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.
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.
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.
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.