# How do you write the orbital diagram for phosphate?

Aug 11, 2017

I interpreted this as the molecular orbital diagram... and it became rather complicated (I honestly don't think you want to know the minor details!).

Here's the scratchwork when I was getting the reducible representations ${\Gamma}_{S \left(O\right)}$, ${\Gamma}_{{p}_{y} \left(O\right)}$, ${\Gamma}_{{p}_{x} \left(O\right)}$, and ${\Gamma}_{{p}_{z} \left(O\right)}$, and the decomposed irreducible representations (${A}_{1} + {T}_{2}$, $E + {T}_{1} + {T}_{2}$), for the four oxygen atoms transforming as a group:

The whole point of that was to see how the oxygen orbital energies split up (and which were two-fold or three-fold degenerate). It was also to figure out which orbitals on phosphorus interact with which orbitals on the oxygen atoms.

The resultant MO diagram was:

Takeaways:

• This is only qualitative, so take it with a grain of salt.
• Phosphate has all electrons paired.
• The nonbonding orbitals are nonbonding either due to mismatched symmetries, overly different inner-atom/outer-atom orbital energies, or interactions between both high-lying and low-lying orbitals.
• There are $22$ nonbonding electrons, which accounts for all $11$ lone pairs of electrons that belong to the oxygen atoms. The remaining $2$ nonbonding electrons account for the $\boldsymbol{\pi}$ bond that is delocalized.
• The $\sigma$ bond order that this MO diagram seems to account for is $1$ for all four $\boldsymbol{\sigma}$ bonds, as there are four clearly-bonding MOs (${a}_{1} + {t}_{2}$), one for each bond.