Question

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

Answer 1

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

`"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 `"PO"_4^(3-)`, and has the following resonance structure (`+3` other permutations):

Assign the labels `"O"_((1))`, . . . , `"O"_((4))` starting from the top oxygen and going clockwise. Then, `"O"_((2) - (4))` are identical in formal charge, as they are equivalent oxygens in this representation.

• `"O"_((1))` would bring in `bb6` valence electrons, and it owns `overbrace("2 lone pairs" xx "2 electrons each")^("lone pairs")` `+` `overbrace("2 bonds" xx "1 electron")^("bonding electrons") = bb6` of them.

Thus, its formal charge is `0`.

• `"O"_((2) - (4))` would each bring in `bb6` valence electrons, and they each own `overbrace("3 lone pairs" xx "2 electrons each")^("lone pairs")` `+` `overbrace("1 bond" xx "1 electron")^("bonding electrons") = bb7` of them.

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

• And lastly, the `"P"` in the middle would bring in `5` valence electrons, but own `5 xx 1 = 5` of them, having a formal charge of `0`.

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

`overbrace((0))^(O_((1))) + 3 xx overbrace((-1))^(O_((2) - (4))) + overbrace((0))^(P) = overbrace(3^-)^(PO_4^(3-))`