 Solution

$\mathrm{OP}(\mathrm{OMe}{)}_3$ also has approximately tetrahedral molecular geometry at phosphorus, now with three P–O–C sigma linkages and a terminal phosphoryl O. Only the latter is being compared with the terminal O of Me${}_3$PO; the three P–O–C bonds are chemically distinct.

The electronegative alkoxy substituents withdraw electron density from P and improve its capacity to accept donation from terminal O into the surrounding antibonding framework. The elementary substituent argument therefore predicts a stronger, stiffer terminal P–O bond in $\mathrm{OP}(\mathrm{OMe}{)}_3$ than in $\mathrm{M}{\mathrm{e}}_3\mathrm{PO}$. A description invoking polar bonding and negative hyperconjugation avoids assigning a literal localized d-orbital double bond solely because the formula is written P=O.

Measure the terminal P–O stretching vibration by infrared or Raman spectroscopy under comparable, non-hydrogen-bonding conditions. In an approximate local-mode comparison the reduced mass is the same, so a larger terminal stretching wavenumber indicates a larger force constant, $k\propto {\tilde{\nu }}^2$. Mode assignment can be checked by ${}^{18}$O substitution at the terminal oxygen or by a normal-mode calculation; otherwise an alkoxy P–O–C vibration could be compared by mistake. X-ray diffraction can supply a supporting bond-length comparison, but neither bond length nor a single coupled vibrational frequency directly measures an isolated bond-dissociation enthalpy.