Jan 8, 2019 at 18:38
  • $\begingroup$ I guess that since the bond primarily exists as $\ce{P+-O-}$, then the interaction between both the atoms would consist of a sigma bond and an electrostatic potential energy between the opposite charges similar to an ionic bond. This might be responsible for the higher bond energy $\endgroup$ Apr 11, 2020 at 14:11

1 Answer 1


The phosphorus oxygen double bond is a relatively short bond. The oxygen p-orbital lone pairs are donated to the phosphorus antibonding orbitals further stabilizing and strengthening the P-O bond. This idea is supported by molecular orbital theory and ab inito calculations. Consequently, this suggests there is no d-orbital involvement in the phosphorus(V) atom.


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