= Solution
\b[$\mathrm{OP(OMe)_3}$ 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 \b[a stronger, stiffer terminal P–O bond in $\mathrm{OP(OMe)_3}$ than in $\mathrm{Me_3PO}$]. 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, $k\propto\widetilde\nu^2$. 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.
Back to article page
