What is the formal charge on the phosphate ion in PO_3^-?

Aug 12, 2017

Assigning formal charges assumes 100% covalent character, meaning the electrons in a given chemical bond are assumed to be equally shared.

${\text{FC" = "valence e"^(-) - "owned e}}^{-}$

where:

• owned electrons are found by cleaving each bond homolytically so that one electron goes to each atom that was bonding.
• valence electrons are found from the group number of the main group elements.

Of course, this is meaningless for a molecule as a whole. Furthermore, the ion you've quoted does not exist.

Phosphate is actually ${\text{PO}}_{4}^{3 -}$, and has the following resonance structure ($+ 3$ other permutations): Assign the labels ${\text{O}}_{\left(1\right)}$, . . . , ${\text{O}}_{\left(4\right)}$ starting from the top oxygen and going clockwise. Then, ${\text{O}}_{\left(2\right) - \left(4\right)}$ are identical in formal charge, as they are equivalent oxygens in this representation.

• ${\text{O}}_{\left(1\right)}$ would bring in $\boldsymbol{6}$ valence electrons, and it owns $\overbrace{\text{2 lone pairs" xx "2 electrons each")^("lone pairs}}$ $+$ $\overbrace{\text{2 bonds" xx "1 electron")^("bonding electrons}} = \boldsymbol{6}$ of them.

Thus, its formal charge is $0$.

• ${\text{O}}_{\left(2\right) - \left(4\right)}$ would each bring in $\boldsymbol{6}$ valence electrons, and they each own $\overbrace{\text{3 lone pairs" xx "2 electrons each")^("lone pairs}}$ $+$ $\overbrace{\text{1 bond" xx "1 electron")^("bonding electrons}} = \boldsymbol{7}$ of them.

Thus, their formal charges are all $- 1$.

• And lastly, the $\text{P}$ in the middle would bring in $5$ valence electrons, but own $5 \times 1 = 5$ of them, having a formal charge of $0$.

And we indeed still have a total charge of ${3}^{-}$, i.e.

${\overbrace{\left(0\right)}}^{{O}_{\left(1\right)}} + 3 \times {\overbrace{\left(- 1\right)}}^{{O}_{\left(2\right) - \left(4\right)}} + {\overbrace{\left(0\right)}}^{P} = {\overbrace{{3}^{-}}}^{P {O}_{4}^{3 -}}$