When discussing the Wittig reaction, Clayden's Organic Chemistry cites the strength of the $\ce{P=O}$ formed in triphenylphosphine oxide as a driver of the reaction through enthalpy:

The $\ce{P=O}$ 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 $\ce{P=O}$ bond. No need here for the careful control of an equilibrium necessary when making acetals or imines.

Wittig Reaction

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

However, taking into account the large differences in the p-orbital sizes of phosphorus and oxygen atom, and the sizes of the phosphorus and oxygen atoms, I fail to understand the similarity in the strength of the $\ce{P=O}$ and the 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 $\ce{P=O}$ than in the $\ce{H-F}$ and $\ce{Be-F}$ 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 $\ce{P=O}$ 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. $\endgroup$ – orthocresol Jan 7 '19 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 '19 at 20:47
  • $\begingroup$ The d-orbitals of phosphorus help delocalise the O electrons as well, right? $\endgroup$ – Eashaan Godbole Jan 8 '19 at 18:38
  • $\begingroup$ In your scheme, you missed a $\ce{P}$ in $\ce{Ph3PO}$. And as Ortho said, it’s not a double bond it was just considered one in the past (probably due to Pauling). $\endgroup$ – Jan Oct 11 '19 at 19:17

Firstly, there is an error in the reaction scheme: the triphenylphosphine oxide is missing a phosphorus atom.

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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